Theory question 112
Circle Theorems(a) In the diagram PR is a tangent to the circle centre O at Q. ∠POQ = 56° and PO intersects SQ at V such that ∠SVP = 109°. Calculate: (i) ∠TQP; (ii) ∠QTS.
(b) Simplify: (2n2 − 3n − 2)/(2n2 + 3n + 1) × (n2 − 1)/(n2 − 4)
Show the model answer
Model answer
(a)(i) OT = OQ, so ∠OQT = ∠OTQ = (180° − 56°)/2 = 62°; ∠OQP = 90° (tangent ⊥ radius) ⇒ ∠TQP = 90° − 62° = 28°
(ii) ∠QVO = ∠SVP = 109° (vertically opposite); in triangle QVO, ∠VQO = 180° − 109° − 56° = 15°, so ∠SQP = 90° − 15° = 75°
T lies on the arc on the same side of SQ as P, so ∠QTS = 180° − 75° = 105° (alternate segment theorem)
(b) = (2n + 1)(n − 2)/[(2n + 1)(n + 1)] × (n − 1)(n + 1)/[(n − 2)(n + 2)] = (n − 1)/(n + 2)
Explanation
Factorise every quadratic before cancelling.