# Answer the 20th question."DON'T SEND THE SIMILAR QUERY,KINDLY PERFORM THE CORRECT AND STEP BY STEP CALCULATION".

Here is the proof to your question.

The given information can be represented diagrammatically as:

Let ∠POB =

*x*°It is known that diagonals of a square bisect the angles.

∴ ∠OBP = ∠OAP = (90°/2) = 45°

Using exterior angle property for ∆OPB,

∠OPA = ∠OBP + ∠POB = 45° +

*x*°In OAP, OA = AP

∴ ∠OPA = ∠AOP

⇒ ∠AOP = 45° +

*x*° … (1)It is known that diagonals of square are perpendicular to each other.

∴ ∠AOP + ∠POB = 90°

⇒ 45° +

*x*° +*x*° = 90° [Using (1)]⇒ 2

*x*= 90 – 45 = 45⇒

*x*= 22.5∴ ∠POB = 22.5° and ∠AOP = 45° + 22.5° = 67.5° = 3 × 22.5°

⇒ ∠AOP = 3∠POB

Hope! You got the proof.

Cheers!

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