If the three angles of a triangle are: $${\left(x + 15 \right)^ \circ },$$ $${\left({\frac{{6x}}{5} + 6} \right)^ \circ }$$ and $${\left({\frac{{2x}}{3} + 30} \right)^ \circ }$$ then the triangle is:
A. Isosceles
B. Equilateral
C. Right angled
D. Scalene
Answer: Option B
Solution(By Examveda Team)
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
Related Questions on Triangles
If ABC and PQR are similar triangles in which ∠A = 47° and ∠Q = 83°, then ∠C is:
A. 50°
B. 70°
C. 60°
D. 80°
In the following figure which of the following statements is true?
A. AB = BD
B. AC = CD
C. BC + CD
D. AD < Cd
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