If sin A = -frac{24}{25},cos B = frac{15}{17} | vedclass

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

Question

(C) Given $\sin A = -\frac{24}{25}$. Since $A$ is not in the $4^{	ext{th}}$ quadrant and $\sin A < 0$,$A$ must be in the $3^{	ext{rd}}$ quadrant. Th

Vedclass · Accepted Answer

$3^{	ext{rd}}$ quadrant

Vedclass · Answer

(C) Given $\sin A = -\frac{24}{25}$. Since $A$ is not in the $4^{	ext{th}}$ quadrant and $\sin A Given $\cos B = \frac{15}{17}$. Since $B$ is not in the $1^{	ext{st}}$ quadrant and $\cos B > 0$,$B$ must be in the $4^{	ext{th}}$ quadrant. Thus,$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - (\frac{15}{17})^2} = -\sqrt{1 - \frac{225}{289}} = -\sqrt{\frac{64}{289}} = -\frac{8}{17}$.Now,$\sin(A+B) = \sin A \cos B + \cos A \sin B = (-\frac{24}{25})(\frac{15}{17}) + (-\frac{7}{25})(-\frac{8}{17}) = -\frac{360}{425} + \frac{56}{425} = -\frac{304}{425} $\cos(A+B) = \cos A \cos B - \sin A \sin B = (-\frac{7}{25})(\frac{15}{17}) - (-\frac{24}{25})(-\frac{8}{17}) = -\frac{105}{425} - \frac{192}{425} = -\frac{297}{425} Since both $\sin(A+B) < 0$ and $\cos(A+B) < 0$,$(A+B)$ lies in the $3^{	ext{rd}}$ quadrant.

If $\sin A = -\frac{24}{25}$,$\cos B = \frac{15}{17}$,$A$ does not belong to the $4^{\text{th}}$ quadrant,and $B$ does not belong to the $1^{\text{st}}$ quadrant,then $(A+B)$ lies in which quadrant?

Similar Questions

If $\cos (A-B)=3/5$ and $\tan A \tan B=2$,then which one of the following is true?

$\sin \left(\frac{\pi}{3}+x\right)-\cos \left(\frac{\pi}{6}+x\right) = $

If $\sin A = \frac{1}{\sqrt{10}}$ and $\sin B = \frac{1}{\sqrt{5}}$,where $A$ and $B$ are positive acute angles,then $A + B = $

The value of $\frac{\tan 70^o - \tan 20^o}{\tan 50^o} = $

Prove that $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \left(\frac{x-y}{2}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module