A) Scalene
B) Equilateral
C) Isosecles
D) Right angled
Correct Answer: B
Solution :
We have, \[b+c=2a\] .....(i) \[\cos 60{}^\circ =\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\] Þ \[\frac{1}{2}=\frac{4{{a}^{2}}-2bc-{{a}^{2}}}{2bc}\]Þ \[\frac{1}{2}=\frac{3{{a}^{2}}}{2bc}-1\] Þ \[\frac{3}{2}=\frac{3{{a}^{2}}}{2bc}\]Þ \[bc={{a}^{2}}\] ......(ii) From (i) and (ii), \[b=c=0\] i.e., triangle is equilateral.
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