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 \(\overrightarrow{OX}\).

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,

OQ

⇒ a

⇒ PQ

⇒ PQ

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.

**●** **Trigonometric Functions**

**Basic Trigonometric Ratios and Their Names****Restrictions of Trigonometrical Ratios****Reciprocal Relations of Trigonometric Ratios****Quotient Relations of Trigonometric Ratios****Limit of Trigonometric Ratios****Trigonometrical Identity****Problems on Trigonometric Identities****Elimination of Trigonometric Ratios****Eliminate Theta between the equations****Problems on Eliminate Theta****Trig Ratio Problems****Proving Trigonometric Ratios****Trig Ratios Proving Problems****Verify Trigonometric Identities****Trigonometrical Ratios of 0°****Trigonometrical Ratios of 30°****Trigonometrical Ratios of 45°****Trigonometrical Ratios of 60°****Trigonometrical Ratios of 90°****Trigonometrical Ratios Table****Problems on Trigonometric Ratio of Standard Angle****Trigonometrical Ratios of Complementary Angles****Rules of Trigonometric Signs****Signs of Trigonometrical Ratios****All Sin Tan Cos Rule****Trigonometrical Ratios of (- θ)****Trigonometrical Ratios of (90° + θ)****Trigonometrical Ratios of (90° - θ)****Trigonometrical Ratios of (180° + θ)****Trigonometrical Ratios of (180° - θ)****Trigonometrical Ratios of (270° + θ)****Trigonometrical Ratios of (270° - θ)****Trigonometrical Ratios of (360° + θ)****Trigonometrical Ratios of (360° - θ)****Trigonometrical Ratios of any Angle****Trigonometrical Ratios of some Particular Angles****Trigonometric Ratios of an Angle****Trigonometric Functions of any Angles****Problems on Trigonometric Ratios of an Angle****Problems on Signs of Trigonometrical Ratios**

**11 and 12 Grade Math**

**From Trigonometrical Ratios of 60° to HOME PAGE**

**Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.