Question

In a $\Delta ABC,\angle C=3\angle B=2(\angle A+\angle B)$.

Choose the correct options.

• A. One of the angles is ${20}^{\circ }$
• B. One of the angles is ${70}^{\circ }$
• C. One of the angles is ${40}^{\circ }$
• D. One of the angles is ${120}^{\circ }$

Answer 1

Answer: (A), (C), (D)

The correct options are
A

One of the angles is ${20}^{\circ }$

C

One of the angles is ${40}^{\circ }$

D

One of the angles is ${120}^{\circ }$

Let $\angle A=x^{\circ }$ and $\angle B=y^{\circ }$.

According to the given condition,

$\angle C=3\angle B=(3y{)}^{\circ }.$

We know that,

$\angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow x+y+3y=180$

$\Rightarrow x+4y=180\dots \dots \left(i\right)$

Also,

$\angle C=2(\angle A+\angle B)$.

$\Rightarrow 3y=2(x+y)$

$\Rightarrow 2x-y=0\dots \dots \left(ii\right)$

Solving (i) and (ii) we get,

$x=180-4y$ [From (i)]

Substituting value of x in (ii),

$2(180-4y)-y=0$

$360-8y-y=0\Rightarrow y=40$

Now, $x=180-4y$

$x=180-4\left(40\right)\Rightarrow x=20$

$\Rightarrow x=20,y=40$

$\Rightarrow \angle A={20}^{\circ },\angle B={40}^{\circ }\angle C=(3\times 40{)}^{\circ }={120}^{\circ }$