Question

mcr3u trigonometry assignment t /29

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.
7t
5

Explanation:

Step1: Use the Law of Cosines

The Law of Cosines formula is \(c^{2}=a^{2}+b^{2}-2ab\cos C\). Let \(a = 14.2\), \(b=17.1\), \(C = 47^{\circ}\). First, find the distance from the monument to point \(M\) (let's call it \(d\)). Since the monument is halfway between \(K\) and \(L\), we can use the Law of Cosines in \(\triangle KML\) to find the side opposite the \(47^{\circ}\) angle. But actually, we can also use the formula for the distance from \(M\) to the mid - point (let's assume the distance from \(M\) to the base of the monument is \(d\)). Using the formula \(d=\sqrt{\frac{2a^{2}+2b^{2}-c^{2}}{4}}\) (derived from the median formula in a triangle \(c^{2}=a^{2}+b^{2}-2ab\cos C\), and for a median \(m\), \(4m^{2}=2a^{2}+2b^{2}-c^{2}\)). But another way: using the Law of Cosines in \(\triangle KML\), \(KL^{2}=14.2^{2}+17.1^{2}-2\times14.2\times17.1\times\cos47^{\circ}\).
\(KL^{2}=14.2^{2}+17.1^{2}-2\times14.2\times17.1\times0.682\)
\(KL^{2}=201.64 + 292.41-2\times14.2\times17.1\times0.682\)
\(KL^{2}=494.05-(484.44\times0.682)\)
\(KL^{2}=494.05 - 330.49\)
\(KL^{2}=163.56\), \(KL\approx12.8\) km. The distance from \(M\) to the base of the monument \(d=\frac{KL}{2}\approx6.4\) km \( = 6400\) m.

Step2: Use the tangent function

The tangent of the angle of elevation \(x\) is given by \(\tan x=\frac{\text{opposite}}{\text{adjacent}}\). Here, the opposite side is the height of the monument (\(h = 2351\) m) and the adjacent side is the distance from \(M\) to the base of the monument (\(d = 6400\) m).
\(\tan x=\frac{2351}{6400}\)
\(x=\tan^{- 1}(\frac{2351}{6400})\)
\(x=\tan^{-1}(0.3673)\)

Answer:

\(x\approx20.2^{\circ}\)