Question 2
- In a right-angled triangle, sinX=3/5. Evaluate, leaving the answer as a fraction, 5(cosX)2 – 3.
- The base of a pyramid, 12 cm high, is a rectangle with dimensions 42 cm by 11 cm. if the pyramid is filled with water and emptied into a conical container of equal height and volume, calculate, leaving the answer in surd form (radicals), the base radius of the container. [Take π=22/7]
Observation
The Chief Examiner reported that performance in this question was quite low but the Candidates did better in part (a) than (b).
In part (a), by the use of Pythagoras theorem, the adjacent side of the right – angled triangle is 4. Then, cos(X=4/5) and substituting to get 5(cos X)2 – 3 = 5(4/5)^2-3=16/5-3 = (16 - 15)/5=1/5. Majority of the Candidates were able to evaluate this correctly. However, in part (b), most of the Candidates could not equate the volume of the pyramid to the conical container. They were expected to find Volume of pyramid = 1/3×42×11×12=1,848 (cm)^2 Volume of conical container = 1/3×22/7×r^2×12=1,848 and simplifying to get r2 = 147 so that r=7√3 cmas required.