A boat observes the top of a cliff to be at an angle of elevation of 25 degrees. The cliff is also at a bearing of W 30 degrees N from the boat. The boat then travels 100 meters in a straight path. The boat then observes the cliff to be due north, with an angle of elevation of 15 degrees. Using this sufficient information, calculate the height of the cliff.
The line from the old position of the boat to the sea-level point under the cliff forms one side of a triangle with the line from the new position of the boat to that same sub-cliff point and the line between the old and new positions of the boat. This triangle is the base of a tetrahedron with the cliff base and cliff top as a vertical edge.
Call the initial location of the boat A, its new position as B and the base of the cliff as C, so you have triangle ABC. The measure of the side opposite A will be called a, etc., so that the 100 m distance the boat traveled is c. Call the height of the cliff x.
Side a extends from the new position of the boat to the base of the cliff, and a=x/tan 15. Side b extends from the old position of the boat to the base of the cliff, and b=x/tan 25.
As the cliff was 30 degrees north of west seen from the first position of the boat, that position of the boat was 30 degrees south of east seen from the cliff. The new position of the boat was directly south of the cliff, so angle C is 60 degrees.
By the law of cosines:
100^2 = a^2 + b^2 - 2ab cos 60, or
100^2 = a^2 + b^2 - ab
a/b = tan 25 / tan 15
a = b tan 25 / tan 15
100^2 = b^2 (tan 25 / tan 15)^2 + b^2 - b^2 tan 25 / tan 15
100^2 = b^2 ((tan25/tan15)^2 + 1 - tan25/tan15)
b^2 = 100^2 / ((tan25/tan15)^2 + 1 - tan25/tan15)
b = 66.1063429262889
a = b tan25 / tan 15 = 115.043802444275
x = b tan25 = 30.825893959149 meters
So the answer is about 30.83 meters, and even that probably exceeds the accuracy of the measurements going into it.
Posted by Charlie
on 2006-05-20 12:55:55