For certain specific angles such as 30°, 45° and 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.

Then,

BD = DC = a/2

∠BAD = ∠DAC = 30°

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

In right triangle ADB, by Pythagorean theorem,

AB^{2} = AD^{2} + BD^{2}

a^{2} = AD^{2} + (a/2)^{2}

a^{2} - (a^{2}/4) = AD^{2}

3a²/4 = AD^{2}

√(3a^{2}/4) = AD

√3 ⋅ a/2 = m AD

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,

AC^{2} = AB^{2} + BC^{2}

AC^{2} = a^{2} + a^{2}

AC^{2} = 2a^{2}

Take square root on each side.

AC = a√2

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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