# Calculus Problem

How should this problem be solved? I am not sure where to start.

Thanks!

Note by Asher Joy
4 years, 1 month ago

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Let the angle from the top of the blackboard to the ground be $$\alpha_1$$, and the angle from the bottom of the blackboard to the ground be $$\alpha_2$$.

We can see, by the definition of tangent, that $$\tan \alpha_2 = \dfrac{3}{x}$$ and $$\tan \alpha_1 = \dfrac{15}{x}$$.

Thus, $$\alpha_2=\tan^{-1}\dfrac{3}{x}$$ and $$\alpha_1 = \tan^{-1}\dfrac{15}{x}$$.

Therefore, $$\alpha=\alpha_1-\alpha_2 =\boxed{\tan^{-1}\dfrac{15}{x}- \tan^{-1}\dfrac{3}{x}}$$.

The above can also be rewritten to $$\boxed{\alpha=\cot^{-1}\dfrac{x}{15}- \cot^{-1}\dfrac{x}{3}}$$ which is what you wanted.

- 4 years, 1 month ago

Oh ok, that's what I had originally thought out of, but then I thought it would be harder. oopsies

- 4 years, 1 month ago