Question

given △abc
m∠a =
a = 11.1
m∠b =
b = 12.2
m∠c = 90°
c =

Explanation:

Step1: Find $\sin B$

In a right - triangle, $\sin B=\frac{b}{c}$. First, find $c$ using the Pythagorean theorem $c = \sqrt{a^{2}+b^{2}}$. Given $a = 11.1$ and $b = 12.2$, then $c=\sqrt{11.1^{2}+12.2^{2}}=\sqrt{123.21 + 148.84}=\sqrt{272.05}\approx16.5$.

Step2: Calculate $\sin B$

$\sin B=\frac{b}{c}=\frac{12.2}{16.5}\approx0.7394$. Then $m\angle B=\sin^{- 1}(0.7394)\approx48.1^{\circ}$.

Step3: Calculate $\angle A$

Since the sum of angles in a triangle is $180^{\circ}$ and $m\angle C = 90^{\circ}$, $m\angle A=180^{\circ}-m\angle B - m\angle C$. So $m\angle A=180^{\circ}-90^{\circ}-48.1^{\circ}=41.9^{\circ}$.

Step4: Confirm $c$ value

$c=\sqrt{11.1^{2}+12.2^{2}}=\sqrt{123.21+148.84}=\sqrt{272.05}\approx16.5$.

Answer:

$m\angle A\approx41.9^{\circ}$, $m\angle B\approx48.1^{\circ}$, $c\approx16.5$