Question

In Fig. 10.70, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS.

Answer 1

Step-by-step explanation:

Tangents drawn from an external point to a circle are equal

Hence, PQ = PR

And PQR is an isosceles triangle

ThereforeRQP =QRP

RQP +QRP +RPQ = 180[Angles in a triangle]

2RQP + 30 = 180°

2RQP = 150°

RQP =QRP = 75°

RQP =RSQ= 75 °[ Angles in alternate Segment Theoremstates that angle between chord and tangent is equal to the angle in the alternate segment]

RS II PQ

Therefore RQP = SRQ= 75° [They are alternate angles]

RSQ = SRQ = 75°

Therefore QRS is also an isosceles triangle

RSQ + SRQ + RQS = 180° [Angles in a triangle]

75° + 75° + RQS = 180°

150 °+ RQS = 180°

RQS = 30°

Required Answer: RQS = 30°

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Answer 2

The value of ∠RQS is 30°

Explanation:

Given that the tangents PQ and PR are drawn from an exrernal point P to a circle with centre O, such that ∠RPQ = 30°.

Also, given that a chord RS is drawn parallel to the tangent PQ.

Now, we need to find the value of ∠RQS.

Since, we know that the tangents drawn from an external point to a circle are equal.

Thus, we have,

$P Q=P R$

Hence, PQR is an isosceles triangle.

Thus, we have,

$\angle \mathrm{RQP}=\angle \mathrm{QRP}$

Let us consider the triangle PQR

Applying the triangle sum property, we have,

$\angle {RQP}+\angle {QRP}+\angle {RPQ}=180^{\circ}$

$\angle {RQP}+\angle{RQP}+\angle {RPQ}=180^{\circ}$

                   $2 \angle \mathrm{RQP}+30^{\circ}=180^{\circ}$

                             $2 \angle \mathrm{RQP}=150^{\circ}$

                               $\angle \mathrm{RQP}=75^{\circ}$

Thus, $\angle \mathrm{RQP}=\angle \mathrm{QRP}=75^{\circ}$

The alternate segment theorem states that "angle between chord and tangent is  equal to the angle in the alternate segment"

Thus, we have,

$\angle \mathrm{RQP}=\angle \mathrm{RSQ}=75^{\circ}$

From the figure, we can see that RS is parallel to PQ

Then, the alternate angles $\angle \mathrm{RQP}=\angle \mathrm{SRQ}=75^{\circ}$

$\angle \mathrm{RSQ}=\angle \mathrm{SRQ}=75^{\circ}$

Since, QRS is an isosceles triangle.

By angle sum property of the triangle.

We have,

$\angle R S Q+\angle S R Q+\angle R Q S=180^{\circ}$

          $75^{\circ}+75^{\circ}+\angle \mathrm{RQS}=180^{\circ}$

                  $150^{\circ}+\angle \mathrm{RQS}=180^{\circ}$

                             $\angle R Q S=30^{\circ}$

Thus, the value of ∠RQS is 30°

Learn more:

(1) In figure tangents PQ and PR are drawn to a circle such that angle RPQ=30 deg.A chord RS is drawn parallel to the tangent PQ .Find angle RQS

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