Question

in figure the circle s with centers p and qtouch each otherat r a line passing through r meet s the circle at a and b respectively​

Answer 1

Step-by-step explanation:

The circles with centres P and Q touch each other at R. ∴ By theorem of touching circles, P – R – Q i. In ∆PAR, seg PA = seg PR [Radii of the same circle] ∴ ∠PRA ≅ ∠PAR (i) [Isosceles triangle theorem] Similarly, in ∆QBR, seg QR = seg QB [Radii of the same circle] ∴ ∠RBQ ≅ ∠QRB (ii) [Isosceles triangle theorem] But, ∠PRA ≅ ∠QRB (iii) [Vertically opposite angles] ∴ ∠PAR ≅ ∠RBQ (iv) [From (i) and (ii)] But, they are a pair of alternate angles formed by transversal AB on seg AP and seg BQ. ∴ seg AP || seg BQ [Alternate angles test] ii. In ∆APR and ∆RQB, ∠PAR ≅ ∠QRB [From (i) and (iii)] ∠APR ≅ ∠RQB [Alternate angles] ∴ ∆APR – ∆RQB [AA test of similarity] iii. ∠PAR = 35° [Given] ∴ ∠RBQ = ∠PAR= 35° [From (iv)] In ∆RQB, ∠RQB + ∠RBQ + ∠QRB = 180° [Sum of the measures of angles of a triangle is 180°] ∴ ∠RQB + ∠RBQ + ∠RBQ = 180° [From (ii)] ∴ ∠RQB + 2 ∠RBQ = 180° ∴ ∠RQB + 2 × 35° = 180° ∴ ∠RQB + 70° = 180° ∴ ∠RQB = 110°Read more on Sarthaks.com - https://www.sarthaks.com/851355/adjoining-figure-circles-with-centres-touch-each-other-line-passing-through-meets-circles?show=851359#a851359