Question

PQ is a chord of length 6cm of a circle of radius 6cm. TP and TQ are tangents of circle at point P and Q. find angle PTQ​

Answer 1

Given:-

• PQ = 6cm
• OP = OQ = 6cm

In △OPQ

• PQ = 6cm
• OP = 6cm
• OQ = 6cm

In △OPQ all sides are equal

Hence, it is an equilateral triangle

∠OPQ = ∠PQO = ∠QOP = 60°

{In an equilateral triangle all angles are equal and Each angle measures 60°}

∠QOP + ∠PTQ = 180°

{The angle between twotangents draw from an external point to a circle is supplementary to the angle subtended by the line Segment joining the point of contact at the center}

60° + ∠PTQ = 180°

∠PTQ = 180 - 60

∠PTQ = 120°

HOPE IT HELPS

#NAWABZAADI

Answer 2

Given:-

• PQ = 6cm
• OP = OQ = 6cm

In △OPQ,

• PQ = 6cm
• OP = OQ = 6cm

All sides are eequal hence, △OPQ is an equilateral triangle.

∠OPQ = ∠QOP = ∠POQ = 60°

(In an equilateral triangle all angles are equal to 60°)

∠POQ + ∠PTQ = 180°

(The angles between two tangents drawn from an external point to a circle is supplementary to the angle subtended bye the line Segment joining the points of contact at the center)

60° + ∠PTQ = 180°

∠PTQ = (180 - 60)°

∠PTQ = 120°

#NAWABZAADA