Question

solve question number 3 and 5...pl answer it is very urgent

Answer 1

$\red{\huge{\underline{Answer \:3}}}$

$\angle AOB + \angle BOC = 180 \\ 3x+8+x-2=180°\\ 4x+6=180° \\ 4x = \frac{180\degree-6}{4}\\ x= 43.5\degree$

$\angle AOB \\= 3(43.5)+8 \\= 138.5\degree$

$\angle BOC \\= x-2 \\= 43.5-2\\=41.5\degree$

$\pink{\huge{\underline{Answer \:5}}}$

We have two parallel lines with one transversal

Now relation for all angles

$\red{\underline{\sf{By \:Vertically\: Opposite \:Angles \:Property\colon}}}$

$\angle 1 = \angle 3\\ \angle 2 = \angle 4\\ \angle 5 = \angle 7\\ \angle 8 = \angle 6$

$\red{\underline{\sf{By \:Alternate\: Interior\: Angle \:Property\colon}}}$

$\angle 3 = \angle 5\\ \angle 4 = \angle 6$

$\red{\underline{\sf{By \:Corresponding \:Angle \:Property\colon}}}$

$\angle 1 = \angle 5\\ \angle 2 = \angle 6\\ \angle 3 = \angle 7\\ \angle 4 = \angle 8$