∠COB = 50°
∠AOD = 50° (vertically opposite angles)
Now ∠AOC and ∠COB form a linear pair
Thus ∠AOC + ∠COB = 180°
⇒ ∠AOC + 50° = 180°.
∠AOC = 180° - 50° = 130°
Also ∠AOC and ∠BOD are vertically opposite angles
∠BOD = ∠AOC = 130°
Thus the three angles are
∠AOD = 50°
∠AOC = 130°
∠BOD = 130°

Step 1: Draw the given angle ∠ABC with the measure 80° using protractor.

Step 2: With B as center and convenient radius, draw an arc to cut BA and BC. Mark the points of intersection as E on BA and F on BC.

Step 3: With the same radius and E as center, draw an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc.

Step 4: Mark the point of intersection as G. Draw a ray BX through G. BG is the required bisector of the given angle ∠ABC

Given l is parallel to m and n is transversal to l and m.
We get, y = 2x [Vertically opposite angles are equal]
y + 4x = 180° [sum of interior angles that lie on the same side of the transversal]
2x + 4x = 180° [since y = 2x]
6x = 180°
Dividing by 6 on both sides
\(\frac{x}{6}=\frac{180^o}{6}\)gives, x = 30°
Now, y = 2(30°) = 60°.

(i) The corresponding angles to b° are d° and 68°.
(ii) The measure of b° is 68° (Because b° corresponds to 68°)
(iii) The angles having measure 68° are b°, d°, g°, i° and k°.
(iv) The angles having measure 112° are a°, c°, e°, f°, h° and j°.

(i) The angle that corresponds to \(\angle\)1 is \(\angle\)5
(ii) The angle that is alternate interior to \(\angle\)3 is \(\angle\)5
(iii) The angle that is alternate exterior to \(\angle\)8 is \(\angle\)2
(iv) The angle that corresponds to \(\angle\)8 is \(\angle\)4
(v) The angle that is alternate exterior to \(\angle\)7 is \(\angle\)1
(vi) The angle that is alternate interior to \(\angle\)6 is \(\angle\)4