A surveying instrument aimed at a location can “shoot” the distance to the location, giving the surveyor a measurement. A surveyor used such an instrument to record the distance to a point on a tree 60 m from his position. After rotating his surveying instrument 57° to the left, he measured the distance from his same position to a fence post 35 m away. a.) Draw the diagram and label the tree as point T, the surveyor as point S, and the fence post as point F.b.) Determine the distance between the point on the tree and the fence post. (Show the appropriate formula, substitutions, and work. Give the distance to the nearest tenth of a meter.)c.) Use the Law of Sines to find the measure of T. (Show the appropriate formula, substitutions, and work. Give the measure of T to the nearest degree.)d.) Find mFPlease Help Me :(

Accepted Solution

a. See attached for diagram. The basic idea is that you put a point S down (for the surveyor) and draw a segment of length 60m that goes from S to T (the tree). Then from S you draw a segment 35m to F such that there is an angle of 57 degrees between ST and SF making sure that SF is to the left of ST if you are looking at T and F from point S as the surveyor would be. Then connect FT.

b. Here we are looking for the distance between the tree (T) and the fence post (F). So we want the length of FT. We have the lengths of two sides and an angle and want the other side. The side we seek is opposite the angle whose measure we know. In this case we can use the law of cosines.
The law of cosines is: [tex] c^{2} = a^{2} + b^{2} -2abcosC[/tex]

I have labeled the side we want FT this c in the diagram. It is opposite the 57 degree angle which we will call C. (Little c is the side and big C is the angle). We call the other two sides of the triangle a and b. It doesn't matter which is which. So we have the following:
a= FS = 35
b= ST = 60
c = FT
angle C = angle S = 57 degrees
We plug the above values into the law of cosines and solve for c as follows:
[tex]c^{2} = a^{2} + b^{2} -2abcosC[/tex]
[tex]c^{2} = 35^{2} + 60^{2} -2(35)(60)cos57[/tex]
[tex]c^{2} = 1225+ 3600-(4200)(.544639)[/tex]
[tex] c^{2} =2537.51606[/tex]
You are asked to round to the nearest tenth (one decimal place) which gives the answer 50.4 meters

c. You are asked to find the measure of angle T. Further you are asked to do so using the Law of Sines. The law of sines is: [tex] \frac{a}{sinA} = \frac{B}{sinB} = \frac{c}{sinC} [/tex]. Even though there are three fractions set equal to eahc other, when we use the Law of Sines to find an angle (or a side) we need only use two fractions in most cases. Using the labels we used in part B we have the following:
A = T = the angle we are looking for
a = FS = 35 meters
C = S = 57 degrees
c = FT = 50.4 meters
We substitute these into the Law of Sines and solve for A as follows:
[tex] \frac{a}{sinA} = \frac{c}{sinC} [/tex]
[tex] \frac{35}{sinA} = \frac{50.4}{sin57} [/tex]
[tex] \frac{35}{sinA} = \frac{50.4}{.8386705} [/tex]
Here we use the arc sin (on a calculator it looks like sin to the negative 1 but it's called arcsin)
[tex] sin^{-1}(.58241011662)=A [/tex]
You are asked to round this to the nearest degree which means to the nearest whole number. The answer is 36 degrees.

d. You are asked for the measure of angle F. Recall that the sum of the measures of the angles in a triangle is 180 degrees. We know that S = 57 degrees and T = 36 degrees (from part c). So to find the missing angle we add these and subtract their sum from 180.
[tex]F = 180-(36+57)[/tex]
The measure of angle F is 87 degrees

NOTE: Angles can be measure in degrees and in radians. This problem uses degrees so it is important that your calculator is set to degrees when doing this problem.