The point \(T\) is the peak of a mountain and the point \(O\) is directly below the mountain's peak. The point \(Y\) is due east of \(O\) and the angle of elevation of \(T\) from \(Y\) is 60°. The point \(F\) is 4 km south-west of \(Y\). The points \(O, Y\) and \(F\) are on level ground. The angle of elevation of \(T\) from \(F\) is 45°.
- Let the height of the mountain be \(h\).
- Show that \(O Y=\dfrac{h}{\sqrt{3}}\). (1 mark)
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- Hence, or otherwise, find the value of \(h\), correct to 2 decimal places. (3 marks)
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- Find the bearing of point \(O\) from point \(F\), correct to the nearest degree. (3 marks)
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a. \(\text{In}\ \triangle TOY:\)
| \(\tan 60^{\circ}\) | \(=\dfrac{h}{OY}\) | |
| \(OY\) | \(=\dfrac{h}{\tan 60^{\circ}}\) | \(=\dfrac{h}{\sqrt{3}}\) |
b. \(\text{In}\ \triangle TOF:\)
\(\tan 45^{\circ}=\dfrac{h}{OF} \ \Rightarrow \ OF=h\)
\(\text{Since \(Y\) is due east of \(O\) and \(F\) is south-west of \(Y\):}\)
\(\angle OYF =45^{\circ}\)
\(\text{Using cosine rule in} \ \triangle OYF:\)
| \(OY^2+YF^2-2 \times OY \times YF\, \cos 45^{\circ}\) | \(=OF^2\) |
| \(\left(\dfrac{h}{\sqrt{3}}\right)^2+4^2-2 \times \dfrac{h}{\sqrt{3}} \times 4 \times \dfrac{1}{\sqrt{2}}\) | \(=h^2\) |
| \(\dfrac{h^2}{3}+16-\dfrac{8}{\sqrt{6}} h\) | \(=h^2\) |
| \(\dfrac{2}{3} h^2+\dfrac{8}{\sqrt{6}} h-16\) | \(=0\) |
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\(h\) |
\(=\dfrac{\dfrac{-8}{\sqrt{6}}+\sqrt{\frac{64}{6}+4 \times \frac{2}{3} \times 16}}{2 \times \frac{2}{3}}\ \ \ (h>0)\) |
| \(=3.0277 \ldots\) | |
| \(=3.03 \ \text{km (to 2 d.p.)}\) |
c. \(\text {Using sine rule in} \ \ \triangle OYF:\)
| \(\dfrac{\sin\angle FOY}{4}\) | \(=\dfrac{\sin 45^{\circ}}{3.03}\) | |
| \(\sin \angle F O Y\) | \(=\dfrac{4 \times \sin 45^{\circ}}{3.03}=0.93347\) | |
| \(\angle FOY\) | \(=180-\sin^{-1}(0.93347)=180-68.98 \ldots = 111^{\circ} \ \text{(angle is obtuse)}\) |
\(\angle FO\text{S}^{\prime}=111-90=21^{\circ}\)
\(\angle \text{N}^{\prime}FO=21^{\circ}(\text {alternate})\)
\(\therefore \ \text {Bearing of \(O\) from \(F\)}=021^{\circ} \)