Question

If PA and PB are tangents from an outside point P to a circle such that PA = 30 cm and , then find the length of the chord AB.

Answer 1

Answer:

Given, AP = 10 cm and ∠APB = 60° Represented in the figure We know that, A line drawn from centre to point from where external tangents are drawn divides or bisects the angle made by tangents at that point So, ∠APO = ∠OPB = 1/2 × 60° = 30° And, the chord AB will be bisected perpendicularly ∴ AB = 2AM In ∆AMP, AM = AP sin 30° AP/2 = 10/2 = 5cm [As AB = 2AM] So, AP = 2 AM = 10 cm And, AB = 2 AM = 10 cm Alternate method: In ∆AMP, ∠AMP = 90°, ∠APM = 30° ∠AMP + ∠APM + ∠MAP = 180° 90° + 30° + ∠MAP = 180° ∠MAP = 60° In ∆PAB, ∠MAP = ∠BAP = 60°, ∠APB = 60° We also get, ∠PBA = 60° ∴ ∆PAB is equilateral triangle AB = AP = 10 cm

Answer 2

Answer:

Given, AP = 10 cm and ∠APB = 60°

Step-by-step explanation:

We know that, A line drawn from centre to point from where external tangents are drawn divides or bisects the angle made by tangents at that point So, ∠APO = ∠OPB = 1/2 × 60° = 30° And, the chord AB will be bisected perpendicularly ∴ AB = 2AM In ∆AMP, AM = AP sin 30° AP/2 = 10/2 = 5cm [As AB = 2AM] So, AP = 2 AM = 10 cm And, AB = 2 AM = 10 cm Alternate method: In ∆AMP, ∠AMP = 90°, ∠APM = 30° ∠AMP + ∠APM + ∠MAP = 180° 90° + 30° + ∠MAP = 180° ∠MAP = 60° In ∆PAB, ∠MAP = ∠BAP = 60°, ∠APB = 60° We also get, ∠PBA = 60° ∴ ∆PAB is equilateral trianglesAB = AP = 10 cm