Using the Law of Sines, we have:
$\frac{AB}{\sin \angle B} = \frac{AC}{\sin \angle A}$
Plugging in the given values, we get:
$\frac{AB}{\sin 32^\circ} = \frac{9}{\sin 41^\circ}$
Solving for AB, we get:
$AB = \frac{\sin 32^\circ}{\sin 41^\circ} \cdot 9 \approx 13.6$
Therefore, the answer is (b) 13.6 in.
In ΔABC, m∠A = 41°, m∠B = 32°, and AC = 9 in. What is AB to the nearest tenth of an inch?
a. 13.1 in.
b. 13.6 in.
c. 16.2 in.
d. 16.9 in.
1 answer
1 answer