Question

In figure, OE is the bisector of $$\angle BOD$$. If $$\angle 1=70^o$$, find the magnitudes of $$\angle 2, \angle 3$$ and $$\angle 4$$.

Answer 1

Let $$\angle EOB =x$$

Given,

$$\angle EOB = \angle EOD$$ [ ∵ OE is bisector of $$\angle BOD $$]

$$ x = \angle 1$$.

=> $$\angle 1 = x = 70^o$$.

=> $$\angle BOD = \angle 1 + x = (70+70)^o$$.

=>$$\angle BOD = 140^o$$.

According to the figure,

$$\angle COB = \angle AOD $$ and

$$\angle AOC = \angle BOD = 140^o$$ [ ∵ vertically opposite angles]

We know that,

sum of all the angles around a point $$= 360^o $$

$$ \angle BOD + \angle AOD +\angle AOC + \angle COB = 360^o $$

$$ \angle BOD + \angle AOD +\angle BOD + \angle AOD = 360^o $$

$$ 2 \angle BOD + 2 \angle AOD = 360^ o$$

$$ \angle BOD + \angle AOD = 180^ o$$

$$ 140 +\angle AOD = 180^ o$$

$$ \angle AOD = 180^o-140^ o$$

$$ \angle AOD = 20^ o$$

=> $$ \angle COB = 20^o$$

$$\angle AOC = \angle BOD = 140^o$$

$$ \therefore \angle 2 = \angle 4 = 20^o$$ and $$ \angle 3 = 140^o$$.