Right Triangle Side and Angle Calculator

How to find the sides of a right triangle, how to find the angle of a right triangle, how do you solve a right angle triangle with only one side, how to find the missing side of a right triangle how to find the angle example.

Finding the missing side or angle couldn't be easier than with our great tool – right triangle side and angle calculator. Choose two given values, type them into the calculator, and the calculator will determine the remaining unknowns in a blink of an eye! If you are wondering how to find the missing side of a right triangle, keep scrolling, and you'll find the formulas behind our calculator.

There are a few methods of obtaining right triangle side lengths. Depending on what is given, you can use different relationships or laws to find the missing side:

1. Given two sides

If you know two other sides of the right triangle, it's the easiest option; all you need to do is apply the Pythagorean theorem:

a² + b² = c²

If leg a is the missing side, then transform the equation to the form where a is on one side and take a square root:

a = √(c² - b²)

If leg b is unknown, then:

b = √(c² - a²)

For hypotenuse c missing, the formula is:

c = √(a² + b²)

🙋 Our Pythagorean theorem calculator will help you if you have any doubts at this point.

2. Given an angle and the hypotenuse

Right triangle with law of sines formulas. a over sin(α) equals b over sin(β) equals c, because sin(90°) = 1

Apply the law of sines or trigonometry to find the right triangle side lengths:

a = c × sin(α) or a = c × cos(β)

b = c × sin(β) or b = c × cos(α)

🙋 Refresh your knowledge with Omni's law of sines calculator !

3. Given an angle and one leg

Find the missing leg using trigonometric functions:

a = b × tan(α)

b = a × tan(β)

4. Given the area and one leg

As we remember from basic triangle area formula , we can calculate the area by multiplying the triangle height and base and dividing the result by two. A right triangle is a special case of a scalene triangle, in which one leg is the height when the second leg is the base, so the equation gets simplified to:

area = a × b / 2

For example, if we know only the right triangle area and the length of the leg a , we can derive the equation for the other sides:

  • b = 2 × area / a
  • c = √(a² + (2 × area / a)²)

🙋 For this type of problem, see also our area of a right triangle calculator .

If you know one angle apart from the right angle, the calculation of the third one is a piece of cake:

Given β : α = 90 - β

Given α : β = 90 - α

However, if only two sides of a triangle are given, finding the angles of a right triangle requires applying some basic trigonometric functions:

  • sin(α) = a / c so α = arcsin(a / c) (inverse sine);
  • cos(α) = b / c so α = arccos(b / c) (inverse cosine);
  • tan(α) = a / b so α = arctan(a / b) (inverse tangent);
  • cot(α) = b / a so α = arccot(b / a) (inverse cotangent);

and for β :

  • sin(β) = b / c so β = arcsin(b / c) (inverse sine);
  • cos(β) = a / c so β = arccos(a / c) (inverse cosine);
  • tan(β) = b / a so β = arctan(b / a) (inverse tangent);
  • cot(β) = a / b so β = arccot(a / b) (inverse cotangent).

To solve a triangle with one side, you also need one of the non-right angled angles . If not, it is impossible:

If you have the hypotenuse , multiply it by sin(θ) to get the length of the side opposite to the angle.

Alternatively, multiply the hypotenuse by cos(θ) to get the side adjacent to the angle.

If you have the non-hypotenuse side adjacent to the angle, divide it by cos(θ) to get the length of the hypotenuse .

Alternatively, multiply this length by tan(θ) to get the length of the side opposite to the angle.

If you have an angle and the side opposite to it, you can divide the side length by sin(θ) to get the hypotenuse .

Alternatively, divide the length by tan(θ) to get the length of the side adjacent to the angle.

Let's show how to find the sides of a right triangle with this tool:

Assume we want to find the missing side given area and one side. Select the proper option from a drop-down list . It's the third one.

Type in the given values . For example, the area of a right triangle is equal to 28 in² and b = 9 in.

Our right triangle side and angle calculator displays missing sides and angles! Now we know that:

  • a = 6.222 in
  • c = 10.941 in

Now, let's check how finding the angles of a right triangle works:

Refresh the calculator. Pick the option you need . Assume that we have two sides, and we want to find all angles. The default option is the right one.

Enter the side lengths . Our right triangle has a hypotenuse equal to 13 in and a leg a = 5 in.

Missing side and angles appear . In our example, b = 12 in, α = 22.62° and β = 67.38°.

How many lines of symmetry does a right triangle have?

If a right triangle is isosceles (i.e., its two non-hypotenuse sides are the same length), it has one line of symmetry . Otherwise, the triangle will have no lines of symmetry .

Can a right angled triangle have equal sides?

No, a right triangle cannot have all 3 sides equal , as all three angles cannot also be equal. One has to be 90° by definition. A right triangle can, however, have its two non-hypotenuse sides equal in length. This would also mean the two other angles are equal to 45°.

Are all right triangles similar?

Not all right-angled triangles are similar , although some can be. They are similar if all their angles are the same length, or if the ratio of two of their sides is the same.

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Right Triangles

Rules, Formula and more

Pythagorean Theorem

The sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse .

Usually, this theorem is expressed as $$ A^2 + B^2 = C^2 $$ .

Right Triangle Properties

Right triangle picture

A right triangle has one $$ 90^{\circ} $$ angle ($$ \angle $$ B in the picture on the left) and a variety of often-studied formulas such as:

  • The Pythagorean Theorem
  • Trigonometry Ratios (SOHCAHTOA)
  • Pythagorean Theorem vs Sohcahtoa (which to use)

SOHCAHTOA only applies to right triangles ( more here ) .

sohcahtoa

A Right Triangle's Hypotenuse

The hypotenuse is the largest side in a right triangle and is always opposite the right angle.

Hypotenuse

In the triangle above, the hypotenuse is the side AB which is opposite the right angle, $$ \angle C $$.

Online tool calculates the hypotenuse (or a leg) using the Pythagorean theorem.

Practice Problems

Below are several practice problems involving the Pythagorean theorem, you can also get more detailed lesson on how to use the Pythagorean theorem here .

Find the length of side t in the triangle on the left.

5, 12, 13 right triangle

Substitute the two known sides into the Pythagorean theorem's formula : A² + B² = C²

What is the value of x in the picture on the left?

pythagorean theorem

Set up the Pythagorean Theorem : 14 2 + 48 2 = x 2 2,500 = X 2

$$ x = \sqrt{2500} = 50 $$

Diagram, Pythagorean Theorem

$$ x^2 = 21^2 + 72^2 \\ x^2= 5625 \\ x = \sqrt{5625} \\ x =75 $$

Find the length of side X in the triangle on on the left?

3, 4, 5 right triangle

Substitue the two known sides into the pythagorean theorem's formula : $$ A^2 + B^2 = C^2 \\ 8^2 + 6^2 = x^2 \\ x = \sqrt{100}=10 $$

What is x in the triangle on the left?

pythagorean image

x 2 + 4 2 = 5 2 x 2 + 16 = 25 x 2 = 25 - 16 = 9 x = 3

Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!

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Right Triangle Calculator

Please provide 2 values below to calculate the other values of a right triangle. If radians are selected as the angle unit, it can take values such as pi/3, pi/4, etc.

right triangle

Related Triangle Calculator | Pythagorean Theorem Calculator

Right triangle

A right triangle is a type of triangle that has one angle that measures 90°. Right triangles, and the relationships between their sides and angles, are the basis of trigonometry.

In a right triangle, the side that is opposite of the 90° angle is the longest side of the triangle, and is called the hypotenuse. The sides of a right triangle are commonly referred to with the variables a, b, and c, where c is the hypotenuse and a and b are the lengths of the shorter sides. Their angles are also typically referred to using the capitalized letter corresponding to the side length: angle A for side a, angle B for side b, and angle C (for a right triangle this will be 90°) for side c, as shown below. In this calculator, the Greek symbols α (alpha) and β (beta) are used for the unknown angle measures. h refers to the altitude of the triangle, which is the length from the vertex of the right angle of the triangle to the hypotenuse of the triangle. The altitude divides the original triangle into two smaller, similar triangles that are also similar to the original triangle.

If all three sides of a right triangle have lengths that are integers, it is known as a Pythagorean triangle. In a triangle of this type, the lengths of the three sides are collectively known as a Pythagorean triple. Examples include: 3, 4, 5; 5, 12, 13; 8, 15, 17, etc.

Area and perimeter of a right triangle are calculated in the same way as any other triangle. The perimeter is the sum of the three sides of the triangle and the area can be determined using the following equation:

Special Right Triangles

30°-60°-90° triangle:

The 30°-60°-90° refers to the angle measurements in degrees of this type of special right triangle. In this type of right triangle, the sides corresponding to the angles 30°-60°-90° follow a ratio of 1:√ 3 :2. Thus, in this type of triangle, if the length of one side and the side's corresponding angle is known, the length of the other sides can be determined using the above ratio. For example, given that the side corresponding to the 60° angle is 5, let a be the length of the side corresponding to the 30° angle, b be the length of the 60° side, and c be the length of the 90° side.:

Angles: 30°: 60°: 90°

Ratio of sides: 1:√ 3 :2

Side lengths: a:5:c

Then using the known ratios of the sides of this special type of triangle:

As can be seen from the above, knowing just one side of a 30°-60°-90° triangle enables you to determine the length of any of the other sides relatively easily. This type of triangle can be used to evaluate trigonometric functions for multiples of π/6.

45°-45°-90° triangle:

The 45°-45°-90° triangle, also referred to as an isosceles right triangle, since it has two sides of equal lengths, is a right triangle in which the sides corresponding to the angles, 45°-45°-90°, follow a ratio of 1:1:√ 2 . Like the 30°-60°-90° triangle, knowing one side length allows you to determine the lengths of the other sides of a 45°-45°-90° triangle.

Angles: 45°: 45°: 90°

Ratio of sides: 1:1:√ 2

Side lengths: a:a:c

Given c= 5:

45°-45°-90° triangles can be used to evaluate trigonometric functions for multiples of π/4.

Solving Triangles

"Solving" means finding missing sides and angles.

Six Different Types

If you need to solve a triangle right now choose one of the six options below:

Which Sides or Angles do you know already? (Click on the image or link)

... or read on to find out how you can become an expert triangle solver :

Your Solving Toolbox

Want to learn to solve triangles?

Imagine you are " The Solver " ... ... the one they ask for when a triangle needs solving!

In your solving toolbox (along with your pen, paper and calculator) you have these 3 equations:

1. Angles Add to 180° :

A + B + C = 180°

When you know two angles you can find the third.

2. Law of Sines (the Sine Rule):

When there is an angle opposite a side, this equation comes to the rescue.

Note: angle A is opposite side a, B is opposite b, and C is opposite c.

3. Law of Cosines (the Cosine Rule):

This is the hardest to use (and remember) but it is sometimes needed to get you out of difficult situations.

It is an enhanced version of the Pythagoras Theorem that works on any triangle.

Six Different Types (More Detail)

There are SIX different types of puzzles you may need to solve. Get familiar with them:

This means we are given all three angles of a triangle, but no sides.

AAA triangles are impossible to solve further since there are is nothing to show us size ... we know the shape but not how big it is.

We need to know at least one side to go further. See Solving "AAA" Triangles .

This mean we are given two angles of a triangle and one side, which is not the side adjacent to the two given angles.

Such a triangle can be solved by using Angles of a Triangle to find the other angle, and The Law of Sines to find each of the other two sides. See Solving "AAS" Triangles .

This means we are given two angles of a triangle and one side, which is the side adjacent to the two given angles.

In this case we find the third angle by using Angles of a Triangle , then use The Law of Sines to find each of the other two sides. See Solving "ASA" Triangles .

This means we are given two sides and the included angle.

For this type of triangle, we must use The Law of Cosines first to calculate the third side of the triangle; then we can use The Law of Sines to find one of the other two angles, and finally use Angles of a Triangle to find the last angle. See Solving "SAS" Triangles .

This means we are given two sides and one angle that is not the included angle.

In this case, use The Law of Sines first to find either one of the other two angles, then use Angles of a Triangle to find the third angle, then The Law of Sines again to find the final side. See Solving "SSA" Triangles .

This means we are given all three sides of a triangle, but no angles.

In this case, we have no choice. We must use The Law of Cosines first to find any one of the three angles, then we can use The Law of Sines (or use The Law of Cosines again) to find a second angle, and finally Angles of a Triangle to find the third angle. See Solving "SSS" Triangles .

Tips to Solving

Here is some simple advice:

When the triangle has a right angle, then use it, that is usually much simpler.

When two angles are known, work out the third using Angles of a Triangle Add to 180° .

Try The Law of Sines before the The Law of Cosines as it is easier to use.

Free Mathematics Tutorials

Free Mathematics Tutorials

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Right Triangle Questions

Multiple choice questions right triangle problems related to trigonometry with answers at the bottom of the page.

Questions with their Answers

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Question 10

Question 11, question 12, more references and links, popular pages.

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Chapter 4.2: Right Triangle Trigonometry

Using right triangle trigonometry to solve applied problems, using trigonometric functions.

In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.

How To: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides.

  • For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
  • Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
  • Using the value of the trigonometric function and the known side length, solve for the missing side length.

Example 5: Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure 11.

A right triangle with sides a, c, and 7. Angle of 30 degrees is also labeled.

We know the angle and the opposite side, so we can use the tangent to find the adjacent side.

We rearrange to solve for [latex]a[/latex].

We can use the sine to find the hypotenuse.

Again, we rearrange to solve for [latex]c[/latex].

A right triangle has one angle of [latex]\frac{\pi }{3}[/latex] and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer’s eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer’s eye.

Diagram of a radio tower with line segments extending from the top and base of the tower to a point on the ground some distance away. The two lines and the tower form a right triangle. The angle near the top of the tower is the angle of depression. The angle on the ground at a distance from the tower is the angle of elevation.

How To: Given a tall object, measure its height indirectly.

  • Make a sketch of the problem situation to keep track of known and unknown information.
  • Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
  • At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.
  • Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
  • Solve the equation for the unknown height.

Example 6: Measuring a Distance Indirectly

To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of [latex]57^\circ [/latex] between a line of sight to the top of the tree and the ground, as shown in Figure 13. Find the height of the tree.

A tree with angle of 57 degrees from vantage point. Vantage point is 30 feet from tree.

We know that the angle of elevation is [latex]57^\circ [/latex] and the adjacent side is 30 ft long. The opposite side is the unknown height.

The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of [latex]57^\circ [/latex], letting [latex]h[/latex] be the unknown height.

The tree is approximately 46 feet tall.

How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of [latex]\frac{5\pi }{12}[/latex] with the ground? Round to the nearest foot.

  • Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution
  • Example: Determine the Length of a Side of a Right Triangle Using a Trig Equation. Authored by : Mathispower4u. Located at : https://youtu.be/8jU2R3BuR5E . License : All Rights Reserved . License Terms : Standard YouTube License

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Mathematics LibreTexts

1.2: Right Triangle Trigonometry

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Learning Objectives

  • Define the six trigonometric functions in terms of \(x,y,r\) (rectangular coordinates) and language associated with right triangles. 
  • Find function values for 30°(\(\dfrac{\pi}{6}\)),45°(\(\dfrac{\pi}{4}\)),and 60°(\(\dfrac{\pi}{3}\)).
  • Use equal cofunctions of complementary angles.
  • Use the definitions of trigonometric functions of any angle.
  • Use right-triangle trigonometry to solve applied problems.

Mt. Everest, which straddles the border between China and Nepal, is the tallest mountain in the world. Measuring its height is no easy task and, in fact, the actual measurement has been a source of controversy for hundreds of years. The measurement process involves the use of triangles and a branch of mathematics known as trigonometry. In this section, we will define a new group of functions known as trigonometric functions, and find out how they can be used to measure heights, such as those of the tallest mountains.

We begin by drawing a circle centered at the origin with radius \(r\) , and marking the point on the circle indicated by some angle \(\theta\) as in Figure \(\PageIndex{1}\) . This point has coordinates (\(x\), \(y\)). We can also say that (\(x\), \(y\)) is on the terminal side of angle \(\theta\). 

屏幕快照 2019-07-05 上午9.19.17.png

Figure \(\PageIndex{1}\)

If we drop a line segment vertically down from this point to the x axis, we would form a right triangle inside of the circle.

No matter which quadrant our angle \(\theta\) puts us in we can draw a triangle by dropping a perpendicular line segment to the \(x\) axis, keeping in mind that the values of \(x\) and \(y\) may be positive or negative, depending on the quadrant.

Additionally, if the angle \(\theta\) puts us on an axis, we simply measure the radius as the \(x\) or \(y\) with the other value being 0, again ensuring we have appropriate signs on the coordinates based on the quadrant.

Triangles obtained from different radii will all be similar triangles, meaning corresponding sides scale proportionally. While the lengths of the sides may change, the ratios of the side lengths will always remain constant for any given angle.

屏幕快照 2019-07-05 上午9.14.03.png

\(\dfrac{x_{1} }{r_{1} } =\dfrac{x_{2} }{r_{2} }\)

To be able to refer to these ratios more easily, we will give them names. Since the ratios depend on the angle, we will write them as functions of the angle \(\theta\).

The Six Trigonometric Functions

For the point (\(x\), \(y\)) on a circle of radius \(r\) at an angle of \(\theta\), we can define the six trigonometric functions as the ratios of the sides of the corresponding triangle:

The sine function: \(\sin (\theta )=\dfrac{y}{r}\)

The cosine function: \(\cos (\theta )=\dfrac{x}{r}\)

The tangent  function: \(\tan (\theta )=\dfrac{y}{x}\)

The cosecant  function: \(\csc (\theta )=\dfrac{r}{y}\)

The secant  function: \(\sec (\theta )=\dfrac{r}{x}\)

The cotangent  function: \(\cot (\theta )=\dfrac{x}{y}\)

Example \(\PageIndex{1}\)

The point (3, 4) is on the circle of radius 5 at some angle \(\theta\) . Find the six trigonometric function values of \(\theta\)

We have \(x=3\), \(y=4\), and \(r=5\). Using the previously listed definitions we have

\[ \sin (\theta )=\dfrac{y}{r} =\dfrac{4}{5}\\ \cos (\theta )=\dfrac{x}{r} =\dfrac{3}{5}\\ \tan (\theta )=\dfrac{y}{x} =\dfrac{4}{3}\\ \csc (\theta )=\dfrac{r}{y} =\dfrac{5}{4}\\ \sec (\theta )=\dfrac{r}{x} =\dfrac{5}{3}\\ \cot (\theta )=\dfrac{x}{y} =\dfrac{3}{4} \]

 Consider once again the set-up with a right triangle inscribed in a circle with radius \(r\). The two legs of the right triangle are \(x\) and \(y\) and the hypotenuse is \(r\). According to the Pythagorean Theorem we have the following relationship: 

\(x^2+y^2=r^2\)

屏幕快照 2019-07-05 上午9.16.24.png

Using Right Triangles to Evaluate Trigonometric Functions

We now define the six trigonometric functions in terms of the names of the sides of a right triangle. First, let's recreate our right triangle (shown in Figure \(\PageIndex{2}\). If we drop a vertical line segment from the point \((x,y)\) to the x -axis, we have a right triangle whose vertical side has length \(y\) and whose horizontal side has length \(x\). We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.

Figure \(\PageIndex{2}\)

We want to have another way to express the ratios without relying on a set of axes. To be able to use these ratios freely, we will give the sides more general names: Instead of \(x\),we will call the side between the given angle and the right angle the adjacent side to angle \(\theta\). (Adjacent means “next to.”) Instead of \(y\),we will call the side most distant from the given angle the opposite side from angle \(\theta\). And instead of \(r\),we will call the side of a right triangle opposite the right angle the hypotenuse . These sides are labeled in Figure \(\PageIndex{3}\).

3.7 1.png

Figure \(\PageIndex{3}\): The sides of a right triangle in relation to angle \(\theta\).

Understanding Right Triangle Relationships

Given a right triangle with an acute angle of \(\theta\),

\[\begin{align} \sin (\theta) &= \dfrac{\text{opposite}}{\text{hypotenuse}} \label{sindef}\\ \cos (\theta) &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \label{cosdef}\\ \tan (\theta) &= \dfrac{\text{opposite}}{\text{adjacent}} \label{tandef} \\ \csc (\theta) &= \dfrac{\text{hypotenuse}}{\text{opposite}} \label{cscdef}\\ \sec (\theta) &= \dfrac{\text{hypotenuse}}{\text{adjacent}} \label{secdef}\\ \cot (\theta) &= \dfrac{\text{adjacent}}{\text{opposite}} \label{cotdef}\\\end{align}\]

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “ S ine is o pposite over h ypotenuse, C osine is a djacent over h ypotenuse, T angent is o pposite over a djacent.” To get the other three, we remember that cosecant is the reciprocal of sine, secant is the reciprocal of cosine, and cotangent is the reciprocal of tangent. 

how to: Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle

  • Find the sine as the ratio of the opposite side to the hypotenuse.
  • Find the cosine as the ratio of the adjacent side to the hypotenuse.
  • Find the tangent as the ratio of the opposite side to the adjacent side.
  • Find the cosecant as the ratio of the hypotenuse to the opposite side. 
  • Find the secant as the ratio of the hypotenuse to the adjacent side.
  • Find the cotangent as the ratio of the adjacent side to the opposite side.

Also remember that sine and cosecant are reciprocals, cosine and secant and reciprocals, and tangent and cotangent are reciprocals. 

Example \(\PageIndex{2}\): Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in Figure \(\PageIndex{4}\), find the six trigonometric function values of angle \(\alpha\).

A right triangle with side lengths of 8, 15, and 17. Angle alpha also labeled which is opposite to the side labeled 8.

From the perspective of angle \(\alpha\), the adjacent side has length 15, the opposite side has length 8, and the hypotenuse has length 17, so we have 

\[\begin{align*} \sin (\alpha) &= \dfrac{\text{O}}{\text{H}}  = \dfrac{8}{17} \\ \cos (\alpha) &= \dfrac{\text{A}}{\text{H}} = \dfrac{15}{17} \\ \tan (\alpha) &= \dfrac{\text{O}}{\text{A}}  = \dfrac{8}{15} \\ \csc (\alpha) &= \dfrac{\text{H}}{\text{O}} = \dfrac{17}{8} \\ \sec (\alpha) &= \dfrac{\text{H}}{\text{A}} = \dfrac{17}{15}  \\ \cot (\alpha) &= \dfrac{\text{A}}{\text{O}}  = \dfrac{15}{8} \end{align*}\]

Exercise \(\PageIndex{1}\)

Given the triangle shown in Figure \(\PageIndex{5}\), find the value of \(\sin t\).

A right triangle with sides of 7, 24, and 25. Also labeled is angle t which is opposite the side labeled 7.

\(\frac{7}{25}\)

Relating Angles and Their Functions

When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure \(\PageIndex{6}\). The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

Right triangle with angles alpha and beta. Sides are labeled hypotenuse, adjacent to alpha/opposite to beta, and adjacent to beta/opposite alpha.

We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.

how to: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles

  • If needed, draw the right triangle and label the angle provided.
  • Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
  • sine as the ratio of the opposite side to the hypotenuse
  • cosine as the ratio of the adjacent side to the hypotenuse
  • tangent as the ratio of the opposite side to the adjacent side
  • secant as the ratio of the hypotenuse to the adjacent side
  • cosecant as the ratio of the hypotenuse to the opposite side
  • cotangent as the ratio of the adjacent side to the opposite side

Example \(\PageIndex{3}\): Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in Figure \(\PageIndex{7}\), evaluate \( \sin α, \cos α, \tan α, \sec α, \csc α,\) and \( \cot α\).

Right triangle with sides of 3, 4, and 5. Angle alpha is also labeled which is opposite the side labeled 4.

\[ \begin{align*} \sin α &= \dfrac{\text{opposite } α}{\text{hypotenuse}} = \dfrac{4}{5} \\ \cos α &= \dfrac{\text{adjacent to }α}{\text{hypotenuse}}=\dfrac{3}{5} \\ \tan α &= \dfrac{\text{opposite }α}{\text{adjacent to }α}=\dfrac{4}{3} \\ \sec α &= \dfrac{\text{hypotenuse}}{\text{adjacent to }α}= \dfrac{5}{3} \\ \csc α &= \dfrac{\text{hypotenuse}}{\text{opposite }α}=\dfrac{5}{4} \\ \cot α &= \dfrac{\text{adjacent to }α}{\text{opposite }α}=\dfrac{3}{4} \end{align*}\]

Exercise \(\PageIndex{2}\)

Using the triangle shown in Figure \(\PageIndex{8}\), evaluate \( \sin t, \cos t,\tan t, \sec t, \csc t,\) and \(\cot t\).

Right triangle with sides 33, 56, and 65. Angle t is also labeled which is opposite to the side labeled 33.

\[\begin{align*} \sin t &= \frac{33}{65}, \cos t= \frac{56}{65},\tan t= \frac{33}{56}, \\ \\ \sec t &= \frac{65}{56},\csc t= \frac{65}{33},\cot t= \frac{56}{33} \end{align*}\]

Finding Trigonometric Functions of Special Angles Using Side Lengths

We now use our right triangle definitions to evaluate trigonometric values of special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. Our special angles are \(30°, 45°,\) and \(60°\)

Recall from geometry the \(30°,60°,90°\) triangle, which can also be described as a \(\frac{π}{6}, \frac{π}{3},\frac{π}{2}\) triangle. The sides have lengths in the relation \(s,\sqrt{3}s,2s.\) The sides of a \(45°,45°,90° \)triangle, which can also be described as a \(\frac{π}{4},\frac{π}{4},\frac{π}{2}\) triangle, have lengths in the relation \(s,s,\sqrt{2}s.\) These relations are shown in Figure \(\PageIndex{9}\).

Two side-by-side graphs of circles with inscribed angles. First circle has angle of pi/3 inscribed, radius of 2s, base of length s and height of length . Second circle has angle of pi/4 inscribed with radius , base of length s and height of length s.

We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.

Given trigonometric functions of a special angle, evaluate using side lengths.

  • Use the side lengths shown in Figure \(\PageIndex{9}\) for the special angle you wish to evaluate.
  • Use the ratio of side lengths appropriate to the function you wish to evaluate.

Example \(\PageIndex{4}\): Evaluating Trigonometric Functions of Special Angles Using Side Lengths

Find the exact value of the trigonometric functions of \(\frac{π}{3}\), using side lengths.

\[\begin{align*} \sin (\dfrac{π}{3}) &= \dfrac{\text{opp}}{\text{hyp}}=\dfrac{\sqrt{3}s}{2s}=\dfrac{\sqrt{3}}{2} \\ \cos (\dfrac{π}{3}) &= \dfrac{\text{adj}}{\text{hyp}}=\dfrac{s}{2s}=\dfrac{1}{2} \\ \tan (\dfrac{π}{3}) &= \dfrac{\text{opp}}{\text{adj}} =\dfrac{\sqrt{3}s}{s}=\sqrt{3} \\ \sec (\dfrac{π}{3}) &= \dfrac{\text{hyp}}{\text{adj}} = \dfrac{2s}{s}=2 \\ \csc (\dfrac{π}{3}) &= \dfrac{\text{hyp}}{\text{opp}} =\dfrac{2s}{\sqrt{3}s}=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3} \\ \cot (\dfrac{π}{3}) &= \dfrac{\text{adj}}{\text{opp}}=\dfrac{s}{\sqrt{3}s}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3} \end{align*}\]

Exercise \(\PageIndex{3}\)

Find the exact value of the trigonometric functions of \(\frac{π}{4}\) using side lengths.

\( \sin (\frac{π}{4})=\frac{\sqrt{2}}{2}, \cos (\frac{π}{4})=\frac{\sqrt{2}}{2}, \tan (\frac{π}{4})=1,\)

\( \sec (\frac{π}{4})=\sqrt{2}, \csc (\frac{π}{4})=\sqrt{2}, \cot (\frac{π}{4}) =1 \)

Using Equal Cofunction of Complements

If we look more closely at the relationship between the sine and cosine of the special angles relative to the unit circle, we will notice a pattern. In a right triangle with angles of \(\frac{π}{6}\) and \(\frac{π}{3}\), we see that the sine of \(\frac{π}{3}\), namely \(\frac{\sqrt{3}}{2}\), is also the cosine of \(\frac{π}{6}\), while the sine of \(\frac{π}{6}\), namely \(\frac{1}{2},\) is also the cosine of \(\frac{π}{3}\) (Figure \(\PageIndex{10}\)).

\[\begin{align*} \sin \frac{π}{3} &= \cos \frac{π}{6}=\frac{\sqrt{3}s}{2s}=\frac{\sqrt{3}}{2} \\ \sin \frac{π}{6} &= \cos \frac{π}{3}=\frac{s}{2s}=\frac{1}{2} \end{align*}\]

A graph of circle with angle pi/3 inscribed with a radius of 2s, a base with length s and a height of.

This result should not be surprising because, as we see from Figure \(\PageIndex{10}\), the side opposite the angle of \(\frac{π}{3}\) is also the side adjacent to \(\frac{π}{6}\), so \(\sin (\frac{π}{3})\) and \(\cos (\frac{π}{6})\) are exactly the same ratio of the same two sides, \(\sqrt{3} s\) and \(2s.\) Similarly, \( \cos (\frac{π}{3})\) and \( \sin (\frac{π}{6})\) are also the same ratio using the same two sides, \(s\) and \(2s\).

The interrelationship between the sines and cosines of \(\frac{π}{6}\) and \(\frac{π}{3}\) also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to π, π,and the right angle is \(\frac{π}{2}\), the remaining two angles must also add up to \(\frac{π}{2}\). That means that a right triangle can be formed with any two angles that add to \(\frac{π}{2}\)—in other words, any two complementary angles. So we may state a cofunction identity : If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure \(\PageIndex{11}\).

Right triangle with angles alpha and beta. Equivalence between sin alpha and cos beta. Equivalence between sin beta and cos alpha.

Using this identity, we can state without calculating, for instance, that the sine of \(\frac{π}{12}\) equals the cosine of \(\frac{5π}{12}\), and that the sine of \(\frac{5π}{12}\) equals the cosine of \(\frac{π}{12}\). We can also state that if, for a certain angle \(t, \cos t= \frac{5}{13},\) then \( \sin (\frac{π}{2}−t)=\frac{5}{13}\) as well.

COFUNCTION IDENTITIES

The cofunction identities in radians are listed in Table \(\PageIndex{1}\).

how to: Given the sine and cosine of an angle, find the sine or cosine of its complement.

  • To find the sine of the complementary angle, find the cosine of the original angle.
  • To find the cosine of the complementary angle, find the sine of the original angle.

Example \(\PageIndex{5}\): Using Cofunction Identities

If \( \sin t = \frac{5}{12},\) find \(( \cos \frac{π}{2}−t)\).

According to the cofunction identities for sine and cosine,

\[ \sin t= \cos (\dfrac{π}{2}−t). \nonumber\]

\[ \cos (\dfrac{π}{2}−t)= \dfrac{5}{12}. \nonumber\]

Exercise \(\PageIndex{4}\)

If \(\csc (\frac{π}{6})=2,\) find \( \sec (\frac{π}{3}).\)

Using Trigonometric Functions

In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.

how to: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides

  • For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
  • Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
  • Using the value of the trigonometric function and the known side length, solve for the missing side length.

Example \(\PageIndex{6}\): Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure \(\PageIndex{12}\).

A right triangle with sides a, c, and 7. Angle of 30 degrees is also labeled which is opposite the side labeled 7.

We know the angle and the opposite side, so we can use the tangent to find the adjacent side.

\[ \tan (30°)= \dfrac{7}{a} \nonumber\]

We rearrange to solve for \(a\).

\[\begin{align} a &=\dfrac{7}{ \tan (30°)} \\ & =12.1 \end{align} \nonumber\]

We can use the sine to find the hypotenuse.

\[ \sin (30°)= \dfrac{7}{c} \nonumber\]

Again, we rearrange to solve for \(c\).

\[\begin{align*} c &= \dfrac{7}{\sin (30°)} =14 \end{align*}\]

Exercise \(\PageIndex{5}\):

A right triangle has one angle of \(\frac{π}{3}\) and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

\(\mathrm{adjacent=10; opposite=10 \sqrt{3}; }\) missing angle is \(\frac{π}{6}\)

Using Right Triangle Trigonometry to Solve Applied Problems

Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. See Figure \(\PageIndex{13}\).

Diagram of a radio tower with line segments extending from the top and base of the tower to a point on the ground some distance away. The two lines and the tower form a right triangle. The angle near the top of the tower is the angle of depression. The angle on the ground at a distance from the tower is the angle of elevation.

how to: Given a tall object, measure its height indirectly

  • Make a sketch of the problem situation to keep track of known and unknown information.
  • Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
  • At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.
  • Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
  • Solve the equation for the unknown height.

Example \(\PageIndex{7}\): Measuring a Distance Indirectly

To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of 57° between a line of sight to the top of the tree and the ground, as shown in Figure \(\PageIndex{14}\). Find the height of the tree.

A tree with angle of 57 degrees from vantage point. Vantage point is 30 feet from tree.

We know that the angle of elevation is \(57°\) and the adjacent side is 30 ft long. The opposite side is the unknown height.

The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of \(57°\), letting \(h\) be the unknown height.

\[\begin{array}{cl} \tan θ = \dfrac{\text{opposite}}{\text{adjacent}} & \text{} \\ \tan (57°) = \dfrac{h}{30} & \text{Solve for }h. \\ h=30 \tan (57°) & \text{Multiply.} \\ h≈46.2 & \text{Use a calculator.} \end{array} \]

The tree is approximately 46 feet tall.

Exercise \(\PageIndex{6}\):

How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of \(\frac{5π}{12}\) with the ground? Round to the nearest foot.

About 52 ft

Access these online resources for additional instruction and practice with right triangle trigonometry.

  • Finding Trig Functions on Calculator
  • Finding Trig Functions Using a Right Triangle
  • Relate Trig Functions to Sides of a Right Triangle
  • Determine Six Trig Functions from a Triangle
  • Determine Length of Right Triangle Side

Visit this website for additional practice questions from Learningpod.

Key Equations

Cofunction Identities

\[\begin{align*} \cos t &= \sin ( \frac{π}{2}−t) \\ \sin t &= \cos (\frac{π}{2}−t) \\ \tan t &= \cot (\frac{π}{2}−t) \\ \cot t &= \tan (\frac{π}{2}−t) \\ \sec t &= \csc (\frac{π}{2}−t) \\ \csc t &= \sec (\frac{π}{2}−t) \end{align*}\]

Key Concepts

  • We can define trigonometric functions as ratios of the side lengths of a right triangle. See Example .
  • The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle. See Example .
  • We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur. See Example .
  • Any two complementary angles could be the two acute angles of a right triangle.
  • If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa. See Example .
  • We can use trigonometric functions of an angle to find unknown side lengths.
  • Select the trigonometric function representing the ratio of the unknown side to the known side. See Example .
  • Right-triangle trigonometry permits the measurement of inaccessible heights and distances.
  • The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known. See Example .

solve right angled triangle problems

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Right-Triangle Word Problems

What is a right-triangle word problem.

A right-triangle word problem is one in which you are given a situation (like measuring something's height) that can be modelled by a right triangle. You will draw the triangle, label it, and then solve it; finally, you interpret this solution within the context of the original exercise.

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Right Triangle Word Problems

Once you've learned about trigonometric ratios (and their inverses), you can solve triangles. Naturally, many of these triangles will be presented in the context of word problems. A good first step, after reading the entire exercise, is to draw a right triangle and try to figure out how to label it. Once you've got a helpful diagram, the math is usually pretty straightforward.

  • A six-meter-long ladder leans against a building. If the ladder makes an angle of 60° with the ground, how far up the wall does the ladder reach? How far from the wall is the base of the ladder? Round your answers to two decimal places, as needed.

First, I'll draw a picture. It doesn't have to be good or to scale; it just needs to be clear enough that I can keep track of what I'm doing. My picture is:

To figure out how high up the wall the top of the ladder is, I need to find the height h of my triangle.

Since they've given me an angle measure and "opposite" and the hypotenuse for this angle, I'll use the sine ratio for finding the height:

sin(60°) = h/6

6 sin(60°) = h = 3sqrt[3]

Plugging this into my calculator, I get an approximate value of 5.196152423 , which I'll need to remember to round when I give my final answer.

For the base, I'll use the cosine ratio:

cos(60°) = b/6

6×cos(60°) = b = 3

Nice! The answer is a whole number; no radicals involved. I won't need to round this value when I give my final answer. Checking the original exercise, I see that the units are "meters", so I'll include this unit on my numerical answers:

ladder top height: about 5.20 m

ladder base distance: 3 m

Note: Unless you are told to give your answer in decimal form, or to round, or in some other way not to give an "exact" answer, you should probably assume that the "exact" form is what they're wanting. For instance, if they hadn't told me to round my numbers in the exercise above, my value for the height would have been the value with the radical.

Algebra Tutors

  • A five-meter-long ladder leans against a wall, with the top of the ladder being four meters above the ground. What is the approximate angle that the ladder makes with the ground? Round to the nearest whole degree.

As usual, I'll start with a picture, using "alpha" to stand for the base angle:

They've given me the "opposite" and the hypotenuse, and asked me for the angle value. For this, I'll need to use inverse trig ratios.

sin(α) = 4/5

m(α) = sin −1 (4/5) = 53.13010235...

(Remember that m(α) means "the measure of the angle α".)

So I've got a value for the measure of the base angle. Checking the original exercise, I see that I am supposed to round to the nearest whole degree, so my answer is:

base angle: 53°

  • You use a transit to measure the angle of the sun in the sky; the sun fills 34' of arc. Assuming the sun is 92,919,800 miles away, find the diameter of the sun. Round your answer to the nearest whole mile.

First, I'll draw a picture, labelling the angle on the Earth as being 34 minutes, where minutes are one-sixtieth of a degree. My drawing is *not* to scale!:

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Hmm... This "ice-cream cone" picture doesn't give me much to work with, and there's no right triangle.

The two lines along the side of my triangle measure the lines of sight from Earth to the sides of the Sun. What if I add another line, being the direct line from Earth to the center of the Sun?

Now that I've got this added line, I have a right triangle — two right triangles, actually — but I only need one. I'll use the triangle on the right.

(The angle measure , "thirty-four arc minutes", is equal to 34/60 degrees. Dividing this in half is how I got 17/60 of a degree for the smaller angle.)

I need to find the width of the Sun. That width will be twice the base of one of the right triangles. With respect to my angle, they've given me the "adjacent" and have asked for the "opposite", so I'll use the tangent ratio:

tan(17/60°) = b/92919800

92919800×tan(17/60°) = b = 459501.4065...

This is just half the width; carrying the calculations in my calculator (to minimize round-off error), I get a value of 919002.8129 . This is higher than the actual diameter, which is closer to 864,900 miles, but this value will suffice for the purposes of this exercise.

diameter: about 919,003 miles

  • A private plane flies 1.3 hours at 110 mph on a bearing of 40°. Then it turns and continues another 1.5 hours at the same speed, but on a bearing of 130°. At the end of this time, how far is the plane from its starting point? What is its bearing from that starting point? Round your answers to whole numbers.

The bearings tell me the angles from "due north", in a clockwise direction. Since 130 − 40 = 90 , these two bearings create a right angle where the plane turns. From the times and rates, I can find the distances travelled in each part of the trip:

1.3 × 110 = 143 1.5 × 110 = 165

Now that I have the lengths of the two legs, I can set up a triangle:

(The angle θ is the bearing, from the starting point, of the plane's location at the ending point of the exercise.)

I can find the distance between the starting and ending points by using the Pythagorean Theorem :

143 2 + 165 2 = c 2 20449 + 27225 = c 2 47674 = c 2 c = 218.3437657...

The 165 is opposite the unknown angle, and the 143 is adjacent, so I'll use the inverse of the tangent ratio to find the angle's measure:

165/143 = tan(θ)

tan −1 (165/143) = θ = 49.08561678...

But this angle measure is not the "bearing" for which they've asked me, because the bearing is the angle with respect to due north. To get the measure they're wanting, I need to add back in the original forty-degree angle:

distance: 218 miles

bearing: 89°

Related: Another major class of right-triangle word problems you will likely encounter is angles of elevation and declination .

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solve right angled triangle problems

Mr. Mathematics

Trigonometry – Problems with Right-Angled Triangles

March 12, 2023.

Scheme of work: GCSE Foundation: Year 11: Term 2: Trigonometry – Problems with Right-Angled Triangles

Prerequisite Knowledge

  • Express a multiplicative relationship between two quantities
  • understand and use proportion as equality of ratios
  • know the formulae for: Pythagoras’ theorem, a 2  + b 2  = c 2
  • apply angle facts, triangle congruence, similarity and properties of quadrilaterals to conjecture and derive results about angles and sides, including Pythagoras’ Theorem and the fact that the
  • base angles of an isosceles triangle are equal, and use known results to obtain simple proofs

Success Criteria

  • know the trigonometric ratios, Sin x = Opp/Hyp, Cos x = Adj/Hyp, Tan x = Opp/Adj.
  • apply them to find angles and lengths in right-angled triangles and, where possible, general triangles in two dimensional figures.
  • know the exact values of Sin x and Cos Sin x for x = 0°, 30°, 45°, 60°, and 90°; know the exact value of Tan x for 0°, 30°, 45° and 60°.

Key Concepts

  • Sin, Cos and Tan are trigonometric functions that find lengths and angles in right-angled triangles.
  • The hypotenuse is opposite the right angle; the opposite refers to the side opposite the angle in question, and the adjacent side runs adjacent to the angle.
  • The inverse operations of sin, cos and tan are pronounced arcos, arcsin and arctan.

Common Misconceptions

  • Students often have difficulty knowing which trigonometric ratio to apply. Encourage them to label the sides to identify the correct ratio clearly.
  • Use SOHCAHTOA as a memory aid as students often forget the trigonometric ratios.
  • When using trigonometric ratios to calculate angles students often forget to use the inverse functions.
  • Students often try to apply right-angled formulae to non-right-angled triangles.

Trigonometry Problems with Right-Angled Triangles Resources

Video Tutorial (Free for all)

Online Lesson (Lite/Full)

Downloadable Resources (Full)

View Trigonometry Introduction video tutorial on YouTube

Extended Learning

View Finding Lengths using Trigonometry video tutorial on YouTube

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    Right triangles problems are solved and detailed explanations are included. Example - Problem 1: Find sin (x) and cos (x) in the right triangle shown below. Solution to Problem 1: First use the Pythagorean theorem to find the hypotenuse h of the right triangle. h = √ (6 2 + 8 2 ) = √ (36 + 64) = 10

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    Study Guide Topics Solving Right Triangles Right Triangle Review Techniques for Solving Problems Applications Problems Terms and Formulae Problems Previous Next Problem : Solve the following right triangle, in which C = 90o: a = 6, B = 40o . A = 90o - B = 50o. b = a tan (B) 5.0. c = 7.8 .

  9. 1.4: Solving Right Triangles

    To solve a right triangle, you need to find all sides and angles in it. You will usually use sine, cosine, or tangent; inverse sine, ... Here we will solve several problems involving these angles and distances. Finding the angle of elevation. You are standing 20 feet away from a tree, and you measure the angle of elevation to be \(38^{\circ}\). ...

  10. Finding an Angle in a Right Angled Triangle

    Finding an Angle in a Right Angled Triangle Angle from Any Two Sides We can find an unknown angle in a right-angled triangle, as long as we know the lengths of two of its sides. Example The ladder leans against a wall as shown. What is the angle between the ladder and the wall? The answer is to use Sine, Cosine or Tangent! But which one to use?

  11. 2.2: Solving Right Triangles.

    Practice each skill in the Homework Problems listed. 1 Solve a right triangle #1-16, 63-74. 2 Use inverse trig ratio notation #17-34. 3 Use trig ratios to find an angle #17-22, 35-38. 4 Solve problems involving right triangles #35-48. 5 Know the trig ratios for the special angles #49-62, 75-78.

  12. 5.2: Solution of Right Triangles

    Hipparchus assumed the vertex of each angle to be the center of a circle, as \(\angle AOB\) is shown to be in the circle of Figure \(\PageIndex{1}\). Depending on the number of degrees in \(\angle AOB\), his table would give the length of the chord \(AB\) relative to the radius of the circle.

  13. Right Triangle Calculator

    1 2 ab = 1 2 ch Special Right Triangles 30°-60°-90° triangle: The 30°-60°-90° refers to the angle measurements in degrees of this type of special right triangle. In this type of right triangle, the sides corresponding to the angles 30°-60°-90° follow a ratio of 1:√ 3 :2.

  14. Solving for a side in right triangles with trigonometry

    Solution Step 1: Determine which trigonometric ratio to use. Let's focus on angle B since that is the angle that is explicitly given in the diagram. 50 ∘ 6 ? C B A Note that we are given the length of the hypotenuse , and we are asked to find the length of the side opposite angle B .

  15. Solving Triangles

    1. Angles Add to 180°: A + B + C = 180° When you know two angles you can find the third. 2. Law of Sines (the Sine Rule): a sin (A) = b sin (B) = c sin (C) When there is an angle opposite a side, this equation comes to the rescue. Note: angle A is opposite side a, B is opposite b, and C is opposite c. 3. Law of Cosines (the Cosine Rule):

  16. Right Triangle Questions

    a) 9 b) 9 √2 d) 18 Find the length of the hypotenuse in the right triangle below where is a real number. a) 5 b) 10 c) 25 d) √ 5 Find the area of a square whose diagonal is 40 meters. a) 80 m b) 800 m c) 1600 m d) 40 m In the figure below BC is perpendicular to AD, CD = 8, the measure of angle D is 60° and the measure of angle A is 45°.

  17. PDF RIGHT TRIANGLE TRIGONOMETRY

    Right Triangle Trigonometry Finding Missing Angles of Right Triangles 2. Solution: Using the 55o angle as our reference angle, 14 is the length of the opposite leg and x is the length of the hypotenuse. Therefore, we will use the tangent ratio: Opposite sin( )θ= Hypotenuse 14 sin(55 ) x D= x⋅sin(55 ) 14D= 14 sin(55 ) x = D x ≈17.0908

  18. Using Right Triangle Trigonometry to Solve Applied Problems

    Solution Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height.

  19. 1.2: Right Triangle Trigonometry

    Use right-triangle trigonometry to solve applied problems. Mt. Everest, which straddles the border between China and Nepal, is the tallest mountain in the world. ... the side of a right triangle opposite the right angle. This page titled 1.2: Right Triangle Trigonometry is shared under a CC BY 4.0 license and was authored, ...

  20. Solving right-triangle word problems: Learn here!

    MathHelp.com Right Triangle Word Problems Once you've learned about trigonometric ratios (and their inverses), you can solve triangles. Naturally, many of these triangles will be presented in the context of word problems. A good first step, after reading the entire exercise, is to draw a right triangle and try to figure out how to label it.

  21. Right triangle trigonometry word problems

    Math > High school geometry Right triangles & trigonometry Modeling with right triangles Right triangle trigonometry word problems Google Classroom You might need: Calculator Bugs Bunny was 33 meters below ground, digging his way toward Pismo Beach, when he realized he wanted to be above ground.

  22. PDF Applications of Right Triangles and Trig Functions

    3) Isolate the triangle 4) Solve 5) Answer the question Since we have a fight triangle with an angle and hypotenuse we can use the sine function to find the Draw a picture and label the parts Isolate the triangle and Solve Answer the question! 300' 300' Since the triangle is 4 feet off the ground, we need to add 4' to determine the height

  23. Problems with Right-Angled Triangles

    Sin, Cos and Tan are trigonometric functions that find lengths and angles in right-angled triangles. The hypotenuse is opposite the right angle; the opposite refers to the side opposite the angle in question, and the adjacent side runs adjacent to the angle. The inverse operations of sin, cos and tan are pronounced arcos, arcsin and arctan.