Question

In a $△ABC,$ $E$ is the midpoint of side $BC$ and $D$ is a point on side $AC.$ If length of side $AC$ is $1$ unit and $\angle BAC={60}^{\circ },$ $\angle ACB={20}^{\circ },$ $\angle DEC={80}^{\circ },$ then the value of $Area\left(△ABC\right)+2Area\left(△CDE\right)$ is

• A. $\frac{\sqrt{3}}{16}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{\sqrt{3}}{4}$
• D. $\frac{\sqrt{3}}{8}$

Answer 1

The correct option is D $\frac{\sqrt{3}}{8}$

Let $Area\left(△ABC\right)={\Delta }_1$ and $Area\left(△CDE\right)={\Delta }_2$

${\Delta }_1=\frac{1}{2}\times AB\times AC\times \sin {60}^{\circ }\Rightarrow {\Delta }_1=\frac{\sqrt{3}}{4}AB$
Using sine rule in $△ABC$, we get
$\frac{AB}{\sin {20}^{\circ }}=\frac{AC}{\sin {100}^{\circ }}\Rightarrow {\Delta }_1=\frac{\sqrt{3}}{4}\frac{\sin {20}^{\circ }}{\sin {100}^{\circ }}\Rightarrow {\Delta }_1=\frac{\sqrt{3}}{2}\frac{\sin {10}^{\circ }\cos {10}^{\circ }}{\cos {10}^{\circ }}\Rightarrow {\Delta }_1=\frac{\sqrt{3}}{2}\sin {10}^{\circ }\cdots \left(1\right)$

Now, ${\Delta }_2=\frac{1}{2}\times CD\times CE\times \sin {20}^{\circ }$
$△CED$ is isosceles where $CD=CE=\frac{BC}{2}$
Using sine rule in $△ABC$, we get
$BC=\frac{AC\sin {60}^{\circ }}{\sin {100}^{\circ }}=\frac{\sqrt{3}}{2\sin {100}^{\circ }}$
So, ${\Delta }_2=\frac{1}{2}\times {\left(\frac{\sqrt{3}}{4\sin {100}^{\circ }}\right)}^2\sin {20}^{\circ }$
$\Rightarrow {\Delta }_2=\frac{3}{32}\frac{\sin {20}^{\circ }}{{\cos }^2{10}^{\circ }}\Rightarrow {\Delta }_2=\frac{3}{16}\frac{\sin {10}^{\circ }}{\cos {10}^{\circ }}\cdots \left(2\right)$

Now,
$Area\left(△ABC\right)+2Area\left(△CDE\right)$
$={\Delta }_1+2{\Delta }_2=\frac{\sqrt{3}}{2}\sin {10}^{\circ }+\frac{3}{8}\frac{\sin {10}^{\circ }}{\cos {10}^{\circ }}=\frac{\sqrt{3}}{8}(4\sin {10}^{\circ }+\sqrt{3}\frac{\sin {10}^{\circ }}{\cos {10}^{\circ }})=\frac{\sqrt{3}}{8}\left(\frac{2\sin {20}^{\circ }+\sqrt{3}\sin {10}^{\circ }}{\cos {10}^{\circ }}\right)=\frac{\sqrt{3}}{8}\left(\frac{2\sin {20}^{\circ }+2\sin {60}^{\circ }\cdot \sin {10}^{\circ }}{\cos {10}^{\circ }}\right)=\frac{\sqrt{3}}{8}\left(\frac{2\sin {20}^{\circ }+\cos {50}^{\circ }-\cos {70}^{\circ }}{\cos {10}^{\circ }}\right)=\frac{\sqrt{3}}{8}\left(\frac{\sin {20}^{\circ }+\sin {40}^{\circ }}{\cos {10}^{\circ }}\right)=\frac{\sqrt{3}}{8}\left(\frac{2\sin {30}^{\circ }\cos {10}^{\circ }}{\cos {10}^{\circ }}\right)=\frac{\sqrt{3}}{8}$