Question

a telecommunication mast and a pillar stand on the same horizontal level. from the mid - point, o, between the foot of the pillar, s, and the foot of the mast, t, the angle of elevation of the top of the mast, g, is 70°. from the point, o, the angle of elevation of the top of the pillar, h, is 28°. if the pillar is 30 m high:
(a) (i) represent the information in a diagram;
(ii) calculate, correct to the nearest whole number:
(i) the height of the mast;
(ii) |hg|

1. (a) the points m, n and l lie on circle o with centre o. an angle subtended at the centre by the minor arc mn is 48°.

(i) illustrate the information in a diagram;
(ii) if an angle is subtended by the minor arc mn at l, determine the value of the angle subtended by the minor arc mn at l.
(b) a tiler, cofie, is able to lay 400 tiles each day in 8 hours. another tiler, boadye, works at 3/2 times cofies rate when laying some kind of tiles.
(i) how many tiles can boadye lay in 8 hours?
(ii) if both cofie and boadye work together, determine the time required to lay 2000 tiles.
in the diagram, e, f, g and h lie on the circumference of the circle centre o. eg and fh intersect at j such that ∠gfh = 30° and ∠hje = 110°. calculate:
(i) ∠eof;
(ii) ∠jeh.
(b) consider the following statements: most arts students are good at debates. some non - arts students are also good at debates.
(i) illustrate the information in a venn diagram.
(ii) determine from the diagram whether the following conclusions are valid or not valid:
(a) john is a non - arts student so he is not good at debates.
(b) aky is an arts student so she is good at debates.
12
(a) using the class interval 1 - 10, 11 - 20, 21 - 30, ..., construct a cumulative frequency table for the distribution.
(b) draw a cumulative frequency curve for the distribution.

Explanation:

(a)

1. Draw the diagram:

• Draw a horizontal - line to represent the ground level. Mark the mid - point \(O\). On one side of \(O\), mark the foot of the pillar \(S\) and on the other side, mark the foot of the mast \(T\) such that \(OS = OT\). Draw a vertical line from \(S\) to represent the pillar \(SH\) of height \(30\) m and a vertical line from \(T\) to represent the mast \(TG\). Mark the angles of elevation: \(\angle SOH=28^{\circ}\) and \(\angle TOG = 70^{\circ}\).

(b)(i)

Step1: Find the distance \(OS\) in the right - triangle \(SOH\)

In right - triangle \(SOH\), \(\tan\angle SOH=\frac{SH}{OS}\). Given \(SH = 30\) m and \(\angle SOH = 28^{\circ}\), we have \(\tan28^{\circ}=\frac{30}{OS}\). So, \(OS=\frac{30}{\tan28^{\circ}}\). Since \(\tan28^{\circ}\approx0.5317\), \(OS=\frac{30}{0.5317}\approx56.42\) m.

Step2: Find the height of the mast \(TG\) in the right - triangle \(TOG\)

Since \(OT = OS\approx56.42\) m and in right - triangle \(TOG\), \(\tan\angle TOG=\frac{TG}{OT}\). Given \(\angle TOG = 70^{\circ}\) and \(OT\approx56.42\) m, and \(\tan70^{\circ}\approx2.7475\), then \(TG = OT\times\tan70^{\circ}\approx56.42\times2.7475\approx155\) m.

Answer:

The height of the mast is approximately \(155\) m.

(b)(ii)

1. First, find the angles in \(\triangle SOH\) and \(\triangle TOG\):

• In \(\triangle SOH\), \(\angle SOH = 28^{\circ}\), \(\angle OSH=90^{\circ}\), so \(\angle OHS = 62^{\circ}\). In \(\triangle TOG\), \(\angle TOG = 70^{\circ}\), \(\angle OTG = 90^{\circ}\), so \(\angle OGT = 20^{\circ}\).
• Since \(OS = OT\), we can use the law of cosines in \(\triangle HOG\). Let \(a = OH\), \(b = OG\), and \(c = HG\). First, find \(OH=\frac{30}{\sin28^{\circ}}\approx63.89\) m and \(OG=\frac{TG}{\sin70^{\circ}}\). Since \(TG\approx155\) m, \(OG=\frac{155}{\sin70^{\circ}}\approx164.9\) m.
• The angle \(\angle HOG=180^{\circ}-(28^{\circ}+70^{\circ}) = 82^{\circ}\).
• By the law of cosines \(HG^{2}=OH^{2}+OG^{2}-2\cdot OH\cdot OG\cdot\cos\angle HOG\).
• \(OH^{2}\approx63.89^{2}=4082.93\), \(OG^{2}\approx164.9^{2}=27192.01\), \(2\cdot OH\cdot OG\cdot\cos82^{\circ}\approx2\times63.89\times164.9\times0.1392\approx2977.97\).
• \(HG^{2}=4082.93 + 27192.01-2977.97=28296.97\).
• \(HG=\sqrt{28296.97}\approx168\) m.