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Find A and B if \mathrm{\sin (A + 2B) = \frac{\sqrt{3}}{2}} and \mathrm{\cos (A + 4B) = 0}, where A and B are acute angels.

 

 

 

Answers (1)

Given in the question,

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

A and B are acute angles.

A and B = ?

\mathrm{\Rightarrow \sin (A + 2B)=\frac{\sqrt{3}}{2}}

\mathrm{\Rightarrow \sin (A + 2B)=\sin60\degree}\qquad \left[\because \sin 60\degree = \frac{\sqrt3}{2} \right ]

So, \mathrm{A + 2B = 60}\qquad -(i)

\mathrm{\Rightarrow \cos (A + 4B)=0}

\mathrm{\Rightarrow \cos (A + 2B)=\cos90\degree}\qquad \left[\because \cos 90\degree = 0 \right ]

So,     \mathrm{A +4B = 90}\qquad -(ii)

On subtracting equation (i) by equation (ii)

We get    \mathrm{2B = 30\Rightarrow B = 15\degree}

Put the value of B in equation (i)

we get,    \mathrm{A + 2\cdot 15 = 60}

                        \mathrm{A = 30\degree}

Answer;    \mathrm{A = 30\degree}

                   \mathrm{B = 15\degree}

Posted by

Safeer PP

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