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 between town X and town Y

The Law of Cosines formula is \(c^{2}=a^{2}+b^{2}-2ab\cos C\). Let \(a = 8.7\), \(b=11.5\), and \(C = 37^{\circ}\).

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Step2: Find the distance from Jason to the mid - point of \(XY\)

Since the hotel is at the mid - point of \(XY\), let the mid - point be \(M\). Then \(XM=\frac{XY}{2}\approx\frac{6.946}{2}=3.473\space km\)

Step3: Use the tangent function to find the height of the hotel

We know that \(\tan\theta=\frac{h}{d}\), where \(\theta = 1^{\circ}\) and \(d\) is the distance from Jason to the mid - point of \(XY\) (\(d = 3.473\space km=3473\space m\))

$$h = d\times\tan(1^{\circ})$$

$$h=3473\times\tan(1^{\circ})$$

Since \(\tan(1^{\circ})\approx0.017455\)

$$h = 3473\times0.017455\approx60.7\space m$$

Answer:

The height of the hotel is approximately \(60.7\space m\)