How to find the Trigonometrical Ratios of 60°?
Let a rotating line \(\overrightarrow{OX}\) rotates about O in the anticlockwise 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 \(\overrightarrow{OX}\).
Let a rotating line \(\overrightarrow{OX}\) rotates about O in the anticlockwise 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;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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