Question

21.Two tangents TP and TQ are drawn to a circle with centre O from an
external point T. Prove that PTQ = 2 OPQ​

Answer 1

Step-by-step explanation:

We know that length of taughts drawn from an external point to a circle are equal

∴ TP=TQ−−−(1)

4∴ ∠TQP=∠TPQ (angles of equal sides are equal)−−−(2)

Now, PT is tangent and OP is radius.

∴ OP⊥TP (Tangent at any point pf circle is perpendicular to the radius through point of cant act)

∴ ∠OPT=90°

or, ∠OPQ+∠TPQ=90°

or, ∠TPQ=90 ° −∠OPQ−−−(3)

In △PTQ

∠TPQ+∠PQT+∠QTP=180 ° (∴ Sum of angles triangle is 180°)

or, 90 ° −∠OPQ+∠TPQ+∠QTP=180°

or, 2(90° −∠OPQ)+∠QTP=180 ° [from (2) and (3)]

or, 180° −2∠OPQ+∠PTQ=180°

∴ 2∠OPQ=∠PTQ

hence ,proved

Answer 2

PTQ = 2OPQ

PTQ = x

We know that,

Tangents drawn from an external point to a circle are equal.

TP = TQ

Consider ∆TPQ,

TQP = TPQ (TQ = TP)

Angle sum of ∆TPQ,

TQP + TPQ + PTQ = 180°

TPQ + TPQ + x = 180°

2TPQ + x = 180°

2TPQ = 180° - x

TPQ = (180° - x)/2

TPQ = 90° - x/2

OPT = 90°

(angle between tangent and radius)

OPQ = OPT - TPQ

OPQ = 90° - (90° - x/2)

OPQ = 90° - 90° + x/2

OPQ = x/2

2OPQ = x

2OPQ = PTQ

Hence provedd