# In triangle ABC, how do you solve the right triangle given 'a' is 7 m, and the angle 'A' is 22 degrees?

Sep 29, 2015

(Assuming the right angle is at C)
$\angle B = {68}^{\circ}$
$c = \frac{7}{\sin} \left({22}^{\circ}\right) m e t e r s$
$b = \frac{7}{\tan} \left({22}^{\circ}\right) m e t e r s$

#### Explanation:

Everything that follows is under the assumption that the right angle is $\angle C$. If the intent is that $\angle B$ is the right angle, exchange $B$ and $C$ in the explanation below:

Since $\angle A = {22}^{\circ}$ and $\angle C = {90}^{\circ}$
and the sum of interior angles of a triangle is ${180}^{\circ}$
$\rightarrow \angle B = {68}^{\circ}$

For the $\angle A$
the opposite side is $a \left(= 7\right)$m
the adjacent side is $b$ and
the hypotenuse is $c$

$\sin \left({22}^{\circ}\right) = \left(\text{opposite")/("hypotenuse}\right) = \frac{7}{c}$

$\rightarrow c = \frac{7}{\sin} \left({22}^{\circ}\right)$

and

$\tan \left({22}^{\circ}\right) = \left(\text{opposite")/("adjacent}\right) = \frac{7}{b}$

$\rightarrow b = \frac{7}{\tan} \left({22}^{\circ}\right)$

Note:
a calculator could be used to evaluate:
$\textcolor{w h i t e}{\text{XXX}} b = \frac{7}{\tan} \left({22}^{\circ}\right) \cong 17.33$
and
$\textcolor{w h i t e}{\text{XXX}} c = \frac{7}{\sin} \left({22}^{\circ}\right) \cong 18.69$