Question

If $\cos P=\dfrac{1}{7}$ and $\cos Q=\dfrac{13}{14}$, where P and Q both are acute angles. Then the value of $P-Q$is ?
A . ${{30}^{\circ }}$
B . ${{60}^{\circ }}$
C . ${{45}^{\circ }}$
D . ${{75}^{\circ }}$

Answer 1

Hint: In this question, we have given the value of $\cos P$ and $\cos Q$with both are acute angles. We have to find the value of $P-Q$. To solve the value, first we find the values of $\sin P$and $\sin Q$ with the help of $\cos P$ and $\cos Q$then by using the formula of $\cos (P-Q)$and putting all the values in the formula, we are able to get the value of $P-Q$.

Formula Used:
Formula used in this question are:
$\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$
And $\cos (P-Q)=\cos P\cos Q+\sin P\sin Q$

Complete Step- by- step Solution:
Given $\cos P=\dfrac{1}{7}$ and $\cos Q=\dfrac{13}{14}$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Then $\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$ or $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$
Now we will find the value of $\sin P$and $\sin Q$
Therefore $\sin P$ = $\sqrt{1-{{\cos }^{2}}P}$
$\sin P$ = $\sqrt{1-{{\left( \dfrac{1}{7} \right)}^{2}}}$
$\sin P$= $\sqrt{1-\left( \dfrac{1}{49} \right)}$
$\sin P$= $\dfrac{\sqrt{48}}{7}$
Similarly we will find the value of $\sin Q$
$\sin Q$= $\sqrt{1-{{\cos }^{2}}Q}$
$\sin Q$= $\sqrt{1-{{\left( \dfrac{13}{14} \right)}^{2}}}$
$\sin Q$ = $\sqrt{1-\left( \dfrac{169}{196} \right)}$
$\sin Q$= $\dfrac{\sqrt{27}}{14}$
Now we know the formula of $\cos (P-Q)=\cos P\cos Q+\sin P\sin Q$
Now we will put all the values in the above formula, we get
$\cos (P-Q)=\dfrac{1}{7}\times \dfrac{13}{14}+\dfrac{\sqrt{48}}{7}\times \dfrac{\sqrt{27}}{14}$
$\cos (P-Q)=\dfrac{13}{98}+\dfrac{4\sqrt{3}}{7}\times \dfrac{3\sqrt{3}}{14}$
$\cos (P-Q)=\dfrac{13}{98}+\dfrac{36}{98}$
$\cos (P-Q)=\dfrac{49}{98}$
$\cos (P-Q)=\dfrac{1}{2}$
Therefore, $P-Q={{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ = $\dfrac{\pi }{3}$ $\left( as 0 < P,Q <\dfrac{\pi }{2} \right)$
Therefore, $P-Q=\left( \dfrac{\pi }{3} \right)$ or ${{60}^{\circ }}$
Hence, the value of $P-Q$ = ${{60}^{\circ }}$
Thus, Option ( B ) is correct.

Note: Angles play a crucial role not in geometry but also in trigonometry. When we measure the angle and if the angle comes out to be less than ${{90}^{\circ }}$ is called the acute angle and the angles greater than ${{90}^{\circ }}$ are called obtuse angle. An obtuse angle lies in the opposite of an acute angle. Students generally make mistakes by taking direct inverse to find the value of angles which leads to a complex equation.