SOLUTION: 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

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Question 1118349: 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 josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
I can only try to first describe a summary of the drawing based on your description.

Points P and Q are on the sloping line and Q is the lowest labeled point. The top of the rock is point R. Point R is higher than P and Q. PQ as given is 16 meters. The triangle formed has interior angles 30 degrees at Q, 59 degrees at P; and 91 degrees at R. Note that angle at P is made of 54%2B5=59 degrees. Angle at R is 180-59-30=91 degrees.

Question asks for the distance how much higher, vertically only, is R than P.

Draw a horizontal ray through P; and let X be directly below R and on this ray.

You want to find PR and then use it to find PX.

Law of Sines:
PR%2Fsin%2830%29=16%2Fsin%2891%29

PR=%2816%2Asin%2830%29%29%2Fsin%2891%29

Next,
PX%2FPR=sin%2854%29
PX=PR%2Asin%2854%29

highlight%28PX=%28%2816%2Asin%2830%29%29%2Fsin%2891%29%29sin%2854%29%29


(possible error in interpretation)