Can You Use the Law of Sines or Cosines if You Know All Three Angles to Find the Side

The Laws of Cosines and Sines

Nosotros saw in the department on oblique triangles that the law of cosines and the law of sines were useful in solving for parts of a triangle if certain other parts are known. The question here is "why are those laws valid?"

This is an optional section. When learning how to use trigonometry to solve oblique triangles, it is most important to know when and how to employ these 2 laws. If that'south enough for you, and then just skip on to the next section on area of a triangle. But if yous're interested in why they're truthful, then go along on. As usual, we'll use a standard notation for the angles and sides of a triangle. That means the side a is reverse the angle A, the side b is opposite the angle B, and the side c is opposite the angle C.

The police force of sines

It'due south enough to show the concluding equation

since the kickoff version differs simply in the labelling of the triangle.

An explanation of the police of sines is adequately like shooting fish in a barrel to follow, but in some cases we'll have to consider sines of obtuse angles.

Outset, drop a perpendicular line Advertizement from A down to the base BC of the triangle. The human 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, so the foot D will prevarication on BC extended in the direction of B. Yet if angle C is obtuse, and then D will line on BC extended in the management of C. Fortunately, the argument is the same in all three cases.

Allow h denote the length of this line Advertizement, that is, the pinnacle (or altitude) of the triangle.

When bending B is acute, and then sinB =h/c. But this is truthful fifty-fifty when B is an obtuse angle every bit in the tertiary diagram. There, angle ABC is obtuse. But the sine of an obtuse bending is the same as the sine of its supplement. That means sinABC is the same as sinABD, that is, they both equal h/c.

Likewise, information technology doesn't thing whether bending C is acute or obtuse, sinC =h/b in whatsoever case.

These 2 equations tell us that h equals both c sinB and b sinC. But from the equation c sinB =b sinC, we can easily get the police of sines:

The law of cosines

In that location are two other versions of the law of cosines,

a ^two = b ² + c ² – 2bc cos A and b ^two = a ² + c ² – twoac cos B.

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

In gild to meet why these laws are valid, we'll accept to look at three cases. For instance 1, nosotros'll take the angle C to exist birdbrained. In case 2, angle C will be a correct angle. In case 3, bending C will be astute.

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Case one. For this case, we accept bending C to be obtuse. This case has a contraction in it since the cosine of an obtuse angle is negative. Allow's see how that goes.

First, drop a perpendicular line AD from A downwards to the base BC of the triangle. In this case, the pes D of this perpendicular will lie outside the triangle. Let h announce the peak of the triangle, let d denote BD, and permit e denote CD. We can derive the following equations from the effigy: c two  = d ii + h two b 2  = e ii + h ii d  = a + e cos C  = – e/b In general, the cosine of an birdbrained angle is the negation of the cosine of its supplement. In this example that means the cosine of angle C, that is to say angle ACB, is the negation of the cosine of bending ACD. That's why the minus sign appears in the last equation.

These equations and manifestly algebra finish the argument as follows:

c two	=	d ii + h 2
	=	(a + east)2 + h 2
	=	a 2 + 2ae + east 2 + h 2
	=	a 2 + b 2 + 2ae
	=	a ii + b two – twoab cos C

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

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Case 2. Now consider the case when the angle at C is right. The cosine of a correct bending is 0, so the constabulary of cosines, c ⁱⁱ = a ⁱⁱ + b ² – 2ab cosC, simplifies to becomes the Pythagorean identity, c ² = a ^two + b ², for right triangles which nosotros know is valid.

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Instance three. In this case we assume that the angle C is an acute triangle. Drop a perpendicular line Advertising 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 bending B is correct, or (three) lie on the side BC extended if the angle B is obtuse.

Let h denote the pinnacle of the triangle, let d denote BD, and due east denote CD.

Then we tin can read the following relationships from the diagram:

c 2	=	d 2 + h 2
b 2	=	east 2 + h 2
cos C	=	e/b
d ii	=	(e — a)two

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 =eastward –a. In either instance, d ² = (east –a)ⁱⁱ.

These equations and a little algebra cease the proof equally follows:

c ii	=	d 2 + h 2
	=	d ii – e 2 + b 2
	=	(d – e) (d + e) + b 2
	=	(a – twoeastward) a + b ii
	=	a 2 + b two – 2ae
	=	a ii + b 2 – iiab cos C

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

Incidentally, Euclid included in his Elements a couple of propositions, Ii.12 and II.thirteen, that look very much like the constabulary of cosines, but they are non actually the police force of cosines, of class, since trigonometry had not been developed in Euclid's time.

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Source: https://www2.clarku.edu/faculty/djoyce/trig/laws.html

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