If cos P = frac{1}{7} and cos Q = frac{13}{14 | vedclass

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(B) Given: $\cos P = \frac{1}{7}$ and $\cos Q = \frac{13}{14}$.Since $P$ and $Q$ are acute angles,$\sin P = \sqrt{1 - \cos^2 P} = \sqrt{1 - (\frac{1}{

Vedclass · Accepted Answer

$60$

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(B) Given: $\cos P = \frac{1}{7}$ and $\cos Q = \frac{13}{14}$.Since $P$ and $Q$ are acute angles,$\sin P = \sqrt{1 - \cos^2 P} = \sqrt{1 - (\frac{1}{7})^2} = \sqrt{1 - \frac{1}{49}} = \sqrt{\frac{48}{49}} = \frac{\sqrt{48}}{7} = \frac{4\sqrt{3}}{7}$.Similarly,$\sin Q = \sqrt{1 - \cos^2 Q} = \sqrt{1 - (\frac{13}{14})^2} = \sqrt{1 - \frac{169}{196}} = \sqrt{\frac{27}{196}} = \frac{\sqrt{27}}{14} = \frac{3\sqrt{3}}{14}$.Using the formula $\cos(P - Q) = \cos P \cos Q + \sin P \sin Q$:$\cos(P - Q) = (\frac{1}{7} 	imes \frac{13}{14}) + (\frac{4\sqrt{3}}{7} 	imes \frac{3\sqrt{3}}{14})$$= \frac{13}{98} + \frac{12 	imes 3}{98} = \frac{13 + 36}{98} = \frac{49}{98} = \frac{1}{2}$.Since $\cos(P - Q) = \frac{1}{2}$,we have $P - Q = 60^o$.

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....$^o$

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