Question

If $\cos P=1_{/7}$ and $\cos Q={13}_{/14}$, where P and Q both are acute angles. Then, the value of $P-Q$ is?

• A. ${30}^o$
• B. ${60}^o$
• C. ${45}^o$
• D. ${75}^o$

Answer 1

The correct option is B ${60}^o$

We are given $\cos P=\frac{1}{7}$ & $\cos Q=\frac{13}{14}$

So, we know thta ${\sin }^2\theta +{\cos }^2\theta =1$

$\therefore \cos \theta =\sqrt{1-{\sin }^2\theta }$ or

$\sin \theta =\sqrt{1-{\cos }^2\theta }---\left(1\right)$

$\therefore \sin P=\sqrt{1-{\left(\frac{1}{7}\right)}^2}=\sqrt{1-\frac{1}{49}}=\sqrt{48}/7$

$\therefore \sin Q=\sqrt{1-{\left(\frac{13}{14}\right)}^2}=\sqrt{1-\frac{169}{196}}=\sqrt{27}/14$

let $\cos (P-Q)$

$=\cos P\cos Q+\sin P\sin Q$

$=\frac{1}{7}\times \frac{13}{14}+\frac{\sqrt{48}}{7}\times \frac{\sqrt{27}}{14}$

$=\frac{13}{98}+\frac{4\sqrt{3}\times 3\sqrt{3}}{98}$

$=\frac{13+36}{98}=\frac{49}{98}=\frac{1}{2}$

$\therefore P-Q={\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$ (as $0<P,Q<\frac{\pi }{2}$)

$\therefore P-Q=\left(\frac{\pi }{3}\right)$ or $\left({60}^o\right)$