# TRIGONOMETRIC RATIO TABLE

The table provided in this section will help you to find the values of trigonometric ratios for the special angles 0°, 30°, 45°, 60° and 90°.

The values of trigonometric ratios of some special angles are very important to solve many problems in trigonometry.

Therefore, it is important to remember the values of the trigonometric ratios of some special angles.

## Trigonometric Ratios of Some Special Angles Now, let us see, how the above values of trigonometric ratios are determined for the standard angles.

First, let us see how the values are determined for the angles 30° and 60°.

## Trigonometric Ratios of 30° and 60°

Let ABC be an equilateral triangle whose sides have length a (see the figure given below). Draw AD perpendicular to BC, then D bisects the side BC.

So, BD = DC = a/2 and BAD = DAC = 30°. Now, in right triangle ADB, BAD = 30° and BD = a/2.

In right triangle ADB, by Pythagorean theorem,

Hence, we can find the trigonometric ratios of angle 30° from the right triangle ADB.

In right triangle ADB, ABD = 60°. So, we can determine the trigonometric ratios of angle 60°.

Now, let us see how the values are determined for the angle 45°.

## Trigonometric Ratios of 45°

If an acute angle of a right triangle is 45°, then the other acute angle is also 45°.

Thus the triangle is isosceles. Let us consider the triangle ABC with B = 90°, A = C = 45°. Then AB = BC. Let AB = BC = a.

By Pythagorean theorem,

AC2 = AB2 + BC2

AC2 = a2 + a2

AC2 = 2a2

AC = √2a

Hence, we can find the trigonometric ratios of angle 45° from the right triangle ABC.

Now, let us see how the values are determined for the angles 0° and 90°.

## Trigonometric Ratios of 0° and 90°

Consider the figure given below which shows a circle of radius 1 unit centered at the origin.

Let P be a point on the circle in the first quadrant with coordinates (x, y). We drop a perpendicular PQ from P to the x-axis in order to form the right triangle OPQ.

Let POQ = θ, then

sinθ = PQ/OP = y/1 = y (y coordinate of P)

cosθ = OQ/OP = x/1 = x (x coordinate of P)

tanθ = PQ/OQ = y/x

If OP coincides with OA, then angle θ = 0°.

Since, the coordinates of A are (1, 0), we have

If OP coincides with OB, then angle θ = 90°.

Since, the coordinates of B are (0, 1), we have

## Evaluating Trigonometric Expressions

Example 1 :

Solution :

Example 2 :

Solution :

Example 3 :

Solution :

Example 4 :

Solution :

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