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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
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Verified Answer
The correct answer is:
$60^{\circ}$
Given,
$\cos P=\frac{1}{7}, \cos Q=\frac{13}{14}$
$\begin{aligned}
& \therefore \cos (P-Q)=\cos P \cos Q+\sin P \sin Q \\
& =\frac{1}{7} \cdot \frac{13}{14}+\frac{\sqrt{48}}{7} \cdot \frac{\sqrt{27}}{14}=\frac{13+36}{98}=\frac{1}{2}=\cos 60^{\circ} \\
& \Rightarrow P-Q=60^{\circ}
\end{aligned}$
$\cos P=\frac{1}{7}, \cos Q=\frac{13}{14}$
$\begin{aligned}
& \therefore \cos (P-Q)=\cos P \cos Q+\sin P \sin Q \\
& =\frac{1}{7} \cdot \frac{13}{14}+\frac{\sqrt{48}}{7} \cdot \frac{\sqrt{27}}{14}=\frac{13+36}{98}=\frac{1}{2}=\cos 60^{\circ} \\
& \Rightarrow P-Q=60^{\circ}
\end{aligned}$
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