Trigonometrical Ratios of 60°

How to find the Trigonometrical Ratios of 60°?

Let a rotating line \(\overrightarrow{OX}\) rotates about O in the anti-clockwise sense and starting from its initial position \(\overrightarrow{OX}\) traces out ∠XOY = 60° is shown in the above picture.

Take a point P on \(\overrightarrow{OY}\) and draw \(\overline{PQ}\) perpendicular to \(\overrightarrow{OX}\).

Trigonometrical Ratios of 60°

Let a rotating line \(\overrightarrow{OX}\) rotates about O in the anti-clockwise sense and starting from its initial position \(\overrightarrow{OX}\) traces out ∠XOY = 60° is shown in the above picture.

Take a point P on \(\overrightarrow{OY}\) and draw \(\overline{PQ}\) perpendicular to \(\overrightarrow{OX}\).

Now, take a point R on \(\overrightarrow{OX}\) such that \(\overline{OQ}\) = \(\overline{QR}\)  and join \(\overline{PR}\).

From △OPQ and △PQR we get,

\(\overline{OQ}\)  = \(\overline{QR}\),

\(\overline{PQ}\) common

and ∠PQO = ∠PQR (both are right angles)

Thus, the triangles are congruent.

Therefore,  ∠PRO = ∠POQ = 60°

Therefore, ∠OPR

= 180°  - ∠POQ - ∠PRO

= 180°  - 60° - 60°

=  60°

Therefore, the △POR is equilateral triangle

Let, OP = OR = 2a;

Thus, OQ = a.

Now, from pythagoras theorem we get,

OQ2 + PQ2 = OP2

⇒ a2 + PQ2 = (2a)2

⇒ PQ2 = 4a2 – a2

⇒ PQ2 = 3a2

Taking square roots on both the sides we get,

PQ = √3a (since, PQ > 0)

Therefore, from the right angled triangle POQ we get,
sin 60° = \(\frac{\overline{PQ}}{\overline{OP}} = \frac{\sqrt{3} a}{2a} = \frac{\sqrt{3}}{2}\);
cos 60° = \(\frac{\overline{OQ}}{\overline{OP}} = \frac{a}{2a} = \frac{1}{2}\)
And tan 60° = \(\frac{\overline{PQ}}{\overline{OQ}} = \frac{\sqrt{3} a}{a} = \sqrt{3}\)
Therefore, csc 60° = \(\frac{1}{sin  60°} = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}\)
sec 60° = \(\frac{1}{cos  60°} \)= 2
And cot 60° =  \(\frac{1}{tan  60°} = \frac{1}{\sqrt{3}} = \frac{ \sqrt{3}}{3}\)


Trigonometrical Ratios of 60° are commonly called standard angles and the trigonometrical ratios of these angles are frequently used to solve particular angles.

11 and 12 Grade Math

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