The Laws of Sines and Cosines

  One might ask " why are law of cosines and law of sines is valid?". Following should

The law of sines

This can also be interpreted as three equations:

Since the three versions differ only in the labelling of the triangle, it is enough to verify one just one of them, so we'll just consider the version stated first.

An explanation of the law of sines is fairly easy to follow, but in some cases we'll have to consider sines of obtuse angles.

First, drop a perpendicular line AD from A down to the base BC of the triangle. The foot D of this perpendicular will lie on the edge BC of the triangle when both angles B and C are acute. But if angle B is obtuse, then the foot D will lie on BC extended in the direction of B. Yet if angle C is obtuse, then D will line on BC extended in the direction of C. Fortunately, the argument is the same in all three cases.

Let h denote the length of this line AD, that is, the height (or altitude) of the triangle.

When angle B is acute, then sin B = h/c. But this is true even when B is an obtuse angle as in the third diagram. There, angle ABC is obtuse. But the sine of an obtuse angle is the same as the sine of its supplement. That means sin ABC is the same as sin ABD, that is, they both equal h/c.

Likewise, it doesn't matter whether angle C is acute or obtuse, sin C = h/b in any case.

These two equations tell us that h equals both c sin B and b sin C. But from the equation c sin B = b sin C, we can easily get the law of sines:

The law of cosines

c2 = a2 + b2 – 2ab cos C

There are two other versions of the law of cosines,

a2 = b2 + c2 – 2bc cos A

and

b2 = a2 + c2 – 2ac cos B.

Since the three verions differ only in the labelling of the triangle, it is enough to verify one just one of them. We'll consider the   version stated first.

In order to see why these laws are valid, we'll have to look at three cases. For case 1, we'll take the angle C to be obtuse. In case 2, angle C will be a right angle. In case 3, angle C will be acute.

Case 1.

For this case, we take angle C to be obtuse. This case has a wrinkle in it since the cosine of an obtuse angle is negative. Let's see how that goes.

 These equations and plain algebra finish the argument as follows:

  Thus, the law of cosines is valid when C is an obtuse angle.

Case 2.

Now consider the case when the angle at C is right. The cosine of a right angle is 0, so the law of cosines,

c2 = a2 + b2 – 2ab cos C, simplifies to becomes the Pythagorean identity, c2 = a2 + b2, for right triangles which we know is valid.

Case 3.

In this case we assume that the angle C is an acute triangle. Drop a perpendicular line AD from A down to the base BC of the triangle.   The foot D of the perpendicular will (1) lie on the edge BC if angle B is acute, (2) coincide with the point B if the angle B is right, or (3) lie on the side BC extended if the angle B is obtuse.

Let h denote the height of the triangle, let d denote BD, and e denote CD.

Then we can read the following relationships from the diagram:

That last equation requires explanation. If the point D lies on the side BC, then d = a – e, but if D lies on BC extended, then d = e – a. In either case, d2 = (e – a)2.

These equations and a little algebra finish the proof as follows:

Thus, we now know that the law of cosines is valid when both angle C is acute, and we've finished all three cases.