(i) Find angle A, if

2 sin² A - $\sqrt{3}$ sin A = 0 and 0 < ∠A < 90°

(ii) Find angle B, if

2 cos 2B - 1 = 0 and ∠B is an acute angle.

(iii) State the value of cot (A + B).

(i) Solving,

⇒ 2 sin² A - $\sqrt{3}$ sin A = 0

⇒ sin A(2 sin A - $\sqrt{3}$) = 0

⇒ sin A = 0 or 2 sin A - $\sqrt{3}$ = 0

⇒ sin A = sin 0° or 2 sin A = $\sqrt{3}$

⇒ A = 0° or sin A = $\dfrac{\sqrt{3}}{2}$

⇒ A = 0° or sin A = sin 60°

⇒ A = 0° or A = 60°.

Since, 0 < ∠A < 90°

Hence, A = 60°.

(ii) Solving,

⇒ 2 cos 2B - 1 = 0

⇒ 2 cos 2B = 1

⇒ cos 2B = $\dfrac{1}{2}$

⇒ cos 2B = cos 60°

⇒ 2B = 60°

⇒ B = $\dfrac{60°}{2}$ = 30°.

Hence, B = 30°.

(iii) Substituting values of A and B in cot(A + B), we get :

⇒ cot(60° + 30°)

⇒ cot 90°

⇒ 0.

Hence, cot(A + B) = 0.