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°. Given $\angle A=32^{\circ}40'$, convert 40' to degrees: $40\div60=\frac{2}{3}\approx0.67^{\circ}$, so $\angle A = 32.67^{\circ}$. Then $\angle B=90^{\circ}-\angle A$.
$B = 90^{\circ}-32^{\circ}40'= 57^{\circ}20'$

Step2: Find side a

Use the sine function: $\sin A=\frac{a}{c}$, where $c = 967$ m. So $a = c\times\sin A$.
$a=967\times\sin(32^{\circ}40')$. Since $\sin(32^{\circ}40')=\sin(32+\frac{40}{60})^{\circ}\approx\sin(32.67^{\circ})\approx0.54$. Then $a = 967\times0.54\approx522$ m.

Step3: Find side b

Use the cosine function: $\cos A=\frac{b}{c}$, where $c = 967$ m. So $b = c\times\cos A$.
$b = 967\times\cos(32^{\circ}40')$. Since $\cos(32^{\circ}40')=\cos(32+\frac{40}{60})^{\circ}\approx\cos(32.67^{\circ})\approx0.84$. Then $b=967\times0.84\approx812$ m.

Answer:

$B = 57^{\circ}20'$, $a = 522$ m, $b = 812$ m