In the given fig. PQ is a chord of length 6 cm and the radius of the circle is 6 cm. TP and TQ are two tangents drawn from an external point T. Find ∠PTQ.
Attachments:
Answers
Answered by
214
GIVEN:
PQ = 6 cm , OP = OQ = 6 cm
In ∆OPQ,
PQ = 6 cm , OP = OQ = 6 cm
All sides are equal, hence ∆OPQ is an equilateral triangle.
∠OPQ = ∠OQP = ∠POQ = 60°
[In an equilateral ∆ all angles are of 60°]
∠POQ + ∠PTQ = 180°
[The angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre]
60° + ∠PTQ = 180°
∠PTQ = 180° - 60°
∠PTQ = 120°
Hence, ∠PTQ is 120°.
HOPE THIS WILL HELP YOU..
PQ = 6 cm , OP = OQ = 6 cm
In ∆OPQ,
PQ = 6 cm , OP = OQ = 6 cm
All sides are equal, hence ∆OPQ is an equilateral triangle.
∠OPQ = ∠OQP = ∠POQ = 60°
[In an equilateral ∆ all angles are of 60°]
∠POQ + ∠PTQ = 180°
[The angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre]
60° + ∠PTQ = 180°
∠PTQ = 180° - 60°
∠PTQ = 120°
Hence, ∠PTQ is 120°.
HOPE THIS WILL HELP YOU..
Answered by
39
Hey Bro Your Answer Is Attached With This File..
Attachments:
Similar questions