Question

In the given figure, two straight lines AB and CD intersect at O. If ∠AOC=(3x-5)0and ∠BOC=1180, find the value of ‘x’.Also, find ∠BODand ∠AOD.

Answer 1

Answer:

x = 41°, ∠AOC = 62°, ∠BOD = 62°

Step-by-step explanation:

In the figure ∠AOD = ∠COB (vertically opposite anlges) as it is given AB and CD are straight lines

∴3x-5 = 118

3x = 118 +5

3x = 123

x = 123/3

x = 41°

Now, We know ∠COB = 118° and forms a linear pair with ∠AOC

∠AOC + ∠COB = 180° (linear pair)

∠AOC + 118 ° = 180°

∠AOC = 180° - 118°

∠AOC = 62°

Now, We know ∠AOC = 62°

∠AOC = ∠BOD = 62° (linear pair)

Answer 2

Answer:

x = 41°

∠BOD = 62°

Step-by-step explanation:

∠AOD = (3x - 5)° [Given]

∠BOC = 118° [Given]

Now,

∠AOD = ∠BOC [Vertically Opposite Angles are always equal.]

∴ (3x - 5)° = 118°

3x = 118 + 5

3x = 123°

∴ x = $\frac{123}{3}$

  x = 41°

∴ ∠AOD = (3x - 5)°

              = [3(41) - 5]°

              = (123 - 5)°

              = 118°

Now,

∠BOD + ∠BOC = 180° [The sum of all angles on a straight line always

                                     equals 180°]

∠BOD +  118° = 180°

∴ ∠BOD = 180° - 118°

   ∠BOD = 62°