Question

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

• A. $30°$
• B. $60°$
• C. $45°$
• D. $75°$

Answer 1

The correct option is B

$60°$

Step 1. Find the value of $\mathit{P}\mathbf{}\mathbf{-}\mathbf{}\mathit{Q}$:

Given, $\cos P=\frac{1}{7}$

$\Rightarrow$ $\begin{array}{rcl}\sin P& =& \sqrt{\left(1–\frac{1}{49}\right)}\end{array}$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{A}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\begin{array}{rcl}& =& \sqrt{\left(\frac{48}{49}\right)}\\ & =& \frac{\sqrt{48}}{7}\end{array}$

Similarly,

$\cos Q=\frac{13}{14}$

$\Rightarrow$$\begin{array}{rcl}\sin Q& =& \sqrt{\left(1–\frac{169}{196}\right)}\end{array}$

$\begin{array}{rcl}& =& \sqrt{\left(\frac{27}{196}\right)}\\ & =& \frac{\sqrt{27}}{14}\end{array}$

Step 2. Put the values of $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{P}\mathbf{,}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{P}\mathbf{,}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{Q}\mathbf{,}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{Q}$ in given expression, we get

$\begin{array}{rcl}\cos (P–Q)& =& \cos P\cos Q+\sin P\sin Q\\ & =& \left(\frac{1}{7}\right)\left(\frac{13}{14}\right)+\left(\frac{\sqrt{48}}{7}\right)\left(\frac{\sqrt{27}}{14}\right)\\ & =& \left(\frac{13}{98}\right)+\left(\frac{36}{98}\right)\\ & =& \frac{(13+36)}{98}\\ & =& \frac{49}{98}\\ & =& \frac{1}{2}\end{array}$

$\begin{array}{rcl}\therefore P-Q& =& {\cos }^{-1}\left(\frac{1}{2}\right)\\ & =& \mathbf{60}\mathbf{°}\end{array}$

Hence, Option ‘B’ is Correct.