中三數學問題 急

Firstly, draw out this thing. We know that:

A and B are above ground.
C, D and E are on the ground and on the same line.
A is above D
B is above E
Balloon is tied to C. Rope is taut. So AC and BC are of the same length.

So how are C, D and E lined up? Is it from left to right C,D,E?

---------------------B (100m)
-----------A(60m)-----|
----------------|---------|
-----C--------D--------E

Remember that the ballon is tied by a rope to C. This arrangement is not possible! Because the rope length between AC and BC are different!!!

Try left to right D,C,E:

---------------------B (100m)
---A(60m)-------------|
------|-------------------|
-----D--------C--------E

Now AC and BC can be the same--just that on ground DC and EC are of different lengths.

OK. Now you get an isosceles triangle ABC, with BC = AC. Also, you know that angle of elevation of A from C (that is, angle ACD) = 30 degrees. Therefore,

sin (30 deg) = AD / AC = 60m / AC, AC = 120m
Isosceles triangle: BC = AC, BC = 120m (length of the rope!)

Finally, angle of depression at B... if you draw a horizontal line at B, this would be the ACUTE angle between the horizontal line and line BC. But, using "alternate angles, parallel lines" that would = angle BCE

Therefore, angle of depression:
sin(angle BCE) = BE / BC = 100/120
angle BCE = 56.44 degrees.

Therefore, angle of depression = 56.44 degrees