(a) The triangle ABC has sides AB = 17m, BC = 12m and AC = 10m. Calculate the : (i) largest angle of the triangle ; (ii) area of the triangle. (b) From a po...
Question 1 Report
(a) The triangle ABC has sides AB = 17m, BC = 12m and AC = 10m. Calculate the :
(i) largest angle of the triangle ; (ii) area of the triangle.
(b) From a point T on a horizontal ground, the angle of elevation of the top R of a tower RS, 38m high is 63°. Calculate, correct to the nearest metre, the distance between T and S.
Answer Details
(a)(i) Largest angle. The largest angle faces the longest side \(AB=17\) m, so it is angle \(C\), between sides \(CA=10\) and \(CB=12\). By the cosine rule:
\[\cos C=\frac{CA^{2}+CB^{2}-AB^{2}}{2\,(CA)(CB)}=\frac{10^{2}+12^{2}-17^{2}}{2(10)(12)}=\frac{100+144-289}{240}=\frac{-45}{240}=-0.1875\]\(C=\cos^{-1}(-0.1875)\approx 100.8^{\circ}\).
Largest angle \(\approx 100.8^{\circ}\).
(ii) Area.
\[\text{Area}=\frac{1}{2}(CA)(CB)\sin C=\frac{1}{2}(10)(12)\sin 100.8^{\circ}=60\times0.9823\approx 58.9\text{ m}^{2}\](b) The tower \(RS\) is vertical with \(RS=38\) m; from \(T\) on the ground the elevation of the top \(R\) is \(63^{\circ}\). \(TS\) is the horizontal distance sought.
\[\tan 63^{\circ}=\frac{RS}{TS}=\frac{38}{TS}\Rightarrow TS=\frac{38}{\tan 63^{\circ}}=\frac{38}{1.9626}\approx 19.4\text{ m}\]Distance \(TS\approx 19\) m (nearest metre).
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