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.
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.
To find AB, we can use the law of sines, which states that in a triangle, the ratio of the length of a side to the sine of its opposite angle is constant.
In triangle ABC, we have
m∠A = 41°,
m∠B = 32°,
AC = 9 inches, and we need to find AB.
The law of sines is expressed as:
sin(A) / AB = sin(B) / AC
Let's solve this equation step by step.
1. Substitute the known values into the equation:
sin(41°) / AB = sin(32°) / 9
2. Rearrange the equation to isolate AB:
AB = sin(41°) * 9 / sin(32°)
3. Use a calculator to find the values of sin(41°) and sin(32°):
sin(41°) ≈ 0.6561
sin(32°) ≈ 0.5299
4. Substitute the values back into the equation:
AB = 0.6561 * 9 / 0.5299
5. Calculate to find AB:
AB ≈ 11.78 inches
Therefore, AB is approximately 11.78 inches.
The answer is not one of the given options.