# Thread: Trigonometric Related Rate Question

1. ## Trigonometric Related Rate Question

An illuminated billboard 10m tall stands on top of a cliff 12m high. How far from the foot of a cliff should a man stand in order for the sun to subtend the largest possible angle at his eyes (which are 2m above the ground)? How large is the maximum angle?

2. This is an optimization problem, not related rates.

Anyway, we have to find the value of x in order to maximize the angle.

Let $tan({\alpha})=\frac{22}{x}$

$tan({\beta})=\frac{12}{x}$

So, that ${\theta}={\alpha}-{\beta}=tan^{-1}(\frac{22}{x})-tan^{-1}(\frac{12}{x})$

$\frac{d\theta}{dx}=\frac{12}{x^{2}+144}-\frac{22}{x^{2}+484}$

Set to 0 and solve for $x=2\sqrt{66}\approx{16.25}$

The observer should stand 16.25 m from the cliff.

Check me out. Easy to err.

3. How did you come up with the change in theta with respect to time formula?

4. Could someone provide some insight as to how that formula was created?

5. It's the only thing I'm having trouble understanding.