# Chapter 2 - Acute Angles and Right Triangles - Section 2.4 Solving Right Triangles - 2.4 Exercises - Page 73: 31

$\angle A = 24^{\circ} 10'$ $\angle B = 65^{\circ} 50'$ $b\approx 42.3 cm$

#### Work Step by Step

1. We know $a=18.9cm$ and $c = 46.3 cm$ 2. We use inverse sine to find $\angle A$ $\sin ^{-1} \frac{18.9}{46.3} \approx 24.1^{\circ} \approx 24^{\circ}10'$ minutes are rounded for three significant figures 3. We use cosine to find side $b$ $\cos 24.1^{\circ} = \frac{b}{46.3}$ $b\approx 42.3 cm$ 4. Angles in a triangle must add up to $180^{\circ}$, therefore $\angle B = 180^{\circ} -90^{\circ} -24.1^{\circ} = 65.9 ^{\circ} = 65^{\circ} 50'$

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