Question

In a $∆ABC,ifb+c=2a\text{and}\angle A=60°,$ then $∆ABC$ is

• A. Equilateral Triangle
• B. Right-angle Triangle
• C. Isosceles Triangle
• D. Scalene Traingle

Answer 1

The correct option is A

Equilateral Triangle

Determine the type of triangle

Step 1: Determine relation between angles

Given, $b+c=2a$

Using sine rule, we have

$$\begin{gathered} \Rightarrow k\sin B+k\sin C=2\times k\sin A \\ \Rightarrow \sin B+\sin C=2\times \sin A \\ \Rightarrow 2\sin \frac{(B+C)}{2}\times \cos \frac{(B-C)}{2}=2\sin A \\ \Rightarrow 2\sin \frac{\left({\displaystyle \frac{\mathrm{\pi }}{2}}-A\right)}{2}\cos \frac{(B-C)}{2}=2\sin A \\ \Rightarrow 2\cos \left(\frac{A}{2}\right)\times \cos \frac{(B-C)}{2}=2\times 2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right) \\ \Rightarrow \cos \frac{(B-C)}{2}=2\sin \left(\frac{A}{2}\right)\text{[Given},A=60°] \\ \Rightarrow \cos \frac{(B-C)}{2}=2\sin \left(30°\right) \\ \Rightarrow \cos \frac{(B-C)}{2}=2\times \frac{1}{2}[\because \sin 30°=\frac{1}{2}] \\ \Rightarrow \cos \frac{(B-C)}{2}=1 \\ \Rightarrow \cos \frac{(B-C)}{2}=\cos \left(0°\right) \\ \Rightarrow \frac{(B-C)}{2}=0 \\ \Rightarrow B-C=0 \\ \Rightarrow B=C \end{gathered}$$

Step 2: Determine the value of angles.

We know that the internal angles of the triangle are $180°$,

$$\begin{gathered} \Rightarrow \angle A+\angle B+\angle C=180° \\ \Rightarrow \angle B+\angle C=180°-\angle A \\ \Rightarrow \angle B+\angle B=180°-60°[\because \angle B=\angle C,\text{shown above and}\angle A=60°,\text{Given}] \\ \Rightarrow 2\angle B=120° \\ \Rightarrow \angle B=60° \\ \text{Since}\angle B=\angle C,\text{thus}\angle C=60° \end{gathered}$$

Since the internal angles are equal, we can say that the given triangle is an equilateral triangle.

Hence, option (A) is correct.