(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 \(\pi = 3.142\)).
(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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Correct Answer: Option n
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.
(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.