Question

If $\cos P=\frac{1}{7}\mathrm{and}\cos Q=\frac{13}{14}$, where P and Q both are acute angles. Then, the value of P − Q is
(a) $\frac{\mathrm{\pi }}{6}$

(b) $\frac{\mathrm{\pi }}{3}$

(c) $\frac{\mathrm{\pi }}{4}$

(d) $\frac{\mathrm{\pi }}{12}$

Answer 1

(b) 60⁰ = $\frac{\mathrm{\pi }}{3}$
$\cos P=\frac{1}{7},\cos Q=\frac{13}{14}$
<sup>$\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e},\sin P=\sqrt{1-{\displaystyle \frac{1}{49}}}=\frac{4\sqrt{3}}{7}\mathrm{and}\sin Q=\sqrt{1-{\displaystyle \frac{169}{196}}}=\frac{3\sqrt{3}}{14}$
$\mathrm{Hence},\tan P=4\sqrt{3},\tan Q=\frac{3\sqrt{3}}{13}$
$\cos (P-Q)=\cos P\cos Q+\sin P\sin Q$
$$\begin{gathered} =\frac{1}{7}\times \frac{13}{14}+\frac{4\sqrt{3}}{7}\times \frac{3\sqrt{3}}{14} \\ =\frac{13+36}{98} \end{gathered}$$
$=\frac{49}{98}$
$$\begin{gathered} \therefore \cos \left(P-Q\right)=\frac{1}{2} \\ \Rightarrow P-Q={\cos }^{-1}\frac{1}{2} \\ \Rightarrow P-Q=60° \end{gathered}$$

Hence, the correct answer is option B.</sup>