If sin A = frac{1}{sqrt{10}} and sin B = frac | vedclass

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

Question

(D) Given $\sin A = \frac{1}{\sqrt{10}}$ and $\sin B = \frac{1}{\sqrt{5}}$.Since $A$ and $B$ are acute angles,$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1

Vedclass · Accepted Answer

$\pi/4$

Vedclass · Answer

(D) Given $\sin A = \frac{1}{\sqrt{10}}$ and $\sin B = \frac{1}{\sqrt{5}}$.Since $A$ and $B$ are acute angles,$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$.Similarly,$\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.Using the formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$:$\sin(A + B) = \left(\frac{1}{\sqrt{10}}\right) \left(\frac{2}{\sqrt{5}}\right) + \left(\frac{3}{\sqrt{10}}\right) \left(\frac{1}{\sqrt{5}}\right)$$\sin(A + B) = \frac{2}{\sqrt{50}} + \frac{3}{\sqrt{50}} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.Since $\sin(A + B) = \frac{1}{\sqrt{2}}$,we have $A + B = \frac{\pi}{4}$.

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 = $

Similar Questions

$\sin 15^\circ + \cos 105^\circ = $

If $(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma ) = \tan \alpha \tan \beta \tan \gamma $,then $(\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )(\sec \gamma - \tan \gamma ) = $

$\tan 75^\circ - \cot 75^\circ = $

$\tan \frac{2\pi}{5} - \tan \frac{\pi}{15} - \sqrt{3} \tan \frac{2\pi}{5} \tan \frac{\pi}{15}$ is equal to

If $\theta$ is an acute angle and $\sin\theta = \frac{p - 6}{8 - p}$,then $p$ must satisfy:

Vedclass Test Series

Exam Paper Generator

Online Exam Module