Question

solve the right triangle shown in the figure. a = 28.7, c = 50.9
a ≈ □° (round to the nearest tenth as needed.)
b ≈ □° (round to the nearest tenth as needed.)
b ≈ □ (round to the nearest hundredth as needed.)

Explanation:

Step1: Find side b using Pythagorean theorem

$b = \sqrt{c^{2}-a^{2}}=\sqrt{50.9^{2}-28.7^{2}}=\sqrt{(50.9 + 28.7)(50.9 - 28.7)}=\sqrt{79.6\times22.2}=\sqrt{1767.12}\approx42.03$

Step2: Find angle A using sine - function

$\sin A=\frac{a}{c}$, so $A=\sin^{- 1}(\frac{a}{c})=\sin^{-1}(\frac{28.7}{50.9})\approx34.2^{\circ}$

Step3: Find angle B

Since the sum of angles in a triangle is $180^{\circ}$ and it's a right - triangle ($C = 90^{\circ}$), then $B=90^{\circ}-A$. So $B = 90^{\circ}-34.2^{\circ}=55.8^{\circ}$

Answer:

$A\approx34.2^{\circ}$
$B\approx55.8^{\circ}$
$b\approx42.03$