Question

In the given figure, O is the center of the circle. PQ is the tangent to the circle at A. If ∠PAB = 58°, then ∠AQB equals 

(a) 32° 

(b) 26° 

(c) 44° 

(d) None of these

Answer 1

Answer : (b) 26º 

∠BAR = 90° (Angle in a semicircle) 

∠ARB = ∠PAB = 58° (Angles in alternate segments are equal) 

∠ABQ = 180° – (∠BAR + ∠ARB) (∠Sum property of a ∆) 

= 180° – (90° + 58°) 

= 180° – 148° = 32° 

∠QAR = ∠ABR = ∠ABQ = 32° (Angles in alternate segments are equal) 

∠AQB = 180° – (∠ABQ + ∠BAQ) 

= 180° – (32° + ∠BAR + ∠RAQ) 

= 180° – (32° + 90° + 32°) 

= 180° – 154° 

= 26°.