Question

Directions: The following question has four choices, out of which one or more are correct.In a triangle $\mathrm{\Delta }PQR,\cos (P-R)\cos (Q)+\cos (2Q)=0$Which of the following options is/are correct?

• (A) $${\sin }^{2}P+{\sin }^{2}R=2{\sin }^{2}Q$$
• (B) p², q² and r² are in AP
• (C) p², q² and r² are in GP
• (D) $$\sin P\sin R={\sin }^{2}Q$$

Answer 1

The Correct Option is A, D

Explanation:
Consider the expression.$$\begin{array}{rl}\cos (P-R)\cos (Q)+\cos (2Q)& =0\\ \cos (P-R)\cos [{180}^{\circ }-(P+R)]+1-2{\sin }^{2}Q& =0\\ -\cos (P-R)\cos (P+R)+1-2{\sin }^{2}Q& =0\\ -{\cos }^{2}P+{\sin }^{2}R+1-2{\sin }^{2}Q& =0\\ -(1-{\sin }^{2}P)+{\sin }^{2}R+1-2{\sin }^{2}Q& =0\\ {\sin }^{2}P+{\sin }^{2}R& =2{\sin }^{2}Q\end{array}$$We know that$$\frac{\sin P}{p}=\frac{\sin Q}{q}=\frac{\sin R}{r}=k$$Therefore,$$\begin{array}{c}\sin P=pk,\sin Q=qk,\sin R=rk\\ p^2{k}^2+r^2{k}^2=2q^2{k}^2\\ p^2+r^2=2q^2\end{array}$$Therefore,$$p^2,q^2,r^2\text{ are in AP }$$Hence, it is the required solution.