# Trigo/Geom. problem

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## Homework Statement

D is the top of a tower h metres high. A is the foot of the tower. The angles of depression of the points B and C from D are $45+\theta$ and $45-\theta$ respectively. Show that $BC=2h.tan2\theta$

http://img43.imageshack.us/img43/9195/trigo.th.jpg [Broken]

## The Attempt at a Solution

$$\angle ADB=90-(45+\theta)=45-\theta$$

Therefore, $$tan(45-\theta)=\frac{AB}{h}$$

Then, $$AB=h.tan(45-\theta)$$

$$\angle CDA=90-(45-\theta)=45+\theta$$

Therefore, $$tan(45+\theta)=\frac{AB+CD}{h}$$

Then, $$AB=h.tan(45+\theta)-CD$$

So I equated $$h.tan(45-\theta)=h.tan(45+\theta)-CD$$

And finally, through expanding using the tangent sum formula and 'compressing' back into the double tangent formula, I resulted in $BC=2h.tan2\theta$

My problem is that this way to do it took far too long. There must be an easier way but I cannot seem to find it. I need your help Last edited by a moderator:

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HallsofIvy
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Are B and D assumed to be in a straight line with A?

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Yes. The diagram is simply a right-angled triangle with another line going from D to a point B on the opposite line of the triangle.

EDIT: Re-reading your question Hallsofivy, I think I may need to post a diagram to clear things up.

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Bump. To re-iterate: I doubt such a question given to our class to answer would require over a page of algebraic manipulations, and this is why I'm thinking there is a much easier way to use that $2\theta$ in the diagram directly, and not obtain it through the tangent double angle formula.

HallsofIvy
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If the angle of depression to B is 45+$\theta$, then the angle DB makes with DA is 90- (45+ $\theta-$)= 45-$\theta$ and the angle DC makes with DA is 90- (45- $\theta$= 45+$\theta$. CA/h= tan(45+\theta) and BA/h= tan(45-$\theta$) so CB= CA- BA= h(tan(45+$\theta$)- tan(45-$\theta$)). Now use the identities for tan(a+ b) and tan(a-b).

tiny-tim
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Hi Mentallic! (have a theta: θ )
… Then, $$AB=h.tan(45-\theta)$$

My problem is that this way to do it took far too long. There must be an easier way but I cannot seem to find it. I need your help yeees having got your AB = h.tan(45º - θ),

you never actually said AC = h.tan(45º + θ) …

which gives you the answer much quicker! Homework Helper
Oh thats much quicker! Thanks to both of you Still curious though, can this problem be solved without the trig identities? Such as using the $2\theta$ angle and finding its tangent etc.?