Solve Geom/Trigo Problem: Show BC=2h.tan2\theta

[Mentallic]

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

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

 

[HallsofIvy]

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

 

[Mentallic]

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.

 

[Mentallic]

http://img43.imageshack.us/img43/9195/trigo.th.jpg

 

[Mentallic]

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]

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]

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!

 

[Mentallic]

Oh that's 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.?