Question

1. a large monument with height of 2351 m is located exactly halfway between points k and l. michael is standing at point m and finds that he is 14.2 km from l and 17.1 km from k. he also measures the angle between k and l to be 47°. determine the angle of elevation to the top of the monument, measured from point m.

Explanation:

Step1: Use the Law of Cosines to find the distance from \(M\) to the base of the monument

Since the monument is halfway between \(K\) and \(L\), let the distance from \(M\) to the base of the monument be \(d\).
First, convert the height of the monument \(h = 2351m=2.351km\)
By the Law of Cosines in \(\triangle KML\): \(KL^{2}=KM^{2}+LM^{2}-2(KM)(LM)\cos\angle KML\)
But we don't need \(KL\) directly. Instead, since the monument is at the mid - point, we can use another approach. Let's assume the distance from \(M\) to the base of the monument is \(d\).
We can also use the formula for the area of a triangle \(S=\frac{1}{2}ab\sin C\). But a better way is to use the Law of Cosines in the following form:
Let \(a = 14.2\), \(b = 17.1\), \(C = 47^{\circ}\)
The distance from \(M\) to the base of the monument \(d\) can be found using the formula for the length of the median in a triangle. But an alternative is to use the formula \(d=\sqrt{\frac{2a^{2}+2b^{2}-c^{2}}{4}}\) (not necessary here).
Another approach: We know that \(\tan x=\frac{h}{d}\). First, find \(d\) using the Law of Cosines in \(\triangle KML\)

$$d=\sqrt{14.2^{2}+17.1^{2}-2\times14.2\times17.1\times\cos47^{\circ}}$$

$$d=\sqrt{201.64 + 292.41-484.44\times0.682}$$

$$d=\sqrt{494.05-330.4}$$

$$d=\sqrt{163.65}\approx 12.8$$

Step2: Calculate the angle of elevation \(x\)

We know that \(\tan x=\frac{h}{d}\), where \(h = 2.351km\) and \(d\approx12.8km\)
\(\tan x=\frac{2.351}{12.8}\approx0.184\)
\(x=\arctan(0.184)\approx 10.4^{\circ}\)

Answer:

The angle of elevation is approximately \(10.4^{\circ}\)