Question

6. In Fig. 14.29, PQ = RQ, angle RQP = 72°, PC and QC are tangents to
the circle with centre 0. Calculate

(i) the measure of the angle subtended by the chord PQ at the
centre, and
(ii) Angle PCQ.​

Answer 1

Answer:

see the pic i have showed all the steps

Answer 2

The measure of the angle subtended by the chord PQ at the centre is 108° and the angle PCQ is 72°.

Given:

PQ=RQ

Angle RQP=72°

To find:

i. Angle POQ

ii. Angle PCQ

Solution:

The chord PQ subtends the angle POQ at the circle's centre.

We know that in ΔRQP, PQ=RQ.

So, the angles corresponding to these sides will also be equal.

Angle QRP=Angle QPR

Also, angle QRP+angle QPR+angle RQP=180°

Using values,

2(angle QRP)+72°=180°

2(angle QRP)=180°-72°

2(angle QRP)=108°

angle QRP=108°/2

angle QRP=angle QPR=54°

Now, the angle POQ=2× angle QRP. (ANgle at the centre is twice the angle at the circumference)

So, angle POQ=2×54°

Angle POQ=108°

Now, the tangent QC forms the angle PQC with the triangle RQP.

So, angle PQC=angle QRP=54° (angles in the alternate segment)

We know that PC=QC and so, angle PQC=angle QPC.

In ΔPCQ,

angle PQC+angle QPC+angle PCQ=180°

Using values,

54°+54°+angle PCQ=180°

108°+angle PCQ=180°

Angle PCQ=180°-108°

Angle PCQ=72°

Therefore, the measure of the angle subtended by the chord PQ at the centre is 108° and the angle PCQ is 72°.