Tan 40, 50, 60

Geometry Level 2

$\tan(40^\circ)\tan(50^\circ)\tan(60^\circ)=?$

\tan(40^\circ)

\tan(50^\circ)

\tan(45^\circ)

\tan(60^\circ)

2 solutions

Mahdi Raza
Jan 15, 2020

$\tan(40^\circ) = \tan(90^\circ-50^\circ) = \cot(50^\circ) = \frac{1}{\tan(50^\circ)}$

$\tan(40^\circ)\tan(50^\circ)\tan(60^\circ)$ $\frac{1}{\tan(50^\circ)}\tan(50^\circ)\tan(60^\circ)$ $\frac{1}{\cancel{\tan(50^\circ)}}\tan\cancel{(50^\circ)}\tan(60^\circ)$ $=\boxed{\tan(60^\circ)}$

Quite easy....

Nikola Alfredi - 1 year, 3 months ago

Tom Engelsman
Jan 15, 2020

Knowing that $\tan(40) = \cot(50) \Rightarrow \cot(50) \cdot \tan(50) = 1.$ Thus the product is simply $\tan(60).$

You need to put a backslash "\" before function name in LaTex \tan, \cot, \sin, \cos, \int, \sum, \gcd, ....

Chew-Seong Cheong - 1 year, 4 months ago

Will do, Chew-Seong, thanks!

tom engelsman - 1 year, 4 months ago

Tan 40, 50, 60

2 solutions

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