Question

In triangle $ABC,$ $E$ is the mid-point of side $BC$ and $D$ is on side $AC$. Let the length of side $AC$ be unity and $\angle BAC={60}^{\circ },$ $\angle ABC={100}^{\circ },$ $\angle ACB={20}^{\circ },$ $\angle DEC={80}^{\circ }.$ If the area of the triangle $ABC$ plus twice the area of triangle $CDE$ equals $\sqrt{\frac{m}{n}},$ where $m$ and $n$ are coprime, then the value of $(m+n)$ is greater than

• A. $20$
• B. $64$
• C. $78$
• D. $67$

Answer 1

Answer: (A), (B)

$64$


Let ${\Delta }_1=\text{area of}\Delta ABC$
and ${\Delta }_2=\text{area of}\Delta CDE$
${\Delta }_1=\frac{1}{2}\times 1\times c\times \sin {60}^{\circ }=\frac{\sqrt{3}c}{4}$

Applying sine rule in $\Delta ABC,$ we get
$\frac{a}{\sin {60}^{\circ }}=\frac{c}{\sin {20}^{\circ }}=\frac{1}{\sin {100}^{\circ }}$
$\therefore {\Delta }_1=\frac{\sqrt{3}}{4}\times \frac{\sin {20}^{\circ }}{\sin {100}^{\circ }}$
$\Rightarrow {\Delta }_1=\frac{\sqrt{3}}{4}\times \frac{2\sin {10}^{\circ }\cdot \cos {10}^{\circ }}{\cos {10}^{\circ }}$
$\Rightarrow {\Delta }_1=\frac{\sqrt{3}}{2}\sin {10}^{\circ }$

and ${\Delta }_2=\frac{1}{2}\times \frac{a}{2}\times \frac{a}{2}\times \sin {20}^{\circ }$
$\Rightarrow {\Delta }_2=\frac{a^2}{8}\times \sin {20}^{\circ }$
$\Rightarrow {\Delta }_2=\frac{\sin {20}^{\circ }}{8}\times {\left(\frac{\sin {60}^{\circ }}{\sin {100}^{\circ }}\right)}^2$
$\Rightarrow {\Delta }_2=\frac{1}{8}\times \frac{3}{4}\times \frac{\sin {20}^{\circ }}{{\cos }^2{10}^{\circ }}$
$\Rightarrow {\Delta }_2=\frac{3}{16}\times \frac{\sin {10}^{\circ }}{\cos {10}^{\circ }}$

Now, ${\Delta }_1+2{\Delta }_2$
$=\frac{\sqrt{3}}{2}\sin {10}^{\circ }+\frac{3}{8}\frac{\sin {10}^{\circ }}{\cos {10}^{\circ }}$
$=\frac{\sqrt{3}}{8}(4\sin {10}^{\circ }+\sqrt{3}\frac{\sin {10}^{\circ }}{\cos {10}^{\circ }})$
$=\frac{\sqrt{3}}{8}\left(\frac{2\sin {20}^{\circ }+\sqrt{3}\sin {10}^{\circ }}{\cos {10}^{\circ }}\right)$
$=\frac{\sqrt{3}}{8}\left(\frac{2\sin {20}^{\circ }+2\sin {60}^{\circ }\cdot \sin {10}^{\circ }}{\cos {10}^{\circ }}\right)$
$=\frac{\sqrt{3}}{8}\left(\frac{2\sin {20}^{\circ }+\cos {50}^{\circ }-\cos {70}^{\circ }}{\cos {10}^{\circ }}\right)$
$=\frac{\sqrt{3}}{8}\left(\frac{\sin {20}^{\circ }+\sin {40}^{\circ }}{\cos {10}^{\circ }}\right)$
$=\frac{\sqrt{3}}{8}\left(\frac{2\sin {30}^{\circ }\cos {10}^{\circ }}{\cos {10}^{\circ }}\right)$
$=\frac{\sqrt{3}}{8}=\sqrt{\frac{3}{64}}=\sqrt{\frac{m}{n}}$
$\Rightarrow m=3$ and $n=64$
$\therefore m+n=3+64=67$