Question

1. jason is standing 8.7 km from town x and 11.5 km from town y. from where he stands, the angle between the two towns is 37°. a new hotel has just been built on the road connecting town x and town y, exactly halfway between the two towns. from where jason is standing, he sees that the angle of elevation to the top of the hotel is 1°. determine the height of the hotel, to the nearest tenth of a metre. 7t

Explanation:

Step1: Use the Law of Cosines to find the distance from Jason to the mid - point of the road between \(X\) and \(Y\)

Let \(a = 8.7\) km, \(b=11.5\) km, and \(\theta = 37^{\circ}\). The length of the road between \(X\) and \(Y\) is given by the Law of Cosines \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\).

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The distance from Jason to the mid - point \(d\) of the road between \(X\) and \(Y\) (using the formula for the length of the median in a triangle \(m=\frac{1}{2}\sqrt{2a^{2}+2b^{2}-c^{2}}\)) or we can use the Law of Cosines in another way. Let the mid - point be \(M\). We know that \(XM = YM=\frac{c}{2}\approx3.475\space\text{km}\).
Another way: Let's use the formula \(d^{2}=8.7^{2}+(3.475)^{2}-2\times8.7\times3.475\times\cos18.5^{\circ}\) (since the angle at \(X\) for the sub - triangle \(X - M - J\) can be found using the Law of Sines in \(\triangle XJY\): \(\frac{\sin\angle JXY}{11.5}=\frac{\sin37^{\circ}}{6.95}\), \(\sin\angle JXY=\frac{11.5\times\sin37^{\circ}}{6.95}\approx\frac{11.5\times0.6018}{6.95}\approx1\), \(\angle JXY = 90^{\circ}\), and then using the right - triangle properties. But a simpler approach:
We know that if we consider the right - triangle formed by the height \(h\) of the hotel, the line of sight from Jason to the top of the hotel and the horizontal distance from Jason to the hotel.
We use the formula \(h = d\times\sin1^{\circ}\). First, find \(d\) using the Law of Cosines in \(\triangle XJY\) (where \(JX = 8.7\), \(JY = 11.5\), \(\angle XJY=37^{\circ}\)) and then find the median.
Alternatively, using the formula for the length of the median \(m=\frac{1}{2}\sqrt{2a^{2}+2b^{2}-c^{2}}\) where \(a = 8.7\), \(b = 11.5\), \(c\) is found from \(c^{2}=a^{2}+b^{2}-2ab\cos C\) (\(C = 37^{\circ}\)).
After calculating \(c\approx6.95\), the median \(m=\frac{1}{2}\sqrt{2\times8.7^{2}+2\times11.5^{2}-6.95^{2}}\)

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Step2: Calculate the height of the hotel

We know that in the right - triangle (formed by the height of the hotel \(h\), the line of sight from Jason to the top of the hotel and the horizontal distance from Jason to the hotel), if the angle of elevation is \(\alpha=1^{\circ}\) and the adjacent side (horizontal distance from Jason to the hotel) is \(d = 9.585\space\text{km}=9585\space\text{m}\)
Using the formula \(h=d\times\sin\alpha\) (since \(\sin\alpha=\frac{h}{d}\), \(h = d\sin\alpha\))

$$h = 9585\times\sin1^{\circ}$$

Since \(\sin1^{\circ}\approx0.01745\)

$$h=9585\times0.01745\approx167.3\space\text{m}$$

Answer:

\(167.3\space\text{m}\)