Question

Find acute angles A and B, if $\sin (A+2B)=\frac{\sqrt{3}}{2}$ and $\cos (A+4B)=0$, and $A>B.$

Answer 1

Given, $\sin (A+2B)=\frac{\sqrt{3}}{2}$

$\cos (A+4B)=0$

$A>B$,

We know that, $\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$ and $\cos {90}^{\circ }=0$

Consider,

$\sin (A+2B)=\frac{\sqrt{3}}{2}$ and $\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$

$⟹(A+2B)={60}^{\circ }$ ---------------(i)

Consider,

$\cos (A+4B)=0$ and $\cos {90}^{\circ }=0$

$⟹(A+4B)={90}^{\circ }$ ---------------(ii)

Solve (i) and (ii) :

$(A+2B)={60}^{\circ }$

$(A+4B)={90}^{\circ }$

Subtracting (i) from (ii),

$2B={30}^{\circ }$

$B=\frac{{30}^{\circ }}{2}={15}^{\circ }$

From (ii)

$(A+4B)={90}^{\circ }$

Also, $B={15}^{\circ }$

$A={90}^{\circ }-{60}^{\circ }$

$A={30}^{\circ }$

$\therefore A={30}^{\circ },B={15}^{\circ }$