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°.

Trigonometrical Ratios of 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.


11 and 12 Grade Math

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