Question

solve the right triangle shown in the figure. a = 28.5, c = 52.4 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

By the Pythagorean theorem $b=\sqrt{c^{2}-a^{2}}$. Substitute $a = 28.5$ and $c = 52.4$:
$b=\sqrt{52.4^{2}-28.5^{2}}=\sqrt{(52.4 + 28.5)(52.4 - 28.5)}=\sqrt{80.9\times23.9}=\sqrt{1933.51}\approx43.97$

Step2: Find angle A using sine function

$\sin A=\frac{a}{c}$. Substitute $a = 28.5$ and $c = 52.4$, then $A=\sin^{-1}(\frac{28.5}{52.4})\approx32.9^{\circ}$

Step3: Find angle B

Since the sum of angles in a triangle is $180^{\circ}$ and it's a right - triangle ($C = 90^{\circ}$), $B=90^{\circ}-A$. So $B = 90^{\circ}-32.9^{\circ}=57.1^{\circ}$

Answer:

$A\approx32.9^{\circ}$
$B\approx57.1^{\circ}$
$b\approx43.97$