Question

Round your answer to this problem to the nearest degree.

Answer 1

Answer:

∠B = 19°

Step-by-step explanation:

Given : In triangle ABC, if ∠A = 120°, a = 8, and b = 3

We have to find the measure of B that is ∠B

Consider the given triangle ABC,

Using Sine rule ,

For a triangle with measure of angle A, B and C and side  a faces angle A,  

side b faces angle B and  side c faces angle C

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

we have, a = 8 , b = 3 and ∠A = 120°

Consider first two ratios,

$\frac{8}{\sin 120^{\circ}}=\frac{3}{\sin B}$

Solve for B, we have,

$\sin \left(120^{\circ \:}\right)=\frac{\sqrt{3}}{2}$

$\frac{8}{\frac{\sqrt{3}}{2}}=\frac{3}{\sin \left(B\right)}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$8\sin \left(B\right)=\frac{\sqrt{3}}{2}\cdot \:3$

Simplify, we have,

$\sin \left(B\right)=\frac{3\sqrt{3}}{16}$

Taking sine inverse both side, we have,

$B=\sin^{-1}\left(\frac{3\sqrt{3}}{16}\right)$

We have, $B=18.95^{\circ \:}$

Thus, ∠B = 19°

Answer 2

Use the sine law: sinB/b = sinA/a

SinB = b*sinA/a

sinB = 3*sin120° /8

∠ B = 18.94° ≈ 19°