Question

in the given figure AB are the centres of 2 circles. DC is the common tangent. if angle DMP =40 degree find RQC

Answer 1

∠RQC = 50°

Step-by-step explanation:

Given that

$\angle DPM=40^\circ$

$\therefore \angle DAM=2\times 40^\circ=80^\circ$  (∵ The angle subtended by an arc of a circle at its centre is twice of the angle it subtends anywhere on the circle's circumference)

∵ DC is a tangent

∴ AD ⊥ CD

And BC ⊥ CD

$\therefore \angle ADC=\angle BCD=90^\circ$

$ADCB$ will be a quadrilateral

Therefore,

$\angle CBR=360^\circ-90^\circ-90^\circ-80^\circ$

$\implies \angle CBR=100^\circ$

Thus,

$\angle RQC=\frac{1}{2}\angle CBR$

$\implies \angle RQC=\frac{1}{2}\times 100^\circ$

$\implies \angle RQC=50^\circ$

Hope this answer is helpful.

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