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8. Right Triangles & Trigonometry
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Geometry

8.03 Trigonometric ratios

Lesson
Practice

Ratios in right triangles

A trigonometric ratio, or trigonometric function, is a relationship between an angle and a pair of sides in a right triangle.

To talk about the trigonometric ratios, we first label the sides of a right triangle with respect to a particular angle, sometimes called a reference angle:

Right triangle A B C with right angle C. Side A C is labeled Leg adjacent to angle A, side B C labeled Leg opposite angle A, and side A B labeled Hypotenuse.

The adjacent side is the leg of the right triangle that is connected to the angle of reference.

The opposite side is the leg of the right triangle that is across from the angle of reference.

The hypotenuse is the side of the right triangle that is opposite the right angle (the longest side of the right triangle).

Note that in this case \angle A was used as the reference angle. The side labels would be different if \angle B had been used instead.

Exploration

Drag the slider to change the size of the reference angle and drag the triangle to change its size.

Loading interactive...
  1. What do you notice and wonder about the ratios?
  2. Can you justify why the ratios stay the same for a given angle, even when you change the size of the triangle?

Right triangles with the same acute angle are similar. Using the proportionality of similar triangles, we can calculate and estimate unknown side lengths and angles in right triangles.

With the given notation of ratios in mind, we then define the following three trigonometric ratios:

Sine (sin)

The ratio between the length of the side opposite to a given angle and the hypotenuse of the right triangle

Cosine (cos)

The ratio between the length of the side adjacent to a given angle and the hypotenuse of the right triangle

Tangent (tan)

The ratio between the sides opposite and adjacent to a given angle of a right triangle

That is, for a given reference angle \theta, we have:\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}

Examples

Example 1

Write the following ratios for the given triangle:

Right triangle A B C with right angle C. Angle A is labeled alpha, and angle B labeled theta.
a

\sin\theta

Worked Solution
Create a strategy

We want to first label the side lengths of the right triangle as shown in the below diagram.

Right Triangle A B C with right angle C. Side A C is labeled Opposite, side B C labeled Adjacent, and side A B labeled Hypotenuse.

The reference angle is \theta, so the opposite side is \overline{AC} as it is directly across from the angle of reference, \theta. The hypotenuse is \overline{AB} as it is opposite the right angle. The adjacent side is \overline{BC} as the side is connected to the angle of reference and is not the hypotenuse.

Now that we have the sides of the right triangle labeled, we want to write the correct trigonometric ratio for sine. Sine is the ratio between the side opposite to the given angle and the hypotenuse of the right triangle.

Apply the idea

\sin\theta = \dfrac{AC}{AB}

b

\cos \theta

Worked Solution
Create a strategy

We want to reference the sides of the right triangle that we labeled in part (a) to write the correct trigonometric ratio for cosine. Cosine is the ratio between the side adjacent to the given angle and the hypotenuse of a right triangle.

Apply the idea

\cos\theta = \dfrac{BC}{AB}

c

\tan \theta

Worked Solution
Create a strategy

We want to reference the sides of the right triangle that we labeled in part (a) to write the correct trigonometric ratio for tangent. Tangent is the ratio between the sides opposite and adjacent to the given angle of a right triangle.

Apply the idea

\tan\theta = \dfrac{AC}{BC}

Example 2

Consider the triangle in the figure. If \sin\theta=\dfrac{4}{5}:

Right triangle A B C with right angle A. Side A C has a length of x, A B has a length of 4, and B C has a length of 5.
a

Which angle is represented by \theta?

Worked Solution
Create a strategy

In a right triangle, we know that the sine of an angle is equal to the ratio of the side length opposite that angle and the length of the hypotenuse. Since 4 is in the numerator, the side length with length 4 is the opposite side and the hypotenuse has a length of 5. We need to look at the diagram and find the angle that is across from the opposite side with length of 4.

Apply the idea

\angle BCA

Reflect and check

We could have also named this angle \angle ACB, or even just \angle C since there is no ambiguity as to what that represents.

b

Find the value of \cos\theta.

Worked Solution
Create a strategy

From part (a), we labeled the opposite and hypotenuse sides. The adjacent side is labeled x in the diagram. In order to find \cos \theta, we need to find the value of x and then write the trigonometric ratio.

To find the value of x, we can use the Pythagorean theorem.

Apply the idea
\displaystyle a^2+b^2\displaystyle =\displaystyle c^2Pythagorean Theorem
\displaystyle x^2+4^2\displaystyle =\displaystyle 5^2Substitution
\displaystyle x^2+16\displaystyle =\displaystyle 25Simplify
\displaystyle x^2\displaystyle =\displaystyle 9Subtract 16 to both sides
\displaystyle x\displaystyle =\displaystyle 3Square root both sides

Now we want to write the trigonometric ratio for cosine, \dfrac{\text{adjacent}}{\text{hypotenuse}}

\cos\theta = \dfrac{3}{5}

Reflect and check

We could also realize that this a Pythagorean triple of 3, 4, \text{and } 5 at the start and then write the trigonometric ratio for \cos \theta.

c

Find the value of \tan \theta.

Worked Solution
Create a strategy

From part (a), we labeled the opposite and hypotenuse sides. From part (b) we labeled and found the value of the adjacent side. We can use these values to write the tangent ratio, which is the ratio between the sides opposite and adjacent to the given angle of the right triangle.

Apply the idea

\tan\theta = \dfrac{4}{3}

Example 3

Explain why \sin(x) is the same for any of the triangles in the figure.

A right triangle AGF with right angle G. Line segment J E is near side FB perpendicular to AG and forms right angle E. To the left of J E is HC forming right angle C perpendicular to A G. Line segment DB is on the left of HC. D B is perpendicular to A G and forms right angle B. Angle A measures x degrees.
Worked Solution
Create a strategy

Notice that each triangle shares \angle x, and each triangle also has a right angle. Translate each triangle to the right, so that we can see the four triangles from the figure separately.

A series of four right triangles, triangle A D B with right angle B, triangle A C H with right angle H, triangle A E J with right angle E and triangle A G F with right angle G. Angle A measures x degrees for the four right triangles

We can use similar triangles and trigonometric ratios to explain why \sin \left( x \right) is the same for any triangle in the figure.

Apply the idea

Since each triangle has a right angle and shares \angle x , we can say that each triangle is similar by the AA similarity theorem.

\sin \left( x \right)= \dfrac{\text{opposite}}{\text{hypotenuse}} for the first, small triangle. Since the triangles are similar, there is a constant of proportionality, k, between the lengths of opposite sides and between the lengths of the hypotenuses. So, for any other triangle, \sin \left( x \right)= \dfrac{k \cdot \text{opposite}}{k \cdot \text{hypotenuse}}= \dfrac{\text{opposite}}{\text{hypotenuse}}.

Since both the lengths of the opposite and hypotenuse sides change by the same factor, the ratio between them remains the same.

Example 4

Find the height, HC, of the tree.

A tower G has a height of 63 meters. A right triangle T B A is drawn using the height of the tower as T B and A is a point on the ground. A tree is drawn perpendicular to B A. The the top of the tree is label H and the bottom is labeled C. C is between B and A. From the top of the tower T, T H measures 62.29 meters while H A measures 31.14 m. From the bottom of the tower B, The distance from B to C is 46 meters and C A is 23 meters.
Worked Solution
Create a strategy

Notice that the skyscraper and the tree create two right triangles with a common \angle A. So, by the AA similarity theorem, \triangle ATB \sim \triangle AHC. Choose one of the acute angles as the reference angle and set up trigonometric ratios. We can set ratios from the small and large triangles equal to one another, and solve for the height of the tree.

Apply the idea

For \angle A, we have \sin \angle A = \dfrac{HC}{31.14} for the small triangle. We can use the same reference angle for the large triangle so that we have \sin \angle A = \dfrac{63}{62.29+31.14}= \dfrac{63}{93.43}.

Since the triangles are similar, we know that their corresponding side lengths are proportional so \dfrac{HC}{31.14}=\dfrac{63}{93.43}.

\displaystyle \dfrac{HC}{31.14}\displaystyle =\displaystyle \dfrac{63}{93.43}
\displaystyle HC\displaystyle =\displaystyle \dfrac{63}{93.43} \cdot 31.14Multiply both sides of the equation by 31.14
\displaystyle HC\displaystyle =\displaystyle 21Evaluate the division and multiplication

The height of the tree is 21 \text{ m}.

Reflect and check

We can confirm that the height of the tree will lead \triangle {HAC} to be a right triangle by evaluating its side lengths in the Pythagorean theorem:

\displaystyle a^2 + b^2\displaystyle =\displaystyle c^2Pythagorean theorem
\displaystyle 21^2 + 23^2\displaystyle =\displaystyle 31.14^2Substitution
\displaystyle 970\displaystyle =\displaystyle 969.7Evaluate the exponents and addition

Since the sum of the squares of the legs is approximately equal to the square of the hypotenuse, the height of the tree creates a valid right triangle.

Idea summary

Use the following notation for writing the three trigonometric ratios given the acute reference angle \theta in a right triangle:\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}

Relationships between ratios

Exploration

Drag the sliders to change the size of the triangle and measure of the reference angle in the triangle.

Loading interactive...
  1. What is the relationship between m \angle ABC and m \angle BAC?
  2. What do you notice about the trigonometric ratios of the angles?

For the acute angles in a right triangle which are complementary, we can state that:

  • The sine of any acute angle is equal to the cosine of its complement: \sin \left( \theta \right) = \cos \left( 90 - \theta \right)
  • The cosine of any acute angle is equal to the sine of its complement: \cos \left( \theta \right) = \sin \left( 90 - \theta \right)
Triangle A C B with right angle C. Angle A is marked with one arc while angle B is marked with two arcs.

Examples

Example 5

Use the diagram to show that \tan \theta = \dfrac{ \sin \theta}{\cos \theta}.

Triangle A B C with right angle B. Hypotenuse C A is 1 unit. The length of C B is y and the length of A B is x. Angle A is labeled theta.
Worked Solution
Create a strategy

Write the side lengths of the given triangle for their respective trigonometric ratios: \sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}

Apply the idea

Using the given triangle, we have: \sin\theta=\dfrac{y}{1}=y \qquad \cos\theta=\dfrac{x}{1}=x \qquad \tan\theta=\dfrac{y}{x}

If \sin \theta=y, \cos \theta=x, and \tan \theta = \dfrac{y}{x}, we have that:\tan \theta = \dfrac{y}{x} = \dfrac{ \sin \theta}{\cos \theta}

Example 6

Consider the following triangle:

Right triangle A C B wit \h right angle C. A C measures 4, BC measures 3 and A B measures 5. Angle B is labeled theta while angle A is labeled beta.
a

Write a rule to describe the relationship between \theta and \beta.

Worked Solution
Create a strategy

We don't know the angle measures of \theta and \beta, but we can use what we know about the triangle sum theorem to write a rule for their relationship.

Apply the idea

Since m \angle C= 90 \degree, and the triangle sum theorem states that the sum of the angles in a triangle is 180 \degree, we know that the sum of \theta and \beta must be equal to 90 \degree. So \theta = 90 - \beta.

b

Use the rule you found in part (a) to determine the relationship between \sin \theta and \cos \beta.

Worked Solution
Create a strategy

Evaluate \sin \theta and \cos \beta, then compare their values.

Apply the idea

\sin \theta = \dfrac{4}{5} and \cos \beta = \dfrac{4}{5}.

\sin \theta is equal to \cos \beta. This shows that the sine of the acute angle \theta is equal to the cosine of its complement.

Reflect and check

We could also verify that \cos \theta = \sin \beta:

If \cos \theta= \dfrac{3}{5} and \sin \beta = \dfrac{3}{5}, we have shown that \cos \theta = \sin \beta. This shows that the cosine of the acute angle \theta is equal to the sine of its complement.

Example 7

Complete the statement: If \cos \left(30 \degree \right) = \dfrac{\sqrt{3}}{2}, then \sin \left( ? \right) = \dfrac{\sqrt{3}}{2}.

Worked Solution
Apply the idea

The cosine of any acute angle is equal to the sine of its complement, so since \cos \left(30 \degree \right) = \dfrac{\sqrt{3}}{2} = \sin \left( ? \right), we know that the complement of 30 \degree must be 60 \degree.

Reflect and check

We could also draw a diagram with the given information from the trigonometric ratio \cos \left(30 \degree \right) = \dfrac{\sqrt{3}}{2}.

We know that the trigonometric ratio \sin \theta = \dfrac{ \text{opposite}}{\text{hypotenuse}}, so using the diagram we have the side opposite of a missing angle and the hypotenuse of a right triangle.

This means that the reference angle for sine must be the missing angle, which is 60 \degree by the triangle sum theorem.

Idea summary

For the acute angles in a right triangle that are complementary, we can state that:

  • The sine of any acute angle is equal to the cosine of its complement: \sin \left( \theta \right) = \cos \left( 90 - \theta \right)
  • The cosine of any acute angle is equal to the sine of its complement: \cos \left( \theta \right) = \sin \left( 90 - \theta \right)

Outcomes

G.TR.4

The student will model and solve problems, including those in context, involving trigonometry in right triangles and applications of the Pythagorean Theorem.

G.TR.4b

Find and verify trigonometric ratios using right triangles.

G.TR.4c

Model and solve problems, including those in context, involving right triangle trigonometry (sine, cosine, and tangent ratios).

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