Question 1118371: P and Q are two points 16m apart in a hill, which has a slope of 5 degrees. From the base Q, an engineer measured the angle of elevation to the top of a rock further up the hill to be 35 degrees, while from P, the angle of elevation was found to be 54 degrees.
find the perpendicular height of the rock above P.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Let point A be the top of the rock and B be the base of the rock. Extend AB to C, where C is the same elevation as P (so ACP is a right triangle).
When the problem asks for "the perpendicular height of the rock above P", I interpret that as the length of AC.
Angle APB is 54 degrees, and angle BPC is 5 degrees, so angle APC is 59 degrees.
Then 
To find the length of AP we can use the law of sines in triangle APQ. Since angle APB is 54 degrees and angle AQP is 35 degrees, angle PAQ is 19 degrees. Then
And then
which to 3 decimal places is 24.162.
There are other ways to reach this answer. For example, you could extend AB to D, where D is at the same elevation as Q; i.e., QD is perpendicular to AD. Then AD is AQ times sin(40); AQ can be calculated using the law of sines on triangle AQP, and then AC is AD minus CD, which is AD minus 16*sin(5).
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Note a response from another tutor to the same question but on a different thread interprets the 35 degree and 54 degree angles of elevation to be measured from the horizontal, whereas my interpretation above measures those angles of elevation from the 5 degree slope, making the angles with the horizontal 40 degrees and 59 degrees.
I suspect the other tutor's interpretation is correct....
You can use the calculations I described above with the new angles to get the answer to the problem.
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