Question

In triangle ABC, angle A = 74°, a = 126, and b = 84. Find angle B.

Answer 1

Triangle ABC
Angle A = 74°
a = 126
b = 84
Use the sine rule
$\frac{a}{sin A} = \frac{b}{sin B}$
Plug in the given
$\frac{126}{sin(74)} = \frac{84}{sin(B)}$
Cross multiply
126 × sin(B) = 84 × sin(74)
Divide by 126
$\frac{126sin(B)}{126} = \frac{84sin(74)}{126}$
126 and 126 cancels out
sin(B) = $\frac{84sin(74)}{126}$
take sin inverse ($sin^{-1}$)
B = $sin^{-1}$ ($\frac{84sin(74)}{126}$)
B = 39.85° = 39.9°