Q56 / 170 · Manipal (MET) 2019 · Mathematics
If $\cos P=\frac{1}{7}$ and $\cos Q=\frac{13}{14}$, where $P$ and $Q$ both are acute angles. Then the value of $P-Q$ is
- A$30^{\circ}$
- Correct answerB$60^{\circ}$
- C$45^{\circ}$
- D$75^{\circ}$
Solution
Given that, $\cos P=\frac{1}{7}, \cos Q=\frac{13}{14}$
$\because \sin P=\sqrt{1-\cos ^2 P}=\sqrt{1-\frac{1}{49}}=\frac{1}{7} \sqrt{48}$
Now, $\sin Q=\sqrt{1-\cos ^2 Q}=\sqrt{1-\frac{169}{196}}=\frac{\sqrt{27}}{14}$. Now, $\begin{aligned} & \quad \begin{aligned} & \cos (P-Q)= \cos P \cos Q+\sin P \sin Q \\ & \cos (P-Q)=\frac{1}{7} \times \frac{13}{14}+\frac{1}{7} \sqrt{48} \times \frac{1}{14} \sqrt{27} \\ &=\frac{13}{98}+\frac{36}{98}=\frac{49}{98}=\frac{1}{2} \\ & \Rightarrow \quad \cos (P-Q)=\frac{1}{2} \\ & \Rightarrow \quad \cos (P-Q)=\cos 60^{\circ} \\ & \therefore \quad P-Q=60^{\circ} \end{aligned} \end{aligned}$
Now, $\sin Q=\sqrt{1-\cos ^2 Q}=\sqrt{1-\frac{169}{196}}=\frac{\sqrt{27}}{14}$. Now, $\begin{aligned} & \quad \begin{aligned} & \cos (P-Q)= \cos P \cos Q+\sin P \sin Q \\ & \cos (P-Q)=\frac{1}{7} \times \frac{13}{14}+\frac{1}{7} \sqrt{48} \times \frac{1}{14} \sqrt{27} \\ &=\frac{13}{98}+\frac{36}{98}=\frac{49}{98}=\frac{1}{2} \\ & \Rightarrow \quad \cos (P-Q)=\frac{1}{2} \\ & \Rightarrow \quad \cos (P-Q)=\cos 60^{\circ} \\ & \therefore \quad P-Q=60^{\circ} \end{aligned} \end{aligned}$
Hence, the correct answer is option: 'B'.