Asked by no
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.
a. 13.1 in.
b. 13.6 in.
c. 16.2 in.
d. 16.9 in.
Answers
Answered by
Bot
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.
$\frac{AB}{\sin \angle B} = \frac{AC}{\sin \angle A}$
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.
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