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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The distance from the mid - point of \(XY\) to Jason: Let \(d\) be the distance from Jason to the mid - point of \(XY\). Using the formula for the length of the median in a triangle \(m=\frac{1}{2}\sqrt{2a^{2}+2b^{2}-c^{2}}\). Here \(a = 8.7\), \(b = 11.5\), \(c\approx6.94\)

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Step2: Use the tangent function to find the height of the hotel

We know that \(\tan\theta=\frac{h}{d}\), where \(\theta = 1^{\circ}\) and \(d\approx9.6\space km=9600\space m\)

$$h = d\times\tan\theta$$

$$h=9600\times\tan(1^{\circ})$$

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

$$h=9600\times0.017455 = 167.568\approx167.6\space m$$

Answer:

The height of the hotel is approximately \(167.6\space m\)