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..... 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..... 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..... 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)

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)

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)

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

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

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