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In the given figure, PA is tangent to semi-circle SAR. PB is tangent to semi-circle RBT; SRT is a straight line, the lengths of the arcs are indicated in the figure Angle APB is measured by 

(a) \(\frac{1}{2}\) (a − b)   

(b)  a + b 

(c) \(\frac{1}{2}\) (a + b) 

(d)  (a – b)

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Answer : (b) a + b

First, draw the line connecting P and R and denote its other inter-sections with the circles by M and N;see accompanying figure. The arcs MR and NR contain the same number of degrees; so we may denote each arc by x. To verify this, note that we have two isosceles triangle with a base angle of one equal to a base angle of the other. 

∴ ∠NOR = ∠MOR.

∠APR = \(\frac{1}{2}\) {(c + a + c – x) – a} =  \(\frac{1}{2}\) {2c – x} 

∠BPR = \(\frac{1}{2}\) {b + d + d – (b – x)} = \(\frac{1}{2}\) {2d + x} 

and the sum of angles APR and BPR is 

∠BPA = c + d 

The desired angle is 360° – ∠BPA 

= 360° – (c + d) 

= (180° – c) + (180° – d) 

= a + b.

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