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

Solution:

We use the trigonometric table and the trigonometric identities to solve the problem.

Given that, tan (A + B) = √3 and, tan (A - B) = 1/√3

Since, tan 60° = √3 and tan 30° = 1/√3

Therefore, tan (A + B) = tan 60°

(A + B) = 60° ...(i)

tan (A - B) = tan 30°

(A - B) = 30° ...(ii)

On adding both equations (i) and (ii), we obtain:

A + B + A - B = 60° + 30°

2A = 90°

A = 45°

By substituting the value of A in equation (i) we obtain

A + B = 60°

45° + B = 60°

B = 60° - 45° = 15°

Therefore, ∠A = 45° and ∠B = 15° (A > B).

☛ Check: NCERT Solutions Class 10 Maths Chapter 8

Video Solution:

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

Maths NCERT Solutions Class 10 - Chapter 8 Exercise 8.2 Question 3

Summary:

If tan (A + B) = √3 and tan (A - B) = 1/√3; 0° < (A + B) ≤ 90°, A > B, the values of ∠A and ∠B are 45° and 15° respectively.

☛ Related Questions:

• Evaluate the following: (i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan² 45° + cos² 30° - sin² 60° (iii) cos 45°/(sec 30° + cosec 30°) (iv) sin 30° + tan 45° - cosec 60°/(sec 30° + cos 60° - cot 45°)
• Choose the correct option and justify your choice:(i) 2 tan 30°/1 + tan2 30°(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 60°(ii) 1 - tan2 45°/1 + tan2 45°(A) tan 90° (B) 1 (C) sin 45° (D) 0°(iii) sin 2A = 2 sin A is true when A =(A) 0° (B) 30°(C) 45° (D) 60°(iv) 2 tan 30°/1 - tan2 30°(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
• State whether the following are true or false. Justify your answer.(i) sin (A + B) = sin A + sin B.(ii) The value of sin θ increases as θ.(iii) The value of cos θ increases as θ.(iv) sin θ = cos θ for all values of θ.(v) cot A is not defined for A = 0°.