![]() 9 THE LAW OF SINES WE USE THE LAW OF SINES AND THE LAW OF COSINES to solve triangles that are not right-angled. Such triangles are called oblique or scalene triangles. The law of sines states the following: The sides of a triangle are to one another ![]() This means that in the oblique triangle ABC, side a, for example, is to side b as the sine of angle A is to the sine of angle B.
Similarly,
And so on, for any pair of sides and their opposite angles. Strictly, A, B, C are points. They are the vertices of the triangle. "Angle A" is a brief way of saying "The angle at the point A." By "sin A," then, we mean "The sine of the angle at the point A." Example 1. a) The three angles of a triangle are 40°, 75°, and 65°. In what ratio are the three sides? Sketch the figure and place the ratio numbers. Solution. To find the ratios of the sides, we must evaluate the sines of their opposite angles. From the Table, sin 40° = .643 sin 75° = .966 sin 65° = .906 These are the ratios of the sides opposite those angles: ![]() Notice that we may express the ratios as ratios of whole numbers; we may ignore the decimal points. (Equivalently, we have multiplied each side by the same number, namely 1000. The theorem of the same multiple.) b) When the side opposite the 75° angle is 10 cm, how long is the side opposite the 40° angle? ![]() Solution. Let us call that side x. Now, according to the Law of Sines, in every triangle with those angles, the sides are in the ratio 643 : 966 : 906. Therefore,
With the aid of a calculator, x Problem 1. The three angles of a triangle are A = 30°, B = 70°, and C = 80°.
To see the answer, pass your mouse over the colored area. sin 30° = .500 sin 70° = .940 sin 80° = .985
In every triangle with those angles, the sides are in the ratio 500 : 940 : 985. But the side corresponding to 500 has been divided by 100. Therefore, each side will be divided by 100. Side b will equal 9.4 cm, and side c = 9.85 cm. The sine of an obtuse angle The trigonometric functions (sine, cosine, etc.) are defined in a right triangle in terms of an acute angle. What, then, shall mean by the sine of an obtuse angle ABC? ![]() The sine of an obtuse angle is defined to be the sine of its supplement. For example,
(Topic 7.) To see why we make this definition, let ABC be an obtuse angle, and ![]() draw CD perpendicular to AB extended. Then the sine of angle ABC is defined as follows:
But this is the sine of angle CBD -- opposite-over-hypotenuse. And angle CBD is the supplement of angle ABC. So, by the sine of an obtuse angle we mean the sine of its supplement. For example, ![]()
Problem 1. Evaluate the following: a) sin 135°
= sin 45° = ½ b) sin 127°
= sin (180° − 127°) = sin 53° = .799 Problem 2. a) The three angles of a triangle are 105°, 25°, and 50°. In what ratio are
Therefore, the sides opposite those angles are in the ratio 966 : 423 : 766 b) If the side opposite 25° is 10 cm, how long is the side opposite 50°?
With the aid of a calculator, this implies:
The ambiguous case The so-called ambiguous case arises from the fact that an acute angle and an obtuse angle have the same sine. If we had to solve sin x = ½ for example, we would have x = 45° or x = 135°. In the following example, we will see how this ambiguity could arise. In triangle ABC, angle A = 30°, side a = 1.5 cm, and side b = 2 cm. Let us use the law of sines to find angle B.
On inspecting the Table for the angle whose sine is closest to .666, we find B But the sine of an angle is equal to the sine of its supplement. That is, .666 is also the sine of 180° − 42° = 138°. This problem has two solutions. Not only is angle CBA a solution, but so is angle CB'A, which is the supplement of angle CBA. (We can see that it is the supplement in the isosceles triangle CBB'; angle CB'A is the supplement of angle CB'B, which is equal to angle CBA.) Given two sides of a triangle a, b, then, and the acute angle opposite one of them, say angle A, under what conditions will the triangle have two solutions, only one solution, or no solution? Let us first consider the case a < b. Upon applying the law of sines, we arrive at this equation:
then b sin A = h. On replacing this in the right-hand side of equation 1), it becomes
There are now three possibilities:
In the first of these -- h or b sin A < a -- there will be two triangles. In the second -- h or b sin A = a -- there will be one right-angled triangle. And in the third -- h or b sin A > a -- there will be no solution. Example 2. Let a = 2 cm, b = 6 cm, and angle A = 60°. How many solutions are there for angle B? Answer. We must calculate b sin A. If it is less than a, there will be two solutions. If it is equal to a, there will be one solution. And if it is greater than a, there will be no solution.
Since a = 2, then b sin A > a. ( There is no solution. Finally, we will consider the case in which angle A is acute, and a > b. In this case, there is only one solution, namely, the angle B in triangle CBA. For, in triangle CAB', the angle CAB' is obtuse. Problem 3. In each of the following, find the number of solutions. a) Angle A = 45°, a =
Since b) Angle A = 45°, a = 1.8, b = 2.
Again, a < b. b sin A = 2 c) Angle A = 45°, a = 2, b = 1.5. Here, a > b. Therefore there is one solution. d) Angle A = 45°, a = 1.4, b = 2.
a < b. b sin A = 2 Proof of the law of sines The sides of a triangle are to one another in the same ratio as the sines ![]() In triangle ABC, draw CD perpendicular to AB. Then CD is the height h of the triangle. The height now separates triangle ABC into two right triangles, CDA and CDB. (This was necessary, because the trigonometric functions are defined in terms of a right triangle.) We will now show that
Now, in triangle CDA,
While in triangle CDB,
Therefore,
This is what we wanted to prove. In the same way, we could prove that
and so on, for any pair of angles and their opposite sides. Next Topic: The Law of Cosines Please make a donation to keep TheMathPage online. Copyright © 2001-2007 Lawrence Spector Questions or comments? E-mail: themathpage@nyc.rr.com |