Question

If $$\cos P=\frac{1}{7}\text{ and }\cos Q=\frac{13}{14}$$, where P and Q both are acute angles. Then, the value of P − Q is

 

Options

• $$\frac{\pi }{6}$$

   

• $$\frac{\pi }{3}$$

   

• $$\frac{\pi }{4}$$

   

• $$\frac{\pi }{12}$$

   

Answer 1

60⁰ = $$\frac{\pi }{3}$$

$$\cos P=\frac{1}{7},\cos Q=\frac{13}{14}$$

$$\text{ Therefore, }\sin P=\sqrt{1-\frac{1}{49}}=\frac{4\sqrt{3}}{7}\text{ and }\sin Q=\sqrt{1-\frac{169}{196}}=\frac{3\sqrt{3}}{14}$$

Hence, ${\displaystyle \tan p=4\sqrt{3},\tan Q=\frac{3\sqrt{3}}{14}}$

${\displaystyle \cos (P-Q)=\cos P\cos Q+\sin P\sin Q}$

${\displaystyle =\frac{1}{7}\times \frac{13}{14}+\frac{4\sqrt{3}}{7}\times \frac{3\sqrt{3}}{14}}$

${\displaystyle =\frac{13+36}{98}}$

${\displaystyle =\frac{49}{98}}$

${\displaystyle \therefore \cos (P-Q)=\frac{1}{2}}$

${\displaystyle \Rightarrow P-Q={\cos }^{-1} \frac{1}{2}}$

${\displaystyle \Rightarrow P-Q={60}^{\circ }}$

^{Hence, the correct answer is option B.}