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. $2\sin {32}^{°}$
• B. $4\sin {32}^{°}$
• C. $6\sin {32}^{°}$
• D. $8\sin {32}^{°}$

Answer 1

The correct option is D

$8\sin {32}^{°}$

Explanation for the correct option:

Step 1: Finding the measures of angles in triangles $ABC$

Let $A,B,C$be the angles of the triangle that is in A.P.

Given:$2\angle B=\mathbf{\angle }A+\angle C\to \left(i\right)$

We know that,

$$\begin{gathered} \angle A+\angle B+\angle C={180}^{°} \\ \Rightarrow 2\angle B+\angle B={180}^{°}\left[\because 2\angle B=\angle A+\angle C\right] \\ \Rightarrow 3\angle B={180}^{°} \\ \Rightarrow \angle B={60}^{°} \end{gathered}$$

Also, lets say $A$ is the largest angle then,

$\angle A=2\angle C$ (since the largest angle is twice the smallest angle.)

Substitute in equation $\left(i\right)$

$$\begin{gathered} 2\angle B=\mathbf{\angle }A+\angle C \\ =2\angle C+\angle C \\ =3\angle C \\ \Rightarrow 2\angle B=3\angle C \\ \Rightarrow 2\times 60°=3\angle C\left[\because \angle B=60°\right] \\ \Rightarrow 120°=3\angle C \\ \Rightarrow \angle C=40° \end{gathered}$$

Therefore ,

$$\begin{gathered} \angle A=2\angle C \\ =2\times 40° \\ =80° \end{gathered}$$

Step 2: Finding the length of the largest side in the triangle:

The side which is opposite to the largest angle is largest side. So here $BC$ is the side opposite to the largest angle$\angle A$.

Given the median divides $80°$in the ratio$2:3$ [median of the triangle divides the triangle into two equal areas]

that is

$\begin{array}{rcl}2x+3x& =& 80°\\ x& =& 16°\\ & \Rightarrow & 2x=32°\mathrm{and}3x=48°\end{array}$

On dividing, we get:

$\angle BAD={32}^{°}and\angle DAC={48}^{°}$

From $∆ABD,$

$$\begin{gathered} \frac{AD}{\sin {60}^{°}}=\frac{BD}{\sin {32}^{°}} \\ \Rightarrow BD=\frac{2\sqrt{3}}{{\displaystyle \frac{\sqrt{3}}{2}}}\sin {32}^{°}\mathbf{}\left[\because AD=2\sqrt{3}\right] \\ =4\sin {32}^{°} \end{gathered}$$

$$\begin{gathered} BC=2BD \\ \therefore BC=8\sin {32}^{°} \end{gathered}$$ ( since $D$ is a median)

Hence, option (D) is the correct answer.