Question

pls answer this question​

Answer 1

EXPLANATION.

Two tangents TP and TQ are drawn to a circle O from external point.

As we know that,

⇒ TP = TQ.

⇒ ∠TPQ = ∠TQP. ⇒(1).

⇒ PT is a tangent.

⇒ OP is a radius.

⇒ OP ⊥ TP.

As we know that,

In ∠OPT = 90°.

⇒ ∠OPQ + ∠TPQ = 90°.

⇒ ∠TPQ = 90° - ∠OPQ ⇒ (2).

In ΔPTQ.

⇒ ∠TPQ + ∠TQP + ∠PTQ = 180°. [Sum of angle of a triangle = 180°].

⇒ ∠TPQ + ∠TPQ + ∠PTQ = 180°.

⇒ 2∠TPQ + ∠PTQ = 180°.

⇒ 2(90° - ∠OPQ) + ∠PTQ = 180°. [From equation (2) ].

⇒ 180° - 2∠OPQ + ∠PTQ = 180°.

⇒ ∠PTQ = 2∠OPQ.

HENCE PROVED.

Answer 2

Let's understand the question -

★ This question says that there are two tangents TP and TQ and these are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

★ We are able to see in the figure or the attachment that there is a circle drawn to a circle with centre O from an external point T. And there, a triangle like structure is also drawn as angle PTQ. In short it is cleared that the concepts that are used in this question are of circle and triangle both.

Given that -

★ Tangents = TP and TQ

★ Centre of circle = O

★ External point = T

To prove -

★ ∠PTQ = 2∠OPQ.

Using concepts -

★ Angle sum property.

★ Some tangents theorm.

Knowledge required -

• Triangle -:

★ The angle sum property of a triangle says that the three angle of triangle measure 180° always.

★ In a triangle equal sides have equal angles opposite to them.

• Tangent -:

★ Tangent of circle is perpendicular to the radius through the point of the contact.

★ Tangent - It's that line which touch a curve or curved surface but not the point.

Let's prove that ∠PTQ = 2∠OPQ -

~ As according to the figure we are able to see tha and we already know that the length of tangents drawn through a circle is always equal.

∴ TP = TQ

Now in (triangle) ∆TPQ

➟ TP = TQ

∴ TQP = TPQ

Now have to use angle sum property

➟ ∠ (TQP + TPQ + PTQ) = 180°

➟ ∠ (2TPQ + PTQ) = 180°

➟ ∠PTQ = 180° - 2 ∠TPQ ..(1)

Now let's carry on

➟ OP is perpendicular to PT

∴ OPT = 90°

➟ ∠ (OPQ + TPQ) = 90°

➟ ∠OPQ = 90° - ∠TPQ

➟ 2∠OPQ = 2(90° - ∠TPQ) = 180° - 2∠TPQ ..(2)

Now from ..(1) and ..(2)

➟ ∠PTQ = 2∠OPQ.

➟ Henceforth, proved!

Additional information -

➝ PA² = PC × PB (Relationship to tagnet-secant therom).

Or...

➝ PC × PB = PA² (Relationship to tagnet-secant therom).