Question

Directions: The following question has four choices, out of which one or more are correct.In a triangle ΔPQR,cos⁡(P−R)cos⁡(Q)+cos⁡(2Q)=0Which of the following options is/are correct?

• A. (A) sin2⁡P+sin2⁡R=2sin2⁡Q
• B. (B) p2, q2 and r2 are in AP
• C. (C) p2, q2 and r2 are in GP
• D. (D) sin⁡Psin⁡R=sin2⁡Q

Answer 1

Answer: (A), (D)

(A) sin2⁡P+sin2⁡R=2sin2⁡Q

Answer 2

Explanation:
Consider the expression.cos⁡(P−R)cos⁡(Q)+cos⁡(2Q)=0cos⁡(P−R)cos⁡[180∘−(P+R)]+1−2sin2⁡Q=0−cos⁡(P−R)cos⁡(P+R)+1−2sin2⁡Q=0−cos2⁡P+sin2⁡R+1−2sin2⁡Q=0−(1−sin2⁡P)+sin2⁡R+1−2sin2⁡Q=0sin2⁡P+sin2⁡R=2sin2⁡QWe know thatsin⁡Pp=sin⁡Qq=sin⁡Rr=kTherefore,sin⁡P=pk,sin⁡Q=qk,sin⁡R=rkp2k2+r2k2=2q2k2p2+r2=2q2Therefore,p2,q2,r2 are in AP Hence, it is the required solution.