Likewise, c / sin C = a / sin A, so c = a sin C / sin A = 10 sin ( 120o ) / sin ( 30o )
= 10 (
)( 1 / 2 ) = 
.
Thus the solution of the triangle is:
B = 120o
b = 10
c =
Given three non-collinear points, A, B and C, there is only one circle containing those three points. Three non-collinear points also form the vertices of a triangle. According to a theorem from geometry, the ratio of a side of a triangle to the sine of the opposite angle equals the diameter of the circumscribed circle. Since for a given triangle there is only one circumscribed circle, then one may compute its diameter by computing the ratio between any of its three sides and the sine of the angle opposite that side. That is,
d = a / sin A = b / sin B = c / sin C.
This identity is called the Law of Sines.
The Law of Sines, together with the Law of Cosines which we will study shortly, are useful for solving triangles.
There are six pieces of information that must be known before a triangle is 'solved': the size of its three angles and the size of its three sides. In order to solve a triangle, one needs to have at least three of the six pieces of information to begin with. The one exception to this principle is this: knowing only the three angles is insufficient for finding the three sides
Why can't one determine the length of the three sides of a triangle if one knows all three angles?
There are six different ways to classify the three given pieces of information:
SSS: The lengths of all three sides are given.
SAS: The lengths of two sides and the size of the angle between them are given.
SSA: The lengths of two sides and the size of the angle opposite one of them are given.
ASA: The sizes of two angles and the length of the side between them are given.
SAA: The sizes of two angles and the length of the side opposite one of them are given.
AAA: The sizes of all three angles are given.
Note that some combinations of three pieces of information will not have a solution. For example, the sum of the three angles of a triangle must be 180o. Thus in the case of ASA, if the two angles add to more than 180o, there will be no solution. Or in the case of SSS, if the sum of two of the sides is less than the third side, there will be no solution, since the sum of two sides of a triangle is always greater than the third side. SAS is the only case where all possible lengths of the sides and all possible angles greater than 0o and less than 180o will have a solution.
Explain why each of the following 'triangles' has no solution:
(a) SSS: a = 7, b = 2, c = 4
(b) ASA: A = 85o, b = 10, C = 98o
(c) SAA: a = 12, B = 100o, C = 90o
(d) AAA: A = 65o, B = 82o, C = 35o
In order to apply the Law of Sines, it is necessary to know the size of at least one side and the size of the angle opposite that side. That is, one must know at least one side-sine pair. One will know a side-sine pair in the case of SSA or SAA. But if one is given ASA, one can deduce the third angle since the sum all all three angles must be 180o. Thus one will know a side-sine pair given ASA.
So the Law of Sines is used to solve triangles given SSA, SAA or ASA.
SAA and ASA are worked essentially the same: find the third angle, then use the side-sine pair to solve the other two sides.
Example: Given a = 10, A = 30o and C = 120o, solve the triangle.
First find B = 180o - 30o
- 120o = 30o.
Next, use the fact that b / sin B = a / sin A to find b,
since b = a sin B / sin A = 10 sin ( 30o
) / sin ( 30o ) = 10.
Likewise, c / sin C = a / sin A, so c = a
sin C / sin A = 10 sin ( 120o ) / sin (
30o )
= 10 (
)( 1 / 2 ) = .
Thus the solution of the triangle is:
B = 120o
b = 10
c =
Exercise 6.1.3
Given A = 24o, B = 80o
and c = 20, solve the triangle.
It is possible for the case SSA to have no solution for some combinations of lengths of the sides and angles.
Explain why the following SSA 'triangle' has no solution. Hint: the sine of an angle cannot be greater than 1.
A = 60o, a = 2, b = 10
Knowing the sine of one of the angles of a triangle is not sufficient information to uniquely determine the angle. For example, suppose we want to find A and all we know is that sin A = 1 / 2. Then A could be 30o or A could be 150o. We cannot assume that the correct value is 30o unless we can rule out 150o. In general, if sin A = y, then either A = arcsin y or A = 180o - arcsin y . Sometimes this situation arises when solving SSA triangles. When this happens and when we cannot rule out the larger solution, then we must list two separate solutions to the triangle.
Find two solutions to the triangle:
a = 10, b = 4, B = 20o
A ship sailing due north spots a lighthouse 30o to the left of its line of travel. Two miles later, the lighthouse is 50o to the left of its line of travel. How far is the ship from the lighthouse at that point? (Assume the earth is flat.)
Scientists wish to measure the diameter of a large circular meteor crater. Two points A and B on the edge of the pit are 120 feet apart. From a point C on the far side of the pit, the angle between the lines AC and BC is measured to be 8o. Use the Law of Sines to find the diameter of the crater.