Answer: Option B
According to question,
 $$ \Rightarrow \left( {x + {{15}^ \circ }} \right) + {\left( {\frac{{6x}}{5} + 6} \right)^ \circ } + $$      $${\left( {\frac{{2x}}{3} + 30} \right)^ \circ } = $$    $${180^ \circ }$$
 $$\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\angle A + \angle B + \angle C = {{180}^ \circ }} \right\}$$
 $$ \Rightarrow x + \frac{{6x}}{5} + \frac{{2x}}{3} = $$     $${180^ \circ } - \left(15 + 6 + 30\right)$$
$$\eqalign{
& \Rightarrow \frac{{15x + 18x + 10x}}{{15}} = 180 - 51 \cr
& \Rightarrow 43x = 129 \times 15 \cr
& \Rightarrow x = {45^ \circ } \cr
& \Rightarrow {\text{each}}\,{\text{angle}} \cr
& \Rightarrow {\left( {x + 15} \right)^ \circ } = 45 + 15 = {60^ \circ } \cr
& \Rightarrow {\left( {\frac{{6x}}{5} + 6} \right)^ \circ } = {60^ \circ } \cr
& \Rightarrow {\left( {\frac{{2x}}{3} + 30} \right)^ \circ } = {60^ \circ } \cr} $$
∵ All three angles are equal 60°
⇒ Triangle will be equilateral triangle