Question

in the following figure ab is parallel to CD find the measure of BOC ​

Answer 1

Step-by-step explanation:

In fig. AF is a straight line

Therefore,

ABO + FBO = 180 (linear pair)

165 + FBO = 180

FBO (EBO) = 180 - 165 = 15

Now, AF is parallel to CD

Therefore, DCE = CEB (alternate angles)

CEB = 75

Now, CO is a straight line

Therefore,

CEB + OEB = 180 (linear pair)

75 + OEB = 180

OEB = 180 - 75 = 105

Now, in triangle EBO

EBO + BOE + OEB = 180

(angle sum property of triangle)

15 + BOE + 105 = 180

BOE = 180 - 105 + 15

BOE = 180 - 120

BOE = 60

Therefore, BOC = 60

Answer 2

Answer:

60°

Step-by-step explanation:

Given,

CD ║ AB

⇒ CD ║ AF   ( shown in the given diagram ),

Let ∠x and ∠DCE are linear pairs,

⇒ ∠x + m∠DCE= 180°,

⇒ ∠x + 75° = 180°        ( by the diagram ),

⇒ ∠x = 180° - 75° = 105°,

By the alternate interior angle theorem,

∠x ≅ ∠1,

⇒ m∠x = m∠1

⇒ m∠1 = 105°,

Now, by the vertically opposite angle theorem,

∠1 ≅ ∠BEO,

⇒ m∠BEO = 105°,

Now, m∠EBO = 15°,

∵ BEO is a triangle,

⇒ m∠EBO + m∠BEO  + m∠BOE = 180°,

⇒  15° + 105° + m∠BOE = 180°,

⇒  120° + m∠BOE = 180°,

⇒  m∠BOE = 60°,

⇒  m∠BOC = 60°,