Question

chapter 10 example 2 and 3（class 10 maths textbook）

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

Example 2

We get,    $\angle P T Q=2 \angle O P Q$

Example 3

$T P=\frac{20}{3} \mathrm{~cm}$

Step-by-step explanation:

Example 2

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $\angle P T Q=2 \angle O P Q$

Given: A circle with centre $\mathrm{O}$

Two tangents TP and TQ to the circle where P and Q are the point of contact.

To prove   $\angle P T Q=2 \angle O P Q$

Step 1

In ΔTPQ  

$TP=TQ$ (Length of tangents from external point to circle are equal)

$\angle \mathrm{TQP}=\angle \mathrm{TPQ}$ (Angles opposite to equal sides are equal)

Step 2

Now, ΔTPQ  

$\angle P T Q+\angle T P Q+\angle T Q P=180^{\circ}$ (Sum of the triangle)

$\angle P T Q=180^{\circ}-2 \angle T P Q$

$\angle P T Q+2\angle T Q P=180^{\circ}$

$\angle P T Q=180^{\circ}-2\left(90^{\circ}-\angle O P Q\right)$        $\left(\because \angle O P T=90^{\circ}\right. )$

$\therefore \angle P T Q=2 \angle O P Q$

Hence, we get

  $\angle P T Q=2 \angle O P Q$

Example 3

PQ is a chord of length 8cm of a circle of radius of 5cm. The tangent at P and Q intersect at a point T. find the length TP.

Step 1

$PQ=8cm$ and

$PR =4cm$ (Perpendicular from the center to the chord bisects the chord)

In ΔPRO

$\angle R=90^{\circ}$

$P R^{2}+O R^{2}=O P^{2}$

$4^{2}+O R^{2}=5^{2}$

$O R^{2}=9$

$OR=3$

Step 2

In ΔOPT and ΔORP

$\angle O=\angle O$ (Common)

$\angle O P T=\angle O R P=90^{\circ}$

$\triangle O P T \sim \triangle O R P$ (AA test)

$\frac{O P}{O R}=\frac{O T}{O P}=\frac{P T}{R P}$

$\frac{5}{3}=\frac{TP}{4}$

$T P=\frac{20}{3} \mathrm{~cm}$

Therefore, we get

$T P=\frac{20}{3} \mathrm{~cm}$

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