What is the Value of Cot 60 Degrees ?

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.equilateral triangle

\(\therefore\)   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 \(cot 60^{\circ}\) = \(1\over \sqrt{3}\)

Leave a Comment

Your email address will not be published. Required fields are marked *