(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 ...

Answer Details

(a) Height of the building.

Let \(d\) be the horizontal distance between the tower and the building. The observer is at the top of the tower, height \(72\text{ m}\).

Bottom of the building (same level as the foot of the tower) has an angle of depression of \(62^{\circ}\), so the vertical drop is the full \(72\text{ m}\):

\[\tan 62^{\circ}=\frac{72}{d}\Rightarrow d=\frac{72}{\tan 62^{\circ}}=\frac{72}{1.8807}=38.28\text{ m}\]

Top of the building has an angle of depression of \(51^{\circ}\), so the drop from the tower top down to the building top is:

\[\text{drop}=d\tan 51^{\circ}=38.28\times 1.2349=47.27\text{ m}\]

The height of the building is the tower height minus this drop:

\[h=72-47.27=24.73\text{ m}\]

Correct to 2 significant figures, the height of the building is \(25\text{ m}\).

(b) Circle, centre O, radius 30 cm, \(\angle POR=120^{\circ}\), \(\pi=3.142\).

(i) Length of chord PR.

Using the cosine rule in triangle \(POR\) with \(|OP|=|OR|=30\text{ cm}\):

\[|PR|^{2}=30^{2}+30^{2}-2(30)(30)\cos 120^{\circ}\]

\[=900+900-1800(-0.5)=1800+900=2700\]

\[|PR|=\sqrt{2700}=51.96\text{ cm}\approx 52.0\text{ cm}\]

(ii) Length of arc PQR.

From the diagram, \(Q\) lies on the major arc, so arc \(PQR\) uses the reflex angle:

\[\text{Reflex }\angle POR=360^{\circ}-120^{\circ}=240^{\circ}\]

\[\text{Arc }PQR=\frac{240}{360}\times 2\pi r=\frac{240}{360}\times 2\times 3.142\times 30\]

\[=\frac{240}{360}\times 188.52=0.6667\times 188.52=125.7\text{ cm}\]

(iii) Perimeter of the shaded portion.

The shaded region is the major segment, bounded by the chord \(PR\) and the major arc \(PQR\):

\[\text{Perimeter}=|PR|+\text{arc }PQR=51.96+125.7=177.7\text{ cm}\]

Correct to 3 significant figures, the perimeter of the shaded portion is \(178\text{ cm}\).