Depression word problem

Question

A pilot maintains an altitude of 25,000 ft. over level ground. The pilot observes a crater on the ground at an angle of depression of 5 degrees. If the plane continues for 16 more miles, what will be the angle of depression to the crater?

Answer 1

Meg:  I will try to explain this to you.  If the plane is at an altitude of 25,000 ft., that would be an adjacent dimension.  The angle of depression is 5 degrees.  So, the angle from the plane to the 25,000 ft. side would be 85 degrees.  Since we are dealing with a right triangle with an adjacent side of the altitude of 25,000 ft. Since tan (85) = opposite / adjacent, you have the adjacent dimension = 25,000 ft.  The ratio of opposite / adjacent of tan (85) = 11.43.  so tan (85) = opposite / adjacent (or opposite / 25,000 ft.).  When you multiply both sides times 25,000, you end up with a horizontal dimension of 285,751 feet for the distance from where the plane is, to the crater.  Since the plane is traveling 16 more miles, multiply 16 x 5,280 ft. per mile = 84,480 ft. traveled in those 16 miles.  Subtract 84,480 from the 285,751 ft. = 201,271 ft. in the horizontal dimension from where the plane is over the ground, to the crater.  Again, since tan = opp./adj., then the ratio of the new distance to the crater divided by 25,000 ft. = 8.05.  Now the arc tan of 8.05 = 82.92.  That means that the new angle of depression to the crater is 90 minus 82.92 = 7.08 degrees.  If we were sitting at the same table, I could draw it out, but I'm in Amarillo, Texas and this will have to do.  I hope we are correct on this.  If you check the answer, we should come up with a satisfactory check.  Good luck!!