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.and ∠OBP = ∠OPB = 60°
Therefore, ∠POB = 30° + 30° = 60°; which shows that each angel of triangle OPQ is 60° . Hence ∆OPQ is equilateral.
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.
● Trigonometric Functions
11 and 12 Grade Math
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