Hi Rida! Here are the answers to your question. The given information can be represented using a figure as ⇒ ∠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! This will help you. Cheers! Posted by Gopal Mohanty(MeritNation Expert)on 24/2/11 This conversation is already closed by Expert

Hi Rida! Here are the answers to your question. The given information can be represented using a figure as ⇒ ∠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! This will help you. Cheers! Posted by Gopal Mohanty(MeritNation Expert)on 24/2/11 This conversation is already closed by Expert

Hi Rida! Here are the answers to your question. The given information can be represented using a figure as ⇒ ∠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! This will help you. Cheers! Posted by Gopal Mohanty(MeritNation Expert)on 24/2/11 This conversation is already closed by Expert

sir is it necessary to use trigo in this??? i mean the tangents ARE ALWAYS inclined wid each othe at 60degreees rite? so can 't v write dis simply??????? Posted by Divya Kaur(student)on 10/3/11

sir is it necessary to use trigo in this??? i mean the tangents ARE ALWAYS inclined wid each othe at 60degreees rite? so can 't v write dis simply??????? Posted by Divya Kaur(student)on 10/3/11

sir is it necessary to use trigo in this??? i mean the tangents ARE ALWAYS inclined wid each othe at 60degreees rite? so can 't v write dis simply??????? Posted by Divya Kaur(student)on 10/3/11

no ...............m sorry ....................itz not always at 60 degrees ! it can vary !! Posted by Afzal Husain(student)on 10/3/11

no ...............m sorry ....................itz not always at 60 degrees ! it can vary !! Posted by Afzal Husain(student)on 10/3/11

no ...............m sorry ....................itz not always at 60 degrees ! it can vary !! Posted by Afzal Husain(student)on 10/3/11