Question

if sin (A+2B)=√3/2,and cos (A+4B)=
0, A<B, A+B<90°,Find AandB​

Answer 1

Answer:

sin(A + 2B) = √3/2

sin(A + 2B) = sin 60°

⇒ A + 2B = 60° -----(1)

also given that,

cos(A + 4B) = 0

cos(A + 4B) = cos 90°

⇒ A + 4B = 90° -----(2)

Subtracting (1) from (2), we get

A + 4B - A - 2B = 90° - 60°

2B = 30°

⇒ B = 15°

Putting B = 15° in eq (1), we get

A + 2 (15°) = 60°

A + 30° = 60°

⇒ A = 30°

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Answer 2

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