Question

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

Answer 1

Answer:

the answer is as follows

Step-by-step explanation:

an(tpo)=an(tqo)=90 by TRT theorem

an(tpq)=an(tqp) as TP=TQ(Tangent)

in triangle(TPQ),

an(ptq)+ an(tpq)+an(tqp)=180

an(ptq)+2[an(tpq)]=180

an(ptq)+2[an(tpo)-an(opq)]=180

an(ptq) +2(90)-2an(opq)=180

an(ptq)-2an(opq)=0

an(ptq)=2an(opq)

Answer 2

Given:

✰ Two tangents TP and TQ are drawn to a circle with centre O from an external point T.

To Prove:

✠ ∠PTQ = 2∠OPQ

Proof:

Let's understand the concept first!

• In such type of questions, we will use basic tangent theorem.
• Two tangents TP and TQ are drawn to a circle with centre O from an external point T are equal because length of the tangent drawn from the external point are always equal.
• After that we know the property that opposite angles to equal sides are equal. Since the lengths of the tangents drawn from the external point are equal, therefore the opposite angles to equal sides are also equal. So, ∆TPQ, ∆TQP and ∆TPQ. These triangles are equal.

We know that two tangents TP and TQ are drawn to a circle with centre O from an external point T.

So, from the theorem the lengths of the tangents drawn from the external point are equal. So,

➛ TP = TQ

∴ ∠TQP = ∠TPQ ...① [ ∵ Opposite angles to equal sides are equal. ]

We know that TP is the tangent and OP is the radius of the circle. The theorem states that the tangent at any point of the circle is perpendicular to the radius of the circle through the point of contact between two.

∴ OP ⟂ TP

So,

➤ ∠OPT = 90°

➤ ∠OPQ + ∠TPQ = 90°

➤ ∠TPQ = 90° - ∠OPQ ...②

Now,

In ∆PTQ

Sum of all angles of a triangle is equal to 180°.

➛ ∠TPQ + ∠TQP + ∠PTQ = 180°

➛ ∠TPQ + ∠TPQ + ∠PTQ = 180° [ from ① ∠TQP = ∠TPQ ]

➛ 2∠TPQ + ∠PTQ = 180°

➛ 2( 90° - ∠OPQ ) + ∠PTQ = 180° [ from ② ∠TPQ = 90° - ∠OPQ ]

➛ 180° - 2∠OPQ + ∠PTQ = 180°

➛ ∠PTQ = 180° - 180° + 2∠OPQ

➛ ∠PTQ = 2∠OPQ

∴ ∠PTQ = 2∠OPQ

Hence Proved!!