__Trigonometric ratios of some specific angles :__

For certain specific angles such as 30°, 45° an d 60°, which are frequently seen in applications, we can use geometry to determine the trigonometric ratios.

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,

AB² = AD² + BD²

a² = AD² + (a/2)²

AD² = a² - (a²/4)

AD² = 3a²/4

AD = √(3a²/4)

AD = √3 x (a/2)

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°.

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 = √2 x a

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

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

sin θ = PQ / OP = y/1 = y (y coordinate of P)

cos θ = OQ / OP = x/1 = x (x coordinate of P)

tan θ = 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

The six trigonometric ratios of angles 0°, 30°, 45°, 60° and 90° are provided in the following table.

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