Question

If sin(A+B) = √3/2 and cos(A- B) = √3/2, 0°< A + B <90°, find the value of A and B​

Answer 1

Answer:

Value of A is 45° and value of B is 15°.

Step-by-step explanation:

Given :-

• sin(A+B) = √3/2
• cos(A-B) = √3/2
• 0° < (A+B)<90°

To find :-

• The value of A and B.

Solution :-

sin(A+B) = √3/2

→ sin(A+B) = sin60° [ • sin60° = √3/2]

→ A+B = 60°.................(i)

&

cos(A-B) = √3/2

→ cos(A-B) = cos30° [ • cos30° = √3/2]

→ A-B = 30°.................(ii)

Now add both equations (i) and (ii).

A+B + (A-B) = 60°+30°

→ A+B + A - B = 90°

→ 2A = 90°

→ A = 90°/2

→ A = 45°

Now put A = 45° in eq(i).

A+B = 60°

→ 45° + B = 60°

→ B = 60°-45°

→ B = 15°

Therefore,

• A = 45°
• B = 15°

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