Question

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. $30^{\circ }$
• B. $60^{\circ }$
• C. $45^{\circ }$
• D. $75^{\circ }$

Answer 1

Answer: (B)

Given that, $\cos P=\frac{1}{7}$ and $\cos Q=\frac{13}{14}$
$\because \sin P=\sqrt{1-{\cos }^{2}P}=\sqrt{1-\frac{1}{49}}$
$=\frac{1}{7}\sqrt{48}$
Now, $\sin Q=\sqrt{1-{\cos }^{2}Q}=\sqrt{1-\frac{169}{196}}$
$=\frac{\sqrt{27}}{14}$
Now, $\cos (P-Q)=\cos P\cos Q+\sin P\sin Q$
$\cos (P-Q)=\frac{1}{7}\times \frac{13}{14}+\frac{1}{7}\sqrt{48}\times \frac{1}{14}\sqrt{27}$
$=\frac{13}{98}+\frac{36}{98}=\frac{49}{98}=\frac{1}{2}$
$\Rightarrow \cos (P-Q)=\frac{1}{2}$
$\Rightarrow \cos (P-Q)=\cos 60^{\circ }$
$\therefore P-Q=60^{\circ }$