If cos P = 1/7 and cos Q = 13/14, Find the Value of P − Q

Solution

Given,

\[ \cos P=\frac{1}{7} \]

Since \(P\) is acute,

\[ \sin P = \sqrt{1-\cos^2P} \]

\[ = \sqrt{1-\left(\frac{1}{7}\right)^2} \]

\[ = \sqrt{\frac{48}{49}} = \frac{4\sqrt{3}}{7} \]

Similarly,

\[ \cos Q=\frac{13}{14} \]

Therefore,

\[ \sin Q = \sqrt{1-\left(\frac{13}{14}\right)^2} \]

\[ = \sqrt{\frac{27}{196}} = \frac{3\sqrt{3}}{14} \]

Now use the identity:

\[ \cos(P-Q) = \cos P\cos Q+\sin P\sin Q \]

Substituting values,

\[ \cos(P-Q) = \frac{1}{7}\cdot\frac{13}{14} + \frac{4\sqrt{3}}{7}\cdot\frac{3\sqrt{3}}{14} \]

\[ = \frac{13}{98} + \frac{36}{98} \]

\[ = \frac{49}{98} = \frac{1}{2} \]

Since \(P\) and \(Q\) are acute angles,

\[ P-Q \]

is an acute angle whose cosine is

\[ \frac{1}{2} \]

Hence,

\[ P-Q=\frac{\pi}{3} \]

Final Answer

\[ \boxed{ P-Q=\frac{\pi}{3} } \]

Correct Option: (b)