Trig word problem - angle of depression.

Here is the word problem.

A bird on top of a vertical pole looked South at an angle of depression of 35 Degrees, saw Icing, Maya's bunny on the ground. The bird turned and looked East, and at an angle of depression of 40 degrees saw, Notowada, Maya's turtle on the ground. The distance between Icing and Ntwoada is 60m. Find the height of the pole given this information.

I have diagrammed this out and see it perfectly I just don't know how to calculate this given the information. (Doh)

I can't see how i can use Sin Rule, Cos rule or anything...Can someone show me the process of doing this. My diagram looks like a right triangular pyramid.

Please **clearly explain your work** so I can understand **why** you are performing certain operations.

Thank you so much in advance! (Nerd)

Sincerely,

Raymond

adjacent/opposite ?? h/x?

Quote:

Originally Posted by

**skeeter** $\displaystyle \tan(35) = \dfrac{h}{x} \implies x = \dfrac{h}{\tan(35)}$

$\displaystyle \tan(40) = \dfrac{h}{y} \implies y = \dfrac{h}{\tan(40)}$

$\displaystyle x^2 + y^2 = 60^2$

$\displaystyle \left[\dfrac{h}{\tan(35)}\right]^2 + \left[\dfrac{h}{\tan(40)}\right]^2 = 60^2$

$\displaystyle \dfrac{h^2}{\tan^2(35)} + \dfrac{h^2}{\tan^2(40)} = 60^2$

$\displaystyle h^2 \left[\dfrac{1}{\tan^2(35)} + \dfrac{1}{\tan^2(40)}\right] = 60^2$

$\displaystyle h^2[\cot^2(35) + \cot^2(40)] = 60^2$

$\displaystyle h^2 = \dfrac{60^2}{\cot^2(35)+\cot^2(40)}$

$\displaystyle h = \dfrac{60}{\sqrt{\cot^2(35) + \cot^2(40)}}$

Why: $\displaystyle \tan(35) = \dfrac{h}{x} $ ?

That's adjacent/opposite. I thought tan* A* was opposite/adjacent.

Did you mean to say $\displaystyle \tan(55) = \dfrac{h}{x} $ and $\displaystyle \tan(50) = \dfrac{h}{y} $ ??