Trig
posted by Nikki on .
From an observation tower that overlooks a runway, the angles of depression of point A, on one side of the runway, and point B, on the oppisite side of the runway are 6 degrees and 13 degreess respectivel. The points and the tower are in the same vertical plane and the distance from A to B is 1.1km. Determine the height of the tower. Please give me a diagram and step by step instructions. This question has me stumped!!!

Let the base of the tower be C, and the top of the tower be T. So, the height h=CT, and we have
AB=1, so AC=x and BC=1x
h/x = tan6°
h/(1x) = tan13°
so, solve for x:
xtan6° = (1x)tan13°
x = 0.687
h = 0.0722km = 72.2m 
Can you show me the steps you used to isolate x?

x is the distance from A to the base of the tower.
Since AB=1, the distance to B is 1x.
Did you not understand the diagram points? 
oops. Sorry. I see that AB=1.1
So, adjust the equations and solve again. 
Yes I understand the diagram points and can see how you came up with this now but am having trouble coming up with an answer for
xtan6°=(1.1x)tan13°
I know this is basic but I am having trouble isolating the x!
I am getting
x=tan13°tan6°/1.1 which is giving me a negative answer. 
plugging in the trig values, you have
.1051x = .2309(1.1x)
Have you forgotten your algebra I?
.1051x = .2540  .2309x
.3360x = .2540
x = 0.756 
Lol! Apparently I have been working on this question too long! Thanks!