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!!!

Trig 
Steve,
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 
Trig 
Nikki,
Can you show me the steps you used to isolate x?

Trig 
Steve,
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? 
Trig  oops 
Steve,
oops. Sorry. I see that AB=1.1
So, adjust the equations and solve again. 
Trig 
Nikki,
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. 
Trig 
Steve,
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 
Trig 
Nikki,
Lol! Apparently I have been working on this question too long! Thanks!