Ex.6.1 Q1 Lines and Angles Solution - NCERT Maths Class 9


Question

In the given figure lines \(AB\) and \(CD\) intersect at \(O.\) If  \(\begin{align} \angle AOC + \angle BOE = 70 ^ { \circ } \end{align}\) and \(\begin{align} \angle BOD = 40 ^ { \circ }, \end{align}\) find \(\angle BOE\) and reflex \(\angle COE\).

 Video Solution
Lines And Angles
Ex 6.1 | Question 1

Text Solution

What is known?

\(\angle AOC+\angle BOE={70}^{\circ}\) and \(\angle BOD = 40^{\circ}\)

What is unknown?

\(\angle BOE =?,\) Reflex \(\angle COE =?\)

Reasoning:

We know that vertically opposite angles formed when two lines intersect are equal. Also, sum of the adjacent angles is \(180\) degrees.

Steps:

Let \(\Delta AOC = \rm x\) and \(\rm{}\Delta BOE =\rm y.\) 

Then \(x+ y=70^\circ\) \((\because AOC+BOE=70^\circ )\) 

Let Reflex \(\Delta COE = \rm z\)

We can see that \(AB\) and \(CD\) are two intersecting lines so the pair of angles formed are vertically opposite angles and they are equal.

\(\text{i.e. } \angle AOD\) and \(\angle BOC\) and \(\angle AOC = \angle BOD.\)

Since \(\angle AOC = x\)  and \(\angle AOC = \angle BOD = 40^ {\circ},\)

we can say that  \(x = 40^ {\circ}.\)

Also we know that,

\[\begin{align} x + y &= 70 ^ {\circ } \\ 40 ^ { \circ } + y &= 70 ^ {\circ } \\ y &= 70 ^ { \circ } - 40 ^ {\circ } \\ y &= 30 ^ { \circ } \\ \angle BOE &= 30 ^ { \circ } \end{align}\]

If we consider line \(AB\) and ray \(OD\) on it, then \(\angle AOD\) and \(\angle BOD\) are adjacent angles.

\[\begin{align} \angle AOD + \angle BOD &= 180 ^ { \circ } \\ \angle AOD + 40 ^ { \circ } &= 180 ^ { \circ } \\ \angle AOD &= 180 ^ { \circ } - 40 ^ { \circ } \\ &= 140 ^ { \circ } \end{align}\]

\(\begin{align}\text{Reflex }\angle COE &=\begin{pmatrix}\angle AOC +\\ \angle AOD +\\ \angle BOD +\\ \angle BOE \end{pmatrix} \\ &= \begin{pmatrix}40 ^ { \circ } + 140 ^ { \circ } \\+ 40 ^ { \circ } + 30 ^ { \circ } \end{pmatrix}\\ &= 250 ^ { \circ } \end{align}\)

 Video Solution
Lines And Angles
Ex 6.1 | Question 1
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