Question

In a $△ABC,\angle C={90}^0,a=3$ and $b=4$ and $D$ is a point on $AB$, so that $\angle BCD={30}^0$, then $CD$ equals

Answer 1

Let $CD=x$
Area of $(△ACD+△BCD)=$area of $△ABC$
$\Rightarrow \frac{1}{2}\times CA\times CD\sin {60}^0+\frac{1}{2}\times BC\times x\sin {30}^0=\frac{1}{2}\times 12$
$\Rightarrow \frac{1}{2}\times 4\times x\frac{\sqrt{3}}{2}+\frac{1}{2}\times 3x\times \frac{1}{2}=6$
$\Rightarrow 4x\sqrt{3}+3x=24$(on simplification)
$\therefore x=\frac{24}{3+4\sqrt{3}}$