Find b, given that a = 20, angle A = 30°, and angle B = 45° in triangle ABC.

Solution:

It is given that

a = 20, angle A = 30°, and angle B = 45°

We know that

Side b is opposite to angle B

By applying the law of sines

a/sin A = b/sin B

Substituting the values

20/sin 30° = b/sin 45°

By further calculation

20/(½) = b/(√2/2)

So we get

20 × √2/2 = b × ½

20√2/2 = b/2

By cross multiplication

2 × b = 2 × 20√2

2b = 40√2

Divide both sides by 2

b = 20√2

Therefore, the value of b is 20√2.

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Find b, given that a = 20, angle A = 30°, and angle B = 45° in triangle ABC.

Summary:

The value of b, given that a = 20, angle A = 30°, and angle B = 45° in triangle ABC is 20√2.