Can you help me set this up? I am told that a weather balloon is directly west of 2 stations which are 10mi apart. The elevations of the balloon from the 2 stations are 17.6deg and 78.2deg. How high is the balloon?
This one was also posted. And another will be posted by the time you see this.
Here's a copy of the first:
angles of elevation and balloon
Question: Problem 24 is giving me some trouble. The problem said: A weather balloon is directly west of two observing stations that are 10mi. apart. The angles of elevation of the balloon from the two stations are 17.6 degree and 78.2 degree. How high is the balloon?
What I know: All the angle of depression and elevation problems are hard to draw and solve. I know that you have to draw it then set-up one of the trig functions to solve the height of the balloon but my problem is setting up each problem. If I can set it up then I could solve it.
Need: I need help setting up angle of depression and elevation problems.
See also my answer on the sand dune problem, posted at the access page for Mth 164. This problem is identical in structure. The two explanations are worded a little differently, so one might make more sense than the other.
A triangle from the first site to the balloon, then straight down to the ground, then back to the site will have base angle 17.6 deg. The distance along the ground is adjacent to this angle and the altitude of the balloon is opposite the angle. If the distance along the ground is x and the height is h, tan(theta) = opp / adj gives us
tan(17.6 deg) = h / x.
A triangle constructed in a similar manner from the second observation point will have base angle 78.2 deg and its opposite side will also be the altitude h of the balloon. To get the greater angle of elevation, this point must be closer to the balloon than the first, and the given information tells us that it is in fact 10 miles closer. The distance from the first site along the ground is x, so the adjacent side of this triangle must be x - 10 mi. The definition of the tangent therefore gives us
tan(78.2 deg) = h / ( x - 10 mi).
We now have two equations for h and x. We can solve the first for x and substitute this quantity for x in the second equation, ending up with an equation in only the one variable h.