Question

If in a right angled triangle ABC, 4 sinAcosB – 1 = 0 and tanA is real then A,B,C are in-

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

Answer: (A)

A.P.

Answer 2

Since 4sinA cosB = 1, so A and B cannot be 90⁰

(as if B = 90⁰, then cosB = 0 and if A = 90⁰, tanA is not

defined)

∴ C = 90⁰

B = 90⁰ – A ⇒ 4sinAcos(90⁰ – A) = 1

sin²A = $\frac{1}{4}$ ⇒ sinA = $\frac{1}{2}$

⇒ A = $\frac{\pi}{6}$ ⇒ B = $\frac{\pi}{3}$

So angle $\frac{\pi}{6}$, $\frac{\pi}{3}$, $\frac{\pi}{2}$ are in A.P