# Trigonometrical Ratios of 30°

How to find the trigonometrical Ratios of 30°?

Let a rotating line $$\overrightarrow{OX}$$ rotates about O in the anti-clockwise sense and starting from the initial position $$\overrightarrow{OX}$$ traces out ∠XOY = 30°.

Take a point P on $$\overrightarrow{OY}$$ and draw PA perpendicular to $$\overrightarrow{OX}$$ Then, ∠OPA = 60°.

Now, produce PA to B such that PA = MB and join OB.

From ∆PMO and ∆QMO we have,

PA = BA,

OA common

and ∠OBP = ∠OPB = 60°

Therefore, ∠POB = 30° + 30° = 60°; which shows that each angel of triangle OPQ is 60° . Hence ∆OPQ is equilateral.

Let, OP = PB = 2a; therefore, PA = ½ PB = a

Again, OA2 + PA2 = OP2

⇒ OA2 + a2 = (2a)2

⇒ OA2 = 4a2 – a2

⇒ OA2 = 3a2

Therefore, OA = √3a (Since, OA > 0).

Now, from the right-angled ∆OPA we have,

sin 30° = $$\frac{\overline{PA}}{\overline{OP}} = \frac{a}{2a} = \frac{1}{2}$$;

cos 30° = $$\frac{\overline{OA}}{\overline{OP}} = \frac{\sqrt{3}a}{2a} = \frac{\sqrt{3}}{2}$$

And tan 30°  = $$\frac{PA}{OA} = \frac{a}{\sqrt{3}a} = \frac{1}{\sqrt3} = \frac{\sqrt{3}}{3}$$

Therefore, csc 30° = $$\frac{1}{sin 30°}$$  = 2;

Sec 30° = $$\frac{1}{cos 30°} = \frac{2}{\sqrt3} = \frac{2\sqrt{3}}{3}$$

And cot 30° = $$\frac{1}{tan 30°}$$ =  √3.

Trigonometrical Ratios of 30° are commonly called standard angles and the trigonometrical ratios of these angles are frequently used to solve particular angles.