WAEC - Mathematics (2006 - No. 37)

In the diagram, PR is a diameter of the circle centre O. RS is a tangent at R and QPR = 58^o. Find < QRS

112^o

116^o

122^o

148^o

PRQ = 90 - 58 = 32^o(angle in a semi-circle)

Since PRS = 90^o(radius angular to tangent)

QRS = 90 + 32

= 122^o