Question

please solve this question ​

Answer 1

Solution :-

Here , given that line BOA is a straight line so ∠BOA = 180° .

Angles on line BOA

• ∠AOC = 60°
• ∠POQ = ?
• ∠BOC = ?

And given that line OP bisects ∠BOC

Now , using straight line property ,

Straight line property :- Angle on a straight line is always equal to 180° or sum of all angles on a straight line equal to 180°

Lets take only angles AOC & BOC

So now using straight line property,

→ ∠AOC + ∠BOC = 180°

→ ∠BOC = 180° - 60°

→ ∠BOC = 120°

Now here already given that OP bisects angle BOC . So ,

• ∠BOP = ∠POC ....eq(1)

Also ,

• ∠BOP + ∠POC = ∠BOC

Substituting eq(1)

• ∠POC + ∠BOP = ∠BOC

• 2∠POC = ∠BOC

• ∠POC = ∠BOC/2

→ ∠POC = 120°/2

→ ∠POC = 60°

Hence ,

• ∠BOC = 120°
• ∠POC = 60°

Answer 2

Answer:

Given :

• OP bisects ∠AOC, OQ bisects ∠BOC and OP⊥OQ

To prove :

• The points A,O and B are collinear.

Proof :

Since, OP bisects ∠AOC,

∠AOP = ∠COP …… (1)

Since, OQ bisects ∠BOC,

∠BOQ = ∠COQ …… (2)

Now, ∠AOB

= ∠AOP + ∠COP + ∠COQ + ∠BOQ

= ∠COP + ∠COP + ∠COQ + ∠COQ

From (1) and (2), we have

= 2 (∠COP + ∠COQ)

= 2∠POQ

Substitute all values :

= 2 ( 90° )

= 180°

Therefore, points A,O and B are collinear.