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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
- \[\frac{\pi}{6}\]
- \[\frac{\pi}{3}\]
- \[\frac{\pi}{4}\]
- \[\frac{\pi}{12}\]
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Solution
60⁰ = \[\frac{\pi}{3}\]
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.
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