Question

mcr3u
trigonometry assignment t /29

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 distance from Jason to the mid - point be \(d\). We can use the formula for the length of the median in a triangle. Another way is to use the Law of Cosines in the sub - triangle. But since we know \(XY\approx6.94\space km\), the mid - point divides \(XY\) into two equal parts of length \(l=\frac{XY}{2}\approx3.47\space km\).
We can also use the formula \(d^{2}=\frac{2a^{2}+2b^{2}-c^{2}}{4}\) (formula for the length of the median in a triangle, where \(a = 8.7\), \(b = 11.5\), \(c = XY\approx6.94\))

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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 \(d\approx9.59\space km=9590\space m\)

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

$$h=9590\times0.017455$$

$$h\approx167.5\space m$$

Answer:

The height of the hotel is approximately \(167.5\space m\)