Question

If $A+B,A$ are acute angles such that $\sin (A+B)=\frac{24}{25}$ and $\tan A=\frac{3}{4},$ find the value of $\cos B.$

Answer 1

$\tan A=\frac{3}{4}$ $\because =\frac{Perpendicular}{Base}$

Perpendicular $=3$, Base $=$

$\therefore Hypotenus=5$ $\left(\right(3{)}^2+(4{)}^2=x^2)$

$x=5$

$sinA=\frac{3}{5}$ and $\cos A=\frac{4}{5}$.....(i)

We have to calculate $\cos B$

By identity ${\cos }^{2}B+{\sin }^{2}B=1$

$sinB=\sqrt{1-{\cos }^{2}B}$

By equation $sin(A+B)=sinA\cos B+\cos AsinB=\frac{24}{25}$

putting eq from (i)

$sin\Rightarrow \frac{3}{5}\cos B+\frac{4}{5}sinB=\frac{24}{25}$

$\frac{3}{5}\cos B+\frac{4}{5}\sqrt{1-{\cos }^{2}B}B=\frac{24}{25}$

Divide by $5$ on both sides

$3\cos B+4\left(\sqrt{1-{\cos }^{2}B}\right)=\frac{24}{5}$

$15\cos B+20(\sqrt{1-{\cos }^{2}B}=24$

$20\left(\sqrt{1-{\cos }^{2}B}\right)=24-15\cos B$

On squaring both sides

$\Rightarrow \left(20{)}^2\right(1-{\cos }^{2}B)=(27{)}^2+(15{)}^2{\cos }^{2}B-2\times 24\times 15\cos B$

$\Rightarrow 400-400{\cos }^{2}B=576+225{\cos }^{2}B-720\cos B$

$\Rightarrow -625{\cos }^{2}B+720\cos B-176=0$

$\Rightarrow 625{\cos }^{2}B-720\cos B-176=0$

$\Rightarrow 720\pm \sqrt{(720{)}^2-4\times 625\times 176}/2\times 625$

$\Rightarrow \frac{720\pm \sqrt{78400}}{1250}$

$\Rightarrow \frac{720\pm 280}{1250}=\frac{4}{5}$ and $\frac{440}{1250}$

Hence $\cos B=\frac{4}{5}$