# How do you solve the right triangle given a=32m, A=46°?

Sep 28, 2016

The angles are 90°, 46° and 44°
The sides are $32 m , 30.9 m \mathmr{and} 44.5 m$

#### Explanation:

Let's call the triangle ABC, where $\hat{B}$ is the 90° angle.

The sum of the angles in a triangle is 180°

If hatB = 90°, then hatA +hatC=90°

From hatA = 46°, you can calculate $\hat{C}$

hatC = 90°-46° = 44°

Using trigonometry to find the length of $A B \mathmr{and} c$

"opp"/"adj" = c/32 = tan44°

c = 32tan44°3

$c = 30.9 m$

The length of the hypotenuse, AC, can be found by using trig or Pythagoras's Theorem.

Pythagoras$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times}$ Trig

${a}^{2} + {c}^{2} = {b}^{2} \textcolor{w h i t e}{\times \times \times \times \mathcal{\mathcal{c}} \times x \times} \frac{\text{a}}{32} = \frac{1}{\cos 44}$

${32}^{2} + {30.9}^{2} = 1978.9 \textcolor{w h i t e}{\times \times \times \times x} a = \frac{32}{\cos 44}$

$\sqrt{1978.9} = 44.5 m \textcolor{w h i t e}{\times \times \times \times \times x} a = 44.5 m$