Question

solve the right triangle.
b = ☐° ☐ (round to the nearest integer as needed.)
a = ☐ m (round to the nearest integer as needed.)
b = ☐ m (round to the nearest integer as needed.)

Explanation:

Step1: Find angle B

In a right - triangle, the sum of the two non - right angles is 90°.
$B = 90^{\circ}-33^{\circ}34'$
First, convert 90° to 89°60' (since 1° = 60'). Then $B=89^{\circ}60' - 33^{\circ}34'=56^{\circ}26'$

Step2: Find side a

We know that $\sin A=\frac{a}{c}$, where $c = 963$ m and $A = 33^{\circ}34'$. First convert $33^{\circ}34'$ to decimal degrees. $34'=\frac{34}{60}\approx0.567^{\circ}$, so $A\approx33.567^{\circ}$. Then $a = c\times\sin A=963\times\sin(33.567^{\circ})\approx963\times0.553\approx532$ m

Step3: Find side b

We know that $\cos A=\frac{b}{c}$, where $c = 963$ m and $A = 33^{\circ}34'\approx33.567^{\circ}$. Then $b = c\times\cos A=963\times\cos(33.567^{\circ})\approx963\times0.833\approx802$ m

Answer:

$B = 56^{\circ}26'$
$a = 532$ m
$b = 802$ m