Question

The angles of a triangle $ABC$ are in A.P. The largest angle is twice the smallest and the median to the largest side divides the angle at the vertex in the ratio $2:3$.

If the length of the median is $2\sqrt{3}$cm, then the length of the largest side is?

• A. $8\cos {48}^0$
• B. $8\sin {48}^0$
• C. $4\cos {42}^0$
• D. $4\sin {42}^0$

Answer 1

The correct option is B $8\sin {48}^0$

Given $A,B,C$ are in A.P $\Rightarrow 2B=A+C$

Also, $A+B+C={180}^0$

$\Rightarrow B={60}^0$ and $A+C={120}^0$

Let $A=2C\Rightarrow A={80}^0,C={40}^0$

Let $AD$ be the median of length $2\sqrt{3}$ cm

Then, $\angle BAD+\angle DAC={80}^0$

Since $\angle BAD:\angle DAC=3:2$

$\therefore \angle BAD={48}^0;\angle DAC={32}^0$

From $△BAD,\frac{BD}{\sin {48}^0}=\frac{AD}{\sin {60}^0}$

$\Rightarrow BD=\frac{2\sqrt{3}\sin {48}^0}{\frac{\sqrt{3}}{2}}=4\sin {48}^0\therefore DC=8\sin {48}^0$