Question

In a triangle $ABC$, if $b+c=2a$ and $\angle A={60}^{\circ }$, then $\Delta ABC$ is.

• A. Rightangled
• B. Scalene
• C. Equilateral
• D. Isosceles

Answer 1

Answer: (C)

Equilateral

$b+c=2a,\angle A={60}^{\circ }$

By using projection formula

$\begin{array}{c}\Rightarrow b=a\cos c+c\cos A\to \left(i\right)\\ \Rightarrow c=a\cos B+b\cos A\to \left(ii\right)\end{array}$

Adding equation (i) and (ii)

$\begin{array}{cc}\Rightarrow b+c=a\left\{\cos C+\cos B\right\}+\left(b+c\right)\cos A& \\ \Rightarrow \left(b+c\right)=a\left\{2\cos \left(\frac{C+B}{2}\right).\cos \left(\frac{C-B}{2}\right)\right\}+\left(2a\times \cos {60}^{\circ }\right)& \\ \Rightarrow 2a=a\left\{2\cos \left(\frac{\pi }{2}-\frac{A}{2}\right).\cos \left(\frac{C-B}{2}\right)\right\}+\left(2a\times \frac{1}{2}\right)& \left[\begin{array}{c}A+B+C=\pi \\ \frac{A+B}{2}=\frac{\pi }{2}-\frac{A}{2}\end{array}\right]\\ \Rightarrow 2a=a\left\{2.\sin \frac{A}{2}.\cos \left(\frac{C-B}{2}\right)\right\}+a& \\ \Rightarrow a=a\left\{2\times \sin {30}^{\circ }.\cos \left[\frac{C-B}{2}\right]\right\}& \\ 1=1\times \cos \left(\frac{C-B}{2}\right)& \\ \Rightarrow \cos \left(\frac{C-B}{2}\right)=\cos 0^{\circ }& \\ \Rightarrow \therefore \frac{C-B}{2}=0& \\ \therefore C=B& \\ Then,A+B+C={180}^{\circ }& \\ \Rightarrow \therefore {60}^{\circ }+B+B={180}^{\circ }& \\ \Rightarrow 2B={120}^{\circ }& \\ \therefore B={60}^{\circ }& \\ Hence,A=B=C={60}^{\circ }& \end{array}$

(It is equilletral triangle)

Hence, this is the answer.