## Trigonometry (11th Edition) Clone

$\angle A = 70^{\circ} 10'$ $\angle B = 19^{\circ} 50'$ $a\approx 609m$
1. We know $b=219m$ and $c = 647m$ 2. We use inverse sine to find $\angle B$ $\angle B = \sin ^{-1} \frac{219}{647} \approx 19.8^{\circ} \approx 19^{\circ}50'$ minutes are rounded for three significant figures 3. We use cosine to find side $a$ $\cos 19.8^{\circ} = \frac{a}{647}$ $a\approx 609m$ 4. Angles in a triangle must add up to $180^{\circ}$, therefore $\angle A = 180^{\circ} -90^{\circ} -19.8^{\circ} = 70.2 ^{\circ} = 70^{\circ} 10'$