Question

if cos (A-B)=3 and sin (A+B)=3/2,find sinA and sinB, where (A+B) and (A-B) are acute angle​

Answer 1

Answer:

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Answer 2

Correct Question :- if cos (A-B) = √3/2 and sin (A+B) = √3/2, find sinA and sinB, where (A+B) and (A-B) are acute angle ?

Answer :-

we know that,

• cos 30° = √3/2 .
• sin 60° = √3/2 .

so,

→ cos (A - B) = √3 / 2

→ cos (A - B) = cos 30°

→ (A - B) = 30° ---------------- Eqn.(1)

also,

→ sin (A + B) = √3/2

→ sin (A + B) = sin 60°

→ (A + B) = 60° ---------------- Eqn.(2)

adding Eqn.(1) and Eqn.(2) we get,

→ A - B + A + B = 30° + 60°

→ 2A = 90°

→ A = 45° (Ans.)

then, putting value of A in Eqn.(1),

→ 45° - B = 30°

→ B = 45° - 30°

→ B = 15° (Ans.)

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