Question 3
- The angle of depression of a point P on the ground from the top T of a building is 23.60. If the distance from P to the foot of the building is 50 m, calculate, correct to the nearest metre, the height of the building.
In the diagram, PT // SU, QS // TR, /SR/ = 6cm and /RU/ = 10cm. If the area of 
TRU = 45cm2, calculate the area of the trapezium QTUS.
Observation
In part (a), the Chief Examiner reported that Candidates’ had difficulty in representing the question in diagram. They were expected to draw a right-angled triangle and indicate the angle of depression as required which is 23.60 and then use the relevant trigonometry ratio to find the height of the building. They were expected to represent the information in a diagram.
From the diagram, tan 23.60 = 
, where x = height of the building. This then simplifies to x = 50 tan 23.60 = 22 m, correct to the nearest metre.
In part (b), the report stated that majority of the candidates were able to find the height of 
TRU but could not see /SU/ as equal to /QT/ as opposite sides of a parallelogram SRQT. They were supposed to have 45cm2 = 
. Height = 
= 9cm
Area of trapezium QTUS = 
(/QT/ +/SU/) height = 
(6+16) 9 = 99cm2