Question

If tan (A + B) =√3 and tan (A – B) =1/√3,0° < A + B ≤ 90°; A > B, find A and B?

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Answer 1

tan (A + B) = √3

Since √3 = tan 60°

Now substitute the degree value

⇒ tan (A + B) = tan 60°

(A + B) = 60° … (i)

The above equation is assumed as equation (i)

tan (A – B) = 1/√3

Since 1/√3 = tan 30°

Now substitute the degree value

⇒ tan (A – B) = tan 30°

(A – B) = 30° … equation (ii)

Now add the equation (i) and (ii), we get

A + B + A – B = 60° + 30°

Cancel the terms B

2A = 90°

A= 45°

Now, substitute the value of A in equation (i) to find the value of B

45° + B = 60°

B = 60° – 45°

B = 15°

Therefore A = 45° and B = 15°

Answer 2

Answer:

Given tan(A+B)= √3

⟹A+B=60 ⋯(1)

tan(A−B)= 1/√3

⟹A−B=30 ⋯(2)

(1)+(2)⟹2A=90

⟹A=45

Put A value in (1)

⟹B=60 −45 =15