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, `tan p=4sqrt3,tanQ=(3sqrt3)/14`

`cos(P-Q)= cosP cosQ+sinP sinQ`

`=1/7xx13/14+(4sqrt3)/7xx(3sqrt3)/14`

`=(13+36)/98`

`=49/98`

`thereforecos(P-Q)=1/2`

`=>P-Q=cos^(-1)  1/2`

`=>P-Q=60^circ`

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