Posted by **Nikki** on Monday, October 15, 2012 at 3:27pm.

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**, Monday, October 15, 2012 at 3:43pm
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=1-x

h/x = tan6°

h/(1-x) = tan13°

so, solve for x:

xtan6° = (1-x)tan13°

x = 0.687

h = 0.0722km = 72.2m

- Trig -
**Nikki**, Monday, October 15, 2012 at 3:54pm
Can you show me the steps you used to isolate x?

- Trig -
**Steve**, Monday, October 15, 2012 at 3:59pm
x is the distance from A to the base of the tower.

Since AB=1, the distance to B is 1-x.

Did you not understand the diagram points?

- Trig -- oops -
**Steve**, Monday, October 15, 2012 at 4:00pm
oops. Sorry. I see that AB=1.1

So, adjust the equations and solve again.

- Trig -
**Nikki**, Monday, October 15, 2012 at 4:18pm
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°=(1.1-x)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**, Monday, October 15, 2012 at 4:22pm
plugging in the trig values, you have

.1051x = .2309(1.1-x)

Have you forgotten your algebra I?

.1051x = .2540 - .2309x

.3360x = .2540

x = 0.756

- Trig -
**Nikki**, Monday, October 15, 2012 at 4:29pm
Lol! Apparently I have been working on this question too long! Thanks!

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