# Ex.3.7 Q5 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

## Question

In \(\Delta ABC \) ,

\(\angle {C} = 3\angle B = 2(\angle A + \angle B)\).

Find the three angles.

## Text Solution

**What is Known?**

Relation between the angles of the triangle.

**What is Unknown?**

Measurement of each angles of the triangle.

**Reasoning:**

Sum of the measures of all angles of a triangle is \(180^\circ.\)

**Steps:**

Let the measurement of \(\angle A = {x^{\rm{\circ}}}\)

And the measurement of \(\angle B = {y^{\rm{\circ}}}\)

Using the information given in the question,

\(\angle C = 3\angle B = 2\left( {\angle A \!+\! \angle B} \right)\)

\[\begin{align} \Rightarrow 3\angle B &= 2\left( {\angle A + \angle B} \right)\\ \Rightarrow 3y &= 2\left( {x + y} \right)\\ \Rightarrow 3y &= 2x + 2y\\ \Rightarrow \;2x - y &= 0 \qquad \qquad \qquad \left( 1 \right)\end{align}\]

We know that the sum of the measures of all angles of a triangle is \(180^\circ.\)

Therefore,

\[\begin{align}\angle A + \angle B + \angle C &= {180^{\circ}}\\\angle A + \angle B + 3\angle B &= {180^{\circ}} \\&\begin{bmatrix} \because \angle C = 3\angle B \end{bmatrix} \\\angle A + 4\angle B &= {180^{\circ}}\\ x + 4y &= 180 \qquad \quad \left( 2 \right)\end{align}\]

Multiplying equation \((1)\) by \(4,\) we obtain

\[8x - 4y = 0 \qquad \left( 3 \right)\]

Adding equations \((2)\) and \((3),\) we obtain

\[\begin{align}9x &= 180\\x &= 20\end{align}\]

Substituting \(x = 20\) in equation \((1)\), we obtain

\[\begin{align}2 \times 20 - y &= 0\\y &= 40\end{align}\]

Therefore,

\[\begin{align}\angle A &= {x^{\circ}} = {20^{\circ}}\\\angle B &= {y^{\circ}} = {40^{\circ}}\\\angle C &= 3\;\angle B = 3 \times {40^{\circ}} = {120^{\circ}}\end{align}\]