Question

sin a = 1/√5 and cos b = 3/√10, where a and b are positive acute angles, prove that :a+b=45°​

Answer 1

Answer:

If sin A =1/root 5, cos B = 3/root 10 where A and B are positive acute angles , prove that A+B=45.

Answer 2

$\huge\mathtt\pink{Answer}$

${\sf{\underline{\underline{\pink{Step-by-Step \ Explanation:}}}}}$

${\sf{\underline{\underline{\pink{Since \ we \ have \ given \ that}}}}}$

$\begin{gathered}\sin A=\frac{1}{\sqrt{5}}\\\\\cos A=\frac{3}{\sqrt{10}}\end{gathered}$

${\sf{\underline{\underline{\pink{We \ will\ find \ cos \ A \ and \ sin \ B.}}}}}$

$\begin{gathered}\cos A=\sqrt{1-\frac{1}{\sqrt{5}}^2}\\\\\cos A=\sqrt{\frac{5-1}{5}}\\\\\cos A=\sqrt{\frac{4}{5}}\\\\\cos A=\frac{2}{\sqrt{5}}\end{gathered}$

${\sf{\underline{\underline{\pink{Similarly,}}}}}$

$\begin{gathered}\sin A=\sqrt{1-\frac{3}{\sqrt{10}}^2}\\\\\cos A=\sqrt{\frac{10-9}{10}}\\\\\cos A=\sqrt{\frac{1}{10}}\\\\\cos A=\frac{1}{\sqrt{10}}\end{gathered}$

${\sf{\underline{\underline{\pink{Now, \ we \ will \ use}}}}}$

$\begin{gathered}\sin(A+B)= sin \ A \ cos \ B +cos \ A \ sin \ B \\\\\sin(A+B)=\frac{1}{\sqrt{5}}\times \frac{3}{\sqrt{10}}+\frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{10}}\\\\\sin(A+B)=\frac{3}{\sqrt{50}}+\frac{2}{\sqrt{50}}\\\\\sin(A+B)=\frac{5}{\sqrt{50}}\\\\\sin(A+B)=\frac{1}{\sqrt{2}}\\\\A+B=\sin^{-1}(\frac{1}{\sqrt{2}})\\\\A+B=45^\circ\end{gathered}$

$\sf\green{Hence, \ Proved.}$