SINE 60 DEGREES

Before going to our main topic (sine 60 degrees), let’s recall some important points for a better understanding of the concept. We have already studied triangles, and in particular, right triangles. Especially in the concept of trigonometry, the primary ratios like sine, cosine, tangent are normally used to find the angles, distances or heights, and lengths of the right-angled triangle.

Now, consider a right triangle ABC as shown in the figure.

Here, ∠ CAB is an acute angle. Note the position of side BC w.r.t.∠ A. It is the opposite side to ∠ A. AC is the hypotenuse of the right triangle, and the side AB is part of ∠ A. We further work certain ratios involving the sides of a right triangle. These ratios are known as trigonometric ratios.

The trigonometric ratios of ∠ A in right triangle ABC are defined as follows:

Similarly, cosec A, sec A, cot A can also be determined.

VALUE OF SINE 60 DEGREES

Now we are clear with the introduction part, let us now calculate the trigonometric ratio of sine 60o.

Consider an equilateral triangle ABC. Since, each angle in an equilateral triangle is 60 o, ∠ A = ∠ B = ∠ C = 60^o.

Draw a perpendicular AD from A to side BC, as shown in the figure.

Now, Δ ABD is a right triangle, right-angled at D with

∠ ABD = 60^o.

As we know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that AB = 2a.

Then,

Let us know all the trigonometric ratios of sine 0 o, 30 o, 45 o, 60 o and 90^o.

Sin 0^o = 0

Sin 30^o = 1/2

Sin 45^o = 1/√2

Sin 60^o = √3/2

Sin 90^{o= 1}

We can simply memorize the trigonometric ratios of cosine by writing the values of sine from bottom to top, i.e.,

PROBLEM – 1: Sin 60^o Cos30^o + Sin30^o Cos60^o

SOLUTION: We know that,