\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}

If we know an acute angle measure in the right triangle and a side length, we can solve for another side length of the triangle using sine, cosine, or tangent. Make sure to choose the appropriate trigonometric ratio according to where the two sides are located with respect to the angle.

Remember that, for a given angle \theta, the value of each trigonometric ratio stays the same no matter the size of the triangle.

Worked examples

Example 1

Find the length f. Round your answer to two decimal places.

A right teiangle with a base of length f inches and hypotenuse of 8 inches. The angle adjacent the side of length f inches measures 47 degrees.

Approach

We want to start by looking at the figure and determine how the labeled side lengths and angles are related. The triangle is a right triangle. The hypotenuse side length is 8 \text{ in}. With respect to the 47 \degree angle, f is the adjacent side.

We can use this information to write the appropriate trigonometric ratio and then solve for f.

Solution

\displaystyle \cos\theta	\displaystyle =	\displaystyle \dfrac{\text{adjacent}}{\text{hypotenuse}}
\displaystyle \cos(47\degree)	\displaystyle =	\displaystyle \dfrac{f}{8}	Substitution
\displaystyle 8\cos(47\degree)	\displaystyle =	\displaystyle f	Multiply both sides by 8
\displaystyle f	\displaystyle =	\displaystyle 8\cos(47\degree)	Symmetric property of equality

Now we use a calculator to round our answer to two decimal places.

f=5.45

Reflection

The answer for f should be less than the hypotenuse, since the hypotenuse is the longest side of the triangle. This is a good check when solving for missing side lengths.

Example 2

Find the length g. Round your answer to two decimal places.

A right triangle with a hypotenuse of length g feet and a base of length 11 feet. Opposite the base is angle measuring 42 degrees.

Approach

We want to look at the diagram and determine how the given information is related. The hypotenuse is g since the side is opposite the right angle. The side with length of 11 \text{ ft} is the opposite side with respect to the labeled angle of 42 \degree.

We can use this information to write the appropriate trigonometric ratio and then solve for g.

Solution

\displaystyle \sin \theta	\displaystyle =	\displaystyle \dfrac{\text{opposite}}{\text{hypotenuse}}
\displaystyle \sin(42 \degree)	\displaystyle =	\displaystyle \dfrac{11}{g}	Substitution
\displaystyle g\sin(42 \degree)	\displaystyle =	\displaystyle 11	Multiply both sides by g
\displaystyle g	\displaystyle =	\displaystyle \dfrac{11}{\sin(42 \degree)}	Divide both sides by \sin(42 \degree)

Now, we use a calculator to round our answer to two decimal places.

g=16.44

Outcomes

M3.G.SRT.C.5

Solve triangles.*

M3.G.SRT.C.5.A

Know and use the Pythagorean Theorem and trigonometric ratios (sine, cosine, tangent, and their inverses) to solve right triangles in a real-world context.

M3.MP1

Make sense of problems and persevere in solving them.

M3.MP2

Reason abstractly and quantitatively.

M3.MP3

Construct viable arguments and critique the reasoning of others.

M3.MP4

Model with mathematics.

M3.MP5

Use appropriate tools strategically.

M3.MP6

Attend to precision.

M3.MP7

Look for and make use of structure.

M3.MP8

Look for and express regularity in repeated reasoning.

4.04 Solving for sides in right triangles

Concept summary

Worked examples

Example 1

Approach

Solution

Reflection

Example 2

Approach

Solution

Outcomes

M3.G.SRT.C.5

M3.G.SRT.C.5.A

M3.MP1

M3.MP2

M3.MP3

M3.MP4

M3.MP5

M3.MP6

M3.MP7

M3.MP8

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