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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$60$

Vedclass · Answer

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

Similar Questions

If $\cos (\alpha + \beta) = \frac{3}{5}$,$\sin (\alpha - \beta) = \frac{5}{13}$ and $0 < \alpha, \beta < \frac{\pi}{4}$,then $\tan (2\alpha)$ is equal to

If $A+B=\frac{\pi}{4}$,then $\frac{\cos B-\sin B}{\cos B+\sin B}=$

If $\frac{\pi}{2} < \alpha < \pi$ and $\pi < \beta < \frac{3\pi}{2}$,with $\sin \alpha = \frac{15}{17}$ and $\tan \beta = \frac{12}{5}$,then the value of $\sin(\beta - \alpha)$ is (in $/221$)

If $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b},$ then $\frac{\tan x}{\tan y}$ is equal to

If $\tan A=\frac{1}{\sqrt{x(x^2+x+1)}}, \tan B=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and $\tan C=\sqrt{x^{-1}+x^{-2}+x^{-3}}$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module