KnowledgeBoat Logo
|

Mathematics

If tan (A + B) = 3\sqrt{3} and tan (A – B) = 13\dfrac{1}{\sqrt{3}}; 0° < A + B ≤ 90°; A > B, find A and B.

Trigonometric Identities

2 Likes

Answer

Given,

⇒ tan (A + B) = 3\sqrt{3}

⇒ tan (A + B) = tan 60°

⇒ A + B = 60° ……….(1)

Also,

⇒ tan (A - B) = 13\dfrac{1}{\sqrt{3}}

⇒ tan (A - B) = tan 30°

⇒ A - B = 30° ……….(2)

Adding equation (1) and (2), we get :

⇒ A + B + A - B = 60° + 30°

⇒ 2A = 90°

⇒ A = 90°2\dfrac{90°}{2} = 45°.

Substituting value of A in equation (1), we get :

⇒ 45° + B = 60°

⇒ B = 60° - 45° = 15°.

Hence, A = 45° and B = 15°.

Answered By

2 Likes

Related Questions