Question

If tan( A+B ) = √3 and tan (A – B) = 1/√3 such that 0 < A+B <90, A>B
Find the value of A and B​

Answer 1

EXPLANATION.

→ tan ( A + B) = √3 .....(1)

→ tan ( A - B) = 1/√3 ......(2)

→ 0 < A + B < 90 , A > B

→ tan ( A + B) = tan 60° ....... (1)

→ tan ( A - B) = tan 30° .........(2)

→ A + B = 60° ...... (1)

→ A - B = 30° ...... (2)

→ 2A = 90°

→ A = 45°

→ Put the Value of A = 45° in equation (1)

we get,

→ 45° + B = 60°

→ B = 15°

→ Therefore,

→ Value of A = 45° and B = 15°.

Answer 2

Answer :-

• A = 45°
• B = 15°

Given :-

• tan( A+B ) = √3 and tan (A – B) = 1/√3 such that 0 < A+B <90, A>B

To Find :-

• Value of A and B.

Solution :-

Here

• tan (A + B) = √3
• tan (A - B) = 1/√3

→ tan (A + B) = tan 60° .....i)

→ tan (A - B) = tan 30° .....ii)

From i) and ii)

→ A + B = 60° .....iii)

→ A - B = 30° .....iv)

Add iii) and iv)

A + B + (A - B) = 60° + 30°

⇒A + B + A - B = 90°

⇒2A = 90°

⇒A = 90°/2

⇒A = 45°

Put the value of A in iv)

A - B = 30°

⇒45° - B = 30°

⇒B = 45° - 30°

⇒B = 15°

Hence, the value of A and B are 45° and 15° respectively.