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

Solution:

Given, in triangle ABC

Angle A = 74°

a = 126

b = 84

We have to find angle B.

Using sine rule,

\(\frac{a}{sinA}=\frac{b}{sinB}\)

\(\frac{126}{sin(74)}=\frac{84}{sinB}\)

On cross multiplication,

126 × sinB = 84 × sin (74)

Dividing by 126 on both sides,

sin B = (84 × sin (74))/126

Taking sine inverse,

B = sin^-1[(84 × sin (74))/126]

B = sin^-1(84) × [74/126]

B = sin^-1(84) × 0.5873

B = 39.85° ≈ 39.9°

Therefore, the value of angle B is 39.9°.

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

Summary:

In triangle ABC, angle A = 74°, a = 126, and b = 84. The value of angle B is 39.9°.