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Wize High School Geometry Textbook (Common Core) > Trigonometry of Right Triangles

Finding Angle Measures in a Right Triangle

Finding Side Lengths in a Right TrianglePrevious Section
Solving Right TrianglesNext Section
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Finding Angles Using Primary Trig Ratios

Recall

sin⁡θ=OppHyp          cos⁡θ=AdjHyp          tan⁡θ=OppAdj \boxed{\sin\theta=\dfrac{Opp}{Hyp}~~~~~~~~~~ \cos\theta=\dfrac{Adj}{Hyp}~~~~~~~~~~ \tan\theta=\dfrac{Opp}{Adj}}
If we know 1 angle and 1 other side length in a right angle triangle, we can calculate the other missing side lengths.

How about calcuating missing angles?

If we know any 2 side lengths in a right angle triangle, we can use the inverse trig ratios sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1} and tan⁡−1\tan^{-1} to calculate the missing angle. The inverse trig ratio "undoes" the primary trig ratios:
θ=sin⁡−1(OppHyp)          θ=cos⁡−1(AdjHyp)          θ=tan⁡−1(OppAdj) \boxed{\theta=\sin^{-1}\left(\dfrac{Opp}{Hyp}\right)~~~~~~~~~~ \theta=\cos^{-1}\left(\dfrac{Adj}{Hyp}\right)~~~~~~~~~~ \theta=\tan^{-1}\left(\dfrac{Opp}{Adj}\right)}

*In some math books, you'll see some people use arcsin⁡\arcsin instead of sin⁡−1\sin^{-1}, arccos⁡\arccos instead of cos⁡−1\cos^{-1}, and arctan⁡\arctan instead of tan⁡−1\tan^{-1}.

Wize Tip
A trig ratio takes in an angle and produces a ratio, but an inverse trig ratio takes in a ratio and produces an angle.
  • sin⁡30°=12\sin30\degree=\frac{1}{2} means that sin⁡\sin takes in 30°30\degree and returns the ratio 12\frac{1}{2}
  • sin⁡−1(12)=30°\sin^{-1}\left(\frac{1}{2}\right)=30\degree means that sin⁡−1 \sin^{-1\ } takes in the ratio 12\frac{1}{2} and returns 30°30\degree

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Calculator Time!

Watch Out!
*Make sure your calculator is in the correct degree (D) mode.

Use your calculator to evaluate the following:
1.) sin⁡−1(0.5)=\sin^{-1}(0.5)=
30°

2.) cos⁡−1(0.3)=\cos^{-1}(0.3)=
72.54°

3.) tan⁡−1(1)=\tan^{-1}(1)=
45°

4.) cos⁡−1(0.5)=\cos^{-1}(0.5)=
120°

Practice: Inverse Trig Ratios

Use your calculator to evaluate the following.

a) sin⁡−1(0.4)=\sin^{-1}\left(0.4\right)=

b) cos⁡−1(−0.76)=\cos^{-1}\left(-0.76\right)=

c) tan⁡−1(2.3)=\tan^{-1}\left(2.3\right)=



Practice: Using Inverse Trig Ratios to Find an Angle

Solve for the following angles.

a) sin⁡θ=23\sin\theta=\frac{2}{3}

b) cos⁡ϕ=0.9\cos\phi=0.9

c) tan⁡α=79\tan\alpha=\frac{7}{9}

d) tan⁡β=92\tan\beta=\frac{9}{2}
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Example: Solving for Missing Angles

Find the missing angles in this triangle.

Let's start by finding x\bco x

The give sides are the opposite side and hypotenuse side, so we use the sin⁡\sin ratio:
sin⁡x=OppHypsin⁡x=715x=sin⁡−1(715)x≈27.82°\begin{array}{rcl} \sin x&=&\dfrac{Opp}{Hyp}\\[1em] \sin x&=&\dfrac{7}{15}\\[1em] x&=&\sin^{-1}\left(\dfrac{7}{15}\right)\\[1em] x &\approx&27.82\degree \end{array}

Finding y\bco y

Method 1 (easier way)
The interior angles in a triangle must add up to 180°180\degree:
x+y+90°=180°27.82+y+90°=180°y+117.82°=180°y=180°−117.82°y=62.18°\begin{array}{rcl} x+y+90\degree&=&180\degree\\ 27.82+y+90\degree&=&180\degree\\ y+117.82\degree&=&180\degree\\ y&=&180\degree-117.82\degree\\ y&=&62.18\degree \end{array}

Method 2 (using inverse trig ratios again)
We could use the cos⁡\cos ratio:
cos⁡y=AdjHypcos⁡y=715y=cos⁡−1(715)y≈62.18°\begin{array}{rcl} \cos y&=&\dfrac{Adj}{Hyp}\\[1em] \cos y&=&\dfrac{7}{15}\\[1em] y&=&\cos^{-1}\left(\dfrac{7}{15}\right)\\[1em] y &\approx&62.18\degree \end{array}

Practice: Solving for Missing Angles

Find the missing angles.