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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 [itex]45+\theta[/itex] and [itex]45-\theta[/itex] respectively. Show that [itex]BC=2h.tan2\theta[/itex]

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

## The Attempt at a Solution

[tex]\angle ADB=90-(45+\theta)=45-\theta[/tex]

Therefore, [tex]tan(45-\theta)=\frac{AB}{h}[/tex]

Then, [tex]AB=h.tan(45-\theta)[/tex]

[tex]\angle CDA=90-(45-\theta)=45+\theta[/tex]

Therefore, [tex]tan(45+\theta)=\frac{AB+CD}{h}[/tex]

Then, [tex]AB=h.tan(45+\theta)-CD[/tex]

So I equated [tex]h.tan(45-\theta)=h.tan(45+\theta)-CD[/tex]

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

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

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