Question

13. In the diagram, ABT is a circle centre O. PQ is a tangent to the circle at T and ABC is a straight line. TC bisects ∠BTQ. ∠BAT=44° and ∠PTA=60°. Find ∠ACT.
(b) The circumference of the base of a cylindrical tank is 11m. The height of the tank is 3m more than 6 times the base radius. Calculate: (i) radius; (ii) height; (iii) volume of the tank. [Take π=22/7]

Accepted Answer

(a) By the tangent-chord angle theorem, ∠PTA (between tangent PT and chord TA) = angle in the alternate segment = ∠ABT = 60°.
Since ABC is a straight line, in triangle ABT: ∠BAT+∠ABT+∠ATB=180° → 44°+60°+∠ATB=180° → ∠ATB=76°.
Also by the tangent-chord angle theorem, ∠BTQ (between tangent TQ and chord TB) = angle in alternate segment = ∠BAT = 44°.
Since TC bisects ∠BTQ: ∠BTC = ∠CTQ = 22° each.
∠ATC = ∠ATB + ∠BTC = 76° + 22° = 98°.
Since ABC is a straight line, ∠TAC = ∠TAB = 44° (same ray).
In triangle ATC: ∠ACT = 180° − ∠TAC − ∠ATC = 180° − 44° − 98° = 38°.

(b) Circumference = 2πr = 11m → r = 11/(2×22/7) = 11×7/44 = 1.75m.
(i) Radius = 1.75m.
Height = 6r+3 = 6(1.75)+3 = 10.5+3 = 13.5m.
(ii) Height = 13.5m.
(iii) Volume = πr²h = 22/7×(1.75)²×13.5 = 22/7×3.0625×13.5 = 22/7×41.34375 ≈ 129.9m³.

NECO Mathematics 2024 Theory — Question 13

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