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.

As Leming pointed out, the problem statement, “The cliff is also at a bearing of W(est) 30 degrees N(orth) from the boat.”, does seem to be grammatically ambigous.

Of the interpretation chosen by Charlie, the statement should probably be“The cliff is also at a bearing of 30 degrees N(orth) of W(est) from the boat.”.

Determining by ratio the equation for the value of one side of the triangle to the other unknown side, then by substitution into the equation of the law of cosines, the value of the side is determined.

For the side that extends from the cliff to the original position of the boat, this value turns out to be approximately 66.106343.
As the height of the cliff can be given as the distance of the boat divided by the tangent function of the angle of elevation from the boat to the height of the cliff, the height can be calculated.as approximately 30.825894 meters.

For the second interpretation, where the statement may be “ The cliff is also at a bearing of 30 degrees W(est) of N(orth) from the boat.”, the same method may be used, but a different value is the result.

Where b is the distance from the side that extends from the cliff to the original position of the boat…
100² = (b tan 25/ tan 15)² + b² – 2(b tan 25 / tan 15)b cos 30

b^2 = 100²/((tan 25/tan 15)² + 1 – 2*cos 30*tan 25/tan 15)

b = (approx.) 57.598296

Subsituting this value for b into the equation for the height of the cliff ( cliff height = b tan 25), the height is calculated as approximately 26.858527 meters.

Posted by Dej Mar on 2006-05-21 08:44:00