Question

In $∆$ABC, $\angle A=x°,\angle B=3x°and\angle C=y°$. If 3y $-$ 5x = 30. Prove that the triangle is right angled.

Answer 1

$$\begin{gathered} \angle \mathrm{A}=x°,\angle \mathrm{B}=3x°and\angle \mathrm{C}=y° \\ \text{We}\mathrm{know}\mathrm{that}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{angles}\mathrm{of}\text{a}\mathrm{triangle}\mathrm{is}180°. \\ \mathrm{Hence},\angle A+\angle B+\angle C=180° \\ \Rightarrow x+3x+y=180 \\ \Rightarrow 4x°+y°=180°...\left(\mathrm{i}\right) \\ -5x+3y=30...\left(\mathrm{ii}\right)\left(\mathrm{Given}\right) \\ \mathrm{Multiplying}\mathrm{eq}\left(\mathrm{i}\right)\mathrm{by}3,\mathrm{we}\mathrm{get}: \\ 12x+3y=540...\left(\mathrm{iii}\right) \\ \mathrm{Now},\mathrm{subtracting}\left(\mathrm{ii}\right)\mathrm{from}\left(\mathrm{iii}\right),\text{we get:} \\ 12x+3y=540 \\ -5x+3y=30 \\ \underline{+--} \\ 17x=510 \\ \therefore x=30 \\ \mathrm{Substituting}x=30\mathrm{in}4x+y=180,\text{we get:} \\ 4\times 30+y=180 \\ \Rightarrow y=60 \\ \mathrm{Hence},\angle A=30°,\angle B=3\times 30°=90°\text{and}\angle C=60°. \\ \mathrm{Yes},\text{the triangle}\mathrm{is}\mathrm{right}-\mathrm{angled}\mathrm{at}\angle B. \end{gathered}$$