Question

in ABC, A = 61°, a = 25, and b = 27. What are the two possible values for angle B to the nearest tenth of a degree?

Answer 1

Solution:

In ΔABC

- a, b, c are the lengths of its 3 sides, where

# a is opposite to angle A

# b is opposite to angle B

# c is opposite to angle C

- m∠A = 61°

a = 25

and

b = 27

To find m∠B we can use the sin Rule:

$\frac{b}{\sin(B)}=\frac{a}{\sin (A)}$

replacing the data of the problem in the previous equation, we get:

$\frac{27}{\sin (B)}=\frac{25}{\sin (61)}$

by cross-multiplication, we get:

$\sin (B)(25)=\text{ sin(61)(27)}$

solving for sin(B), we get:

$\sin (B)=\frac{\sin (61)(27)}{25}=0.94$

applying the inverse function of sine, we get:

$B=sin^{-1}(0.94)=70.8$

note that the value of sin(B) is positive

∴ Angle B may be in the first quadrant (acute angle) or in the second quadrant (obtuse angle). Thus, the other measure of ∠B would be:

$B\text{ = }180-70.8=\text{ 109}.2$

Then, the two possible values of B are:

$B\text{ = 109}.2$

and

$B\text{ = 70.8}$