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 \(\alpha_1 = \tan^{-1}\dfrac{15}{x}\).

The above can also be rewritten to \(\boxed{\alpha=\cot^{-1}\dfrac{x}{15}- \cot^{-1}\dfrac{x}{3}}\) which is what you wanted.
–
Daniel Liu
·
3 years, 2 months ago

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@Daniel Liu
–
Oh ok, that's what I had originally thought out of, but then I thought it would be harder. oopsies
–
Asher Joy
·
3 years, 2 months ago

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TopNewestLet 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. – Daniel Liu · 3 years, 2 months ago

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oopsies– Asher Joy · 3 years, 2 months agoLog in to reply