Solution :
The value of cot 60 degrees is \(1\over \sqrt{3}\).
Proof :
Consider an equilateral triangle ABC with each side of length of 2a. Each angle of \(\Delta\) ABC is of 60 degrees. Let AD be the perpendicular from A on BC.
\(\therefore\) AD is the bisector of \(\angle\) A and D is the mid-point of BC.
\(\therefore\) BD = DC = a and \(\angle\) BAD = 30 degrees.
In \(\Delta\) ADB, \(\angle\) D is a right angle, AB = 2a and BD = a
By Pythagoras theorem,
\(AB^2\) = \(AD^2\) + \(BD^2\) \(\implies\) \(2a^2\) = \(AD^2\) + \(a^2\)
\(\implies\) \(AD^2\) = \(4a^2\) – \(a^2\) = \(3a^2\) \(\implies\) AD = \(\sqrt{3}a\)
Now, In triangle ADB, \(\angle\) B = 60 degrees
By using trigonometric formulas,
\(cot 60^{\circ}\) = \(base\over perpendicular\) = \(b\over p\)
\(cot 60^{\circ}\) = side adjacent to 60 degrees/side opposite to 60 degrees = \(BD\over AD\) = \(a\over \sqrt{3}a\) = \(1\over \sqrt{3}\)
Hence, the value of \(cot 60^{\circ}\) = \(1\over \sqrt{3}\)
