What is the Value of Cosec 60 Degrees ?

Solution :

The value of cosec 60 degrees is \(2\over \sqrt{3}\).

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,

\(cosec 60^{\circ}\) = \(hypotenuse\over perpendicular\) = \(h\over p\)

\(cosec 60^{\circ}\) = hypotenuse/side opposite to 60 degrees = \(AB\over AD\) = \(2a\over \sqrt{3}\) = \(2\over \sqrt{3}\)

Hence, the value of \(cosec 60^{\circ}\) = \(2\over \sqrt{3}\)

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