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

• Mentallic
In summary, the problem involves a tower D of height h, with foot A and points B and C at angles of depression 45+\theta and 45-\theta respectively. To show that BC=2h.tan2\theta, one can use the trig identities for tangent of sum and difference of angles, or simply use the fact that AB=h.tan(45-\theta) and AC=h.tan(45+\theta) to arrive at the same conclusion.

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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

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

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.

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).

Hi Mentallic!

(have a theta: θ )
Mentallic said:
… 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!

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.?

## What is "Solve Geom/Trigo Problem: Show BC=2h.tan2\theta" all about?

This problem involves using geometry and trigonometry to show that the length of side BC in a right triangle is equal to twice the height of the triangle multiplied by the tangent of twice the angle theta.

## What do BC, h, and theta represent in this problem?

BC represents the length of side BC in the right triangle, h represents the height of the triangle, and theta represents one of the angles in the triangle.

## How do I approach solving this problem?

First, draw a right triangle and label the sides and angles. Then, use the Pythagorean theorem to relate the sides of the triangle. Next, use trigonometric ratios (specifically the tangent function) to relate the sides and angles. Finally, substitute the given values and solve for BC.

## What are the key concepts or formulas needed to solve this problem?

The Pythagorean theorem (a^2 + b^2 = c^2), trigonometric ratios (sin, cos, tan), and the double angle formula for tangent (tan2x = 2tanx / 1-tan^2x).

## How can I check if my solution is correct?

You can check your solution by plugging in the given values for h and theta and solving for BC. Then, use the Pythagorean theorem to check if the sides of the triangle are related correctly. You can also use a calculator to verify your answer.