Question

50 points:-)

Please solve this question:

Q:-P is a point on the arc APB of a circle with centre O. Show that the sum of angles PAB and PBA is constant.

➡ PLEASE GIVE ONLY RELEVANT ANSWERS!!!

Answer 1

⇒ ∠OPA = 30°

Similarly, it can be proved that ∠OPB = 30°.

Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ∆PAB,

PA = PB  [lengths of tangents drawn from an external point to a circle are equal]

⇒∠PAB = ∠PBA   ...(1)  [Equal sides have equal angles opposite to them]

∠PAB + ∠PBA + ∠APB = 180°  [Angle sum property]

⇒∠PAB + ∠PAB = 180° – 60° = 120°  [Using (1)]

⇒2∠PAB = 120°

⇒∠PAB = 60°   ...(2)

From (1) and (2)

∠PAB = ∠PBA = ∠APB = 60°

∴ ∆PAB is an equilateral triangle...
Hope it helps you.....

Answer 2

⇒ ∠OPA = 30°

Similarly, it can be proved that ∠OPB = 30°.

Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ∆PAB,

PA = PB  [lengths of tangents drawn from an external point to a circle are equal]

⇒∠PAB = ∠PBA   ...(1)  [Equal sides have equal angles opposite to them]

∠PAB + ∠PBA + ∠APB = 180°  [Angle sum property]

⇒∠PAB + ∠PAB = 180° – 60° = 120°  [Using (1)]

⇒2∠PAB = 120°

⇒∠PAB = 60°   ...(2)

From (1) and (2)

∠PAB = ∠PBA = ∠APB = 60°

∴ ∆PAB is an equilateral triangle...

Hope it helps you.....