Question

22*. In figure 3.100, two circles intersect
each other at points S and R. Their
common tangent PQ touches the
circle at points P, Q.
Prove that, Z PRQ + Z PSQ = 180°
Fig. 3.100​

Answer 1

Explanation:

Given that two circles intersect each other at points S and R.

Their common tangent PQ touches the circle at points P,Q.

To Prove: $\angle P R Q+\angle P S Q=180^{\circ}$

By inscribed angle, we have,

$\angle S P Q=\frac{1}{2} m({arcPR})$

$\angle S Q R=\frac{1}{2} m({arcPQ})$

By inscribed angles, we have,

The sum of all angles in a triangle add up to 180°

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

Also,  $\angle S Q R=\frac{1}{2} \times m({arcSRQ})$ and

$\angle S P Q=\frac{1}{2} m({arcSR P)$

Since, we know that,

$\angle P S Q=\frac{1}{2} \angle P R Q$

${arcPR}+{arcRQ}=180^{\circ}$

Thus,

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

Hence proved

Learn more:

(1) Two circles intersect each other at points a and b with a common tangent touching them at c andd.find cad+cbd

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(2) Two tangents RQ and RP are drawn from an external point R to the circle with centre O. If angle PRQ=120, then prove that OR=PR+RQ.

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