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

Given:

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

To do:

We have to find angles $A$ and $B$.

Solution:  

$\sin (A+2B)=\frac{\sqrt3}{2}$

$\sin (A+2B)=\sin 60^{\circ}$          (Since $\sin 60^{\circ}=\frac{\sqrt3}{2}$)       

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

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

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

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

$\Rightarrow  A=90^{\circ}-4B$........(ii)

Substituting (ii) in (i), we get,

$90^{\circ}-4B+2B=60^{\circ}$

$\Rightarrow  2B=30^{\circ}$

$\Rightarrow  B=\frac{30^{\circ}}{2}$

$\Rightarrow  B=15^{\circ}$

$\Rightarrow  A=90^{\circ}-4(15^{\circ})$

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

$=30^{\circ}$

The values of $A$ and $B$ are $30^{\circ}$ and $15^{\circ}$ respectively. 

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Updated on: 10-Oct-2022

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