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Example: Finding a Side Length in a Right Triangle

Find the missing side length xx in each of the following triangle:

a)

Since we are given the Hypotenuse and we want to find the Opposite, we must use the sin\sin ratio.
sinθ=OppHypsin40°=x5sin40°×5=x5×5sin40°×5=x0.642×5x3.21xx3.21\begin{array}{rcl} \sin\theta&=&\dfrac{Opp}{Hyp}\\[1em] \sin40\degree&=&\dfrac{x}{5}\\[1em] \sin40\degree\colorTwo{\times5}&=&\dfrac{x}{5}\colorTwo{\times5}\\[1em] \sin40\degree\times5&=&x\\[1em] 0.642\times5&\approx&x\\[1em] 3.21&\approx&x\\[1em] x&\approx&3.21 \end{array}

Therefore, the exact value of xx is 5×sin40°5\times \sin40\degree, which is approximately 3.21.
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b)

Since we are given the Adjacent and we want to find the Hypotenuse, we must use the cos\cos ratio.
cosθ=AdjHypcos35°=4xcos35°×x=4x×xcos35°×x=4cos35°×xcos35°=4cos35°x40.819x4.88\begin{array}{rcl} \cos\theta&=&\dfrac{Adj}{Hyp}\\[1em] \cos35\degree&=&\dfrac{4}{x}\\[1em] \cos35\degree\colorTwo{\times x}&=&\dfrac{4}{x}\colorTwo{\times x}\\[1em] \cos35\degree\times x&=&4\\[1em] \dfrac{\cos35\degree\times x}{\colorTwo{\cos35\degree}}&=&\dfrac{4}{\colorTwo{\cos35\degree}}\\[1em] x&\approx&\dfrac{4}{0.819}\\[1em] x&\approx&4.88\\ \end{array}

Therefore, the exact value of xx is 4cos35°\dfrac{4}{\cos35\degree}, which is approximately 4.88.

Practice: Primary Trig Ratios

If we know C\angle C in the following triangle, which ratio will you use to relate side aa and side bb?

Practice: Finding a Side Length in a Right Triangle

Find the missing side length xx in this triangle

Practice: Finding a Side Length in a Right Triangle


Practice: Finding Side Lengths in a Right Angle Triangle

Find all missing side lengths in the following triangle