Question

Guys pls tell this question answer I am in hurry and pls give a step answer​

Answer 1

Answer:

Step-by-step explanation:

ATQ,

∠x+∠y+∠z=180° (angle-sum property)

=> ∠x=180-∠y-∠z

=> ∠x-180=(-∠y)-∠z

=> ∠y+∠z=180-∠x      ---(1)

Now, we can write ∠CBE as 180-∠y

=> ∠CBO=1/2(180-∠y)

=> ∠CBO=90-1/2∠y

Similarly,

∠BCO=90-1/2∠z

Now, we know that ∠BOC=180-∠CBO-∠BCO(Angle-Sum property)

=> ∠BOC=180-(90-1/2∠y)-(90-1/2∠z)

=> ∠BOC=180-90+1/2∠y-90+1/2∠z

=> ∠BOC=180-180+1/2∠y+1/2∠z

=> ∠BOC=1/2∠y+1/2∠z

=> 2∠BOC=∠y+∠z

Substituting the value of ∠y+∠z from (1), we get,

2∠BOC=180-∠x

=> ∠BOC=90-1/2∠x

Hence Proved

Hope it helps :)