Ex.10.5 Q5 Circles Solution - NCERT Maths Class 9

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Question

In the given figure, \(A, B, C\) and \(D\)  are four points on a circle. \(AC\) and \(BD\)  intersect at a point \(\begin {align}{E} \end {align}\) such that \(\begin {align} \angle {BEC}=130^{\circ} \end {align}\) and \(\begin {align} \angle {ECD}=20^{\circ}. \end {align}\) Find \(\begin {align} \angle {BAC}\end {align}\).

 Video Solution
Circles
Ex 10.5 | Question 5

Text Solution

 

What is known?

Two angles and \(\begin {align} \angle {BEC}=130^{\circ} \text { and } \angle {ECD}=20^{\circ} \end {align}\)

What is unknown?

Value of \(\begin {align} \angle {BAC}\end {align}\)

Reasoning:

  • Sum of angles in a triangle is \(180^{\circ}.\)
  • Angles in the same segment are equal.

Steps:

Consider the straight line \(\begin {align} {BD} \end {align}\). As the line \(\begin {align} {AC} \end {align}\) intersects with the line \(\begin {align} {BD,} \end {align}\) then the sum of two adjacent angles so formed is \(180^{\circ}.\)

Therefore,

\[\begin{align} \angle {BEC}+\angle {DEC} &=180^{\circ} \\ 130^{\circ}+\angle {DEC} &=180^{\circ} \\ \angle {DEC} &=180^{\circ}-130^{\circ} \\ &=50^{\circ} \end{align}\]

Consider the \(\begin{align} \Delta {DEC,} \end{align}\) the sum of all angles will be \(180^{\circ}.\)

\[\begin{align} \angle \mathrm{DEC}+\angle \mathrm{EDC}+\angle \mathrm{ECD} &=180^{\circ} \\ 50^{\circ}+\angle \mathrm{EDC}+20^{0} &=180^{\circ} \\ \angle \mathrm{EDC} &=180^{\circ}-70^{\circ} \\ &=110^{\circ} \\ ∴ \angle \mathrm{BDC}=\angle \mathrm{EDC} &=110^{\circ} \end{align}\]

We know that, angles in the same segment of a circle are equal.

\[\begin {align} ∴ \angle \mathrm{BAC}=\angle \mathrm{BDC}=110^{\circ}\end {align}\]

  
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