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2004 WAEC Mathematics Theory (a) The angles of depression of the top and bottom of a building are 51°...

Mathematics
WAEC 2004

(a) The angles of depression of the top and bottom of a building are 51° and 62° respectively from the top of a tower 72m high. The base of the building is on the same horizontal level as the foot of the tower. Calculate the height of the building correct to 2 significant figures.

(b)  In the diagram, PR is a chord of the circle centre O and radius 30cm, < POR = 120°. Calculate correct to three significant figures : (i) the length of chord PR ; (ii) the length of arc PQR ; (iii) the perimeter of the shaded portion. (Take \(\pi = 3.142\)).

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Explanation

(a) 

ST = QR ; PS = height of tower.

In \(\Delta\) PST,

\(\frac{72}{ST} = \tan 62°\)

\(ST = \frac{72}{\tan 62}\)

= 38.283m \(\approxeq\) 38.3m

In \(\Delta\) PQR, \(\frac{h}{QR} = \tan 51°\)

\(h = QR \times \tan 51°\)

= \(38.3 \times 1.235\)

= \(47.3m\)

QS = RT (height of the building)

QS = PS - PQ

= 72 - 47.3 = 24.7m

(b) 

(i) \(\Delta POR = \Delta QOR\) (isosceles triangle)

\(\frac{PT}{30} = \sin 60\)

\(PT = 30 \sin 60 = 30 \times \frac{\sqrt{3}}{2}\)

= \(15\sqrt{3} cm\)

PT = TR

PR = 2PT 

= \(2 \times 15\sqrt{3} = 30\sqrt{3} cm\)

= \(30 \times 1.732 = 51.96 cm\)

\(\approxeq 52 cm\)

(ii) Length of minor arc, PQR = \(\frac{\theta}{360} \times 2\pi r\)

= \(\frac{120}{360} \times 2 \times 3.142 \times 30\)

= \(62.84 cm\)

(iii) Perimeter of shaded portion = chord PR + Arc PQR

= 51.96 + 62.84

= 114.8 cm \(\approxeq\) 115cm.


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