# Ex.3.7 Q8 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

## Question

\(ABCD\) is a cyclic quadrilateral finds the angles of the cyclic quadrilateral.

## Text Solution

**What is Known?**

Measurement of the angles of the cyclic quadrilateral in terms of *\(x\)* and \(y.\)

**What is Unknown?**

Measurement of the angles of the cyclic quadrilateral.

**Reasoning:**

Pairs of opposite angles of a cyclic quadrilateral are supplementary.

**Steps:**

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is \(180^\circ.\)

Therefore,

\[\begin{align}\angle A + \angle C &= {180^\circ}\\\left( {4y + 20} \right) + \left( { - 4x} \right) &= 180\\

4y + 20 - 4x &= 180\\ - 4\left( {x - y} \right)& = 160\\x - y &= - 40 \qquad \left( 1 \right)\end{align}\]

And

\[\begin{align}\angle B + \angle D &= {180^\circ}\\\left( {3y - 5} \right) + \left( { - 7x + 5} \right) &= 180\\

3y - 5 - 7x + 5 &= 180\\ - 7x + 3y &= 180\\7x - 3y &= - 180 \qquad \left( 2 \right)\end{align}\]

Multiplying equation \((1)\) by \(3,\) we obtain

\[3x - 3y = - 120\qquad \left( 3 \right)\]

Subtracting equation \((3)\) from equation \((2),\) we obtain

\[\begin{align}4x &= - 60\\x &= - 15\end{align}\]

Substituting \(x = - 15\) in equation \((1),\) we obtain

\[\begin{align} - 15 - y &= - 40\\y &= 25\end{align}\]

Therefore,

\[\begin{align}\angle A &= 4 \times 25 + 20 = {120^\circ}\\\angle B &= 3 \times 25 - 5 = {70^\circ}\\

\angle C &= - 4 \times \left( { - 15} \right) = {60^\circ}\\\angle D &= - 7 \times \left( { - 15} \right) + 5 = {110^\circ}\end{align}\]