Question

If sinA=1/√5 and cosB=3/√10,where A,B are positive acute angles prove thatA+B=45°

Answer 1

sin^2a=1/5
cos^a=1-1/5=4/5
cosa=2/root5
cos^2b=9/10
sin^2b=1/10
sinb=1/root10
sin(a+b)=sinacosb+cosasinb=1/root2=sin45
a+b=45(proved)

1

Answer 2

Step-by-Step Explanation:

Since we have given that

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

We will find cos A and sin B.

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

Similarly,

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

Now, we will use :

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

Hence, Proved.