Question

If $\frac{sinA}{sinB}=\frac{\sqrt{3}}{2}and\frac{cosA}{cosB}=\frac{\sqrt{5}}{2},0\le A,B\le \frac{\pi }{2},$ then

• A. $A=\frac{\pi }{3}$
• B. $B=\frac{\pi }{4}$
• C. $tanA=1$
• D. $tanB=\frac{\sqrt{5}}{\sqrt{3}}$

Answer 1

The correct option is B $B=\frac{\pi }{4}$

sin $A=\frac{\sqrt{3}}{2}sinBCosA=\frac{\sqrt{5}}{2}cosB$
Squaring and adding, we get 4 = 3 $si{n}^2$ B + 5 $co{s}^2$ B
$\Rightarrow B=\frac{\pi }{4}$