Question

In a triangle $ABC,\angle C={90}^o,a=3,b=4$ and $D$ is a point on $AB$, so that $\angle BCD={30}^o$. Then the length of $CD$ is

• A. $\frac{18-24\sqrt{3}}{25}$
• B. $\frac{18+24\sqrt{3}}{25}$
• C. $\frac{8-24\sqrt{3}}{25}$
• D. $Noneofthese$

Answer 1

The correct option is D $Noneofthese$

Area of $△BCD$ + Area of $△ACD$ = Area of $△ABC$

$\frac{1}{2}\cdot 3CD\cdot \sin 30+\frac{1}{2}\cdot 4CD\cdot \sin 60=\frac{1}{2}3\times 4$

$3CD\times \frac{1}{2}+4CD\times \frac{\sqrt{3}}{2}=12$

$\frac{CD}{2}(3+4\sqrt{3})=12$

$\therefore CD=\frac{24}{3+4\sqrt{3}}$