Question

In a $\Delta ABC,\angle A=x^o,\angle B=3x^o$ and $\angle C=y^o$. If $3y^o-5x^o={30}^o$, then the triangle is

• A. Obtuse angled triangle
• B. Acute angled triangle
• C. Right angled triangle
• D. None of these

Answer 1

Answer: (C)

Right angled triangle

We know that sum of angles in a triangle $={180}^0$
$=>x+3x+y={180}^0$
$=>4x+y=180$ --- (1)
Also given, $3y-5x=30$ --- (2)

Multiplying equation $\left(1\right)$ with $3$ we get, $12x+3y=540$ ----- equation $\left(3\right)$Subtracting equation $\left(2\right)$from $\left(3\right)$, we get $17x=510=>x=30$

Substituting $x=30$ in the equation

$\left(1\right)$, we get $4\left(30\right)+y=180=>y=60$

Hence, the angles are $A={30}^0;B=3x={90}^0;C={60}^0$

Since angle B is a right angle, ABC is a right angled triangle.