Vertically Opposite Angles
Two angles are called a pair of vertically opposite angles, if their arms form two pairs of opposite rays.
Let two lines AB and CD intersect at a point O. Then, two pairs of vertically opposite angles are formed.
(i) ∠AOC and ∠BOD (ii) ∠AOD and ∠BOC
Theorem 1:
If two lines intersect then the vertically opposite angles are equal.
Given: Two lines AB and CD intersect at a point O.
To prove: (i) ∠AOC = ∠BOD, (ii) ∠AOD = ∠BOC
Proof: Since ray OA stands on line CD, we have:
∠AOC + ∠AOD = 180° [linear pair].
Again, ray OD stands on line AB.
∴ ∠AOD + ∠BOD = 180° [linear pair]
∴ ∠AOC + ∠AOD = ∠AOD + ∠BOD [each equal to 180°]
∴ ∠AOC = ∠BOD
Similarly, ∠AOD = ∠BOC
Vertically Opposite Angles Example Problems With Solutions
Example 1: Two lines AB and CD intersect at O. If ∠AOC = 50°, find ∠AOD, ∠BOD and ∠BOC.
Solution: ∠AOD + ∠AOC = 180° (linear pair)
∠AOD + 50° = 180°
∠AOD = 130°
Also ∠BOD = ∠AOC
(vertically opposite angles)
& ∠BOC = ∠AOD = 130°
(vertically opposite angles)
∵ 130°, 50°, 130°.
Example 2: Two lines AB and CD intersect at a point O such that ∠BOC + ∠AOD = 280°, as shown in the figure. Find all the four angles.
Solution: ∠AOC = ∠BOD = x (Let)
(vertically opposite angles)
∵ ∠AOC + (∠AOD + ∠BOC) + ∠BOD = 360°
⇒ x + 280° + x = 360°
⇒ 2x = 80°
⇒ x = 40°
∵ ∠AOC = ∠BOD = x° = 40°.
and ∠BOC = ∠AOD = 280°/2 = 140°.
Example 3: In Fig., lines l1 and l2 intesect at O, forming angles as shown in the figure. If a = 35º, find the values of b, c, and d.
Solution: Since lines l1 and l2 intersect at O.
Therefore,
∠a = ∠c [Vertically opposite angles]
⇒ ∠c = 35º [∵ ∠a = 35º]
Clearly, ∠a + ∠b = 180º
[Since ∠a and ∠b are angles of a linear pair]
⇒ 35º + ∠b = 180º
⇒ ∠b = 180º – 35º
⇒ ∠b = 145º
Since ∠b and ∠d are vertically opposite angles. Therefore,
∠d = ∠b ⇒ ∠d = 145º [∵ ∠b = 145º]
Example 4: In Fig., determine the the value of y.
Solution: Since ∠COD and ∠EOF are vertically opposite angles. Therefore,
∠COD = ∠EOF ⇒ ∠COD = 5yº
[∵ ∠EOF = 5yº (Given)]
Now, OA and OB are opposite rays.
∵ ∠AOD + ∠DOC + ∠COB = 180º
⇒ 2yº + 5yº + 5yº = 180º
⇒ 12yº = 180º
⇒ yº = 180º/12 = 15.
Thus, yº = 15.
Example 5: In Fig., AB and CD are straight lines and OP and OQ are respectively the bisectors of angles BOD and AOC. Show that the rays OP and OQ are in the same line.
Solution: In order to prove that OP and OQ are in the same line, it is sufficient to prove that
∠POQ = 180º.
Now, OP is the bisector of ∠AOC
⇒ ∠1 = ∠6 …(i)
and, OQ is the bisector of ∠AOC
⇒ ∠3 = ∠4 ….(ii)
Clearly, ∠2 and ∠5 are vertically opposite angles.
∵ ∠2 = ∠5 ….(iii)
We know that the sum of the angles formed at a point is 360º.
Therefore,
∠1 +∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360º
⇒ (∠1 + ∠6) + (∠3 + ∠4) + (∠2 + ∠5) = 360º
⇒ 2∠1 + 2∠3 + 2∠2 = 360º
[Using (i), (ii) and (iii)]
⇒ 2(∠1 + ∠3 + ∠2) = 360º
⇒ ∠1 + ∠2 + ∠3 = 180º ⇒ ∠POQ = 180º
Hence, OP and OQ are in the same straight line.
Example 6: In Fig., two staright lines PQ and RS intersect each other at O. If ∠POT = 75º, find the values of a, b and c.
Solution: Since OR and OS are in the same line. Therefore,
∠ROP + ∠POT + ∠TOS = 180º
⇒ 4bº + 75º + bº = 180º ⇒ 5bº + 75º = 180º
⇒ 5bº = 105º ⇒ bº = 21
Since PQ and RS intersect at O. Therefore,
∠QOS = ∠POR
[Vertically oppsostie angles]
⇒ a = 4b
⇒ a = 4 × 21 = 84 [∵ b = 21]
Now, OR and OS are in the same line. Therefore.
∠ROQ + ∠QOS = 180º [Linear pair]
⇒ 2c + a = 180
⇒ 2c + 84 = 180 [∵ b = 84]
⇒ 2c = 96
⇒ c = 48
Hence, a = 84, b = 21 and c = 48