Question

In a $∆ABC,\angle A=x°,\angle B=\left(3x\right)°\mathrm{and}\angle C=y°.$
If $3y-5x=30,$ show that the triangle is right-angled.

Answer 1

In ∆ ABC, we have:
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°$ (Angle sum property of a triangle)
∴ ​x + 3x + y = 180
4x + y = 180 ....(i)
Again, 3y − 5x = 30 (Given)
⇒ −5x + 3y = 30 ....(ii)
On multiplying (i) by 3, we get:
12x + 3y = 540 ....(iii)
On subtracting (ii) from (iii), we get:
17x = (540 − 30) = 510
⇒ x = 30
On substituting x = 30 in (i), we get:
4 × 30 + y = 180
⇒ 120 + y = 180
⇒ y = (180 − 120) = 60
Thus, we have:
$$\begin{gathered} \angle \mathrm{A}=30° \\ \angle \mathrm{B}=3\times 30°=90° \\ \mathrm{And},\angle \mathrm{C}=60° \end{gathered}$$
Since in the given triangle, one angle is 90°, it is a right-angled triangle.