Question

1. find the measure of angle b.

triangle with vertices a, b, c; side ac = 22, side ab = 63, angle at c is 48°
options: 15.04°, 15.78°, 16.70°, 16.32°

Explanation:

Step1: Apply Law of Sines

$\frac{\sin A}{CD} = \frac{\sin C}{AD}$ → $\frac{\sin A}{CD} = \frac{\sin 48^\circ}{63}$
First, we need to find side $CD$? No, correct Law of Sines for $\triangle ACD$: $\frac{\sin \angle CAD}{CD} = \frac{\sin \angle ACD}{AD}$, but we need angle at $A$? Wait, no, target is angle $B$ (wait, typo, angle at $A$? No, the question says angle $B$, but the triangle is $ACD$. Wait, no, the angle at $D$ is angle $B$? Wait, no, the triangle has sides $AC=22$, $AD=63$, $\angle C=48^\circ$. We need to find $\angle D$ (angle $B$? Maybe typo, $\angle ADC$).

Correct Law of Sines: $\frac{\sin \angle ADC}{AC} = \frac{\sin \angle ACD}{AD}$

Step2: Substitute known values

$\sin \angle ADC = \frac{AC \cdot \sin \angle ACD}{AD}$
$\sin \angle ADC = \frac{22 \cdot \sin 48^\circ}{63}$

Step3: Calculate the value

$\sin 48^\circ \approx 0.7431$
$\sin \angle ADC \approx \frac{22 \times 0.7431}{63} \approx \frac{16.3482}{63} \approx 0.2595$

Step4: Find the angle

$\angle ADC = \arcsin(0.2595) \approx 15.04^\circ$

Answer:

15.04°