carried out with all the accuracy permitted by the tables. Do the given data warrant us in saying that the automobile is either 2.433 miles or 1.009 miles from the fork in the road?

If two sides of a triangle and the angle opposite one of them are given, there can be TWO SOLUTIONS only if the angle is acute and the side opposite it is less than the other given side. Under these conditions, there will be two solutions unless in solving for the second angle (C, in Example 2), it is found that its sine is equal to or greater than unity (sin C = 1). In the first case there is but one solution, a right triangle; and in the second case there is no solution, since the sine of an angle cannot exceed unity. A complete statement of the possible solutions is given in Exercise 3 below.

CASE III. Given two sides and the included angle, the third side is found by the law of cosines, and then the remaining angles by either the law of cosines or the law of sines.

EXAMPLE 3. To find the distance across a bay, AB, a point C is taken whose distances from A and B can be measured. It is found that AC is 70 yards, BC is 120 yards, and ACB is 50°. Find AB and also the angles at A and B.

A triangle is determined whose given parts are a = 120, b = 70, C = 50°. By the law of cosines

c2 = a2 + b2 - 2ab cos C

To find A and B, we use the law of cosines: Substituting the values of

B

whence

a-120

с

FIG. 121.

4900

50°

=

A

b=70

the sides in a2 b2c22bc cos A,

=

Substituting the values of the sides in b2 = c2 + a2 - 2ca cos B, we get

8500 + 14,400 - 2 × 92.2 × 120 x cos B, cos B = 0.8136

Check. The angles having been found without using the fact that the sum of the three angles is two right angles, we have

Having found A as above, B might have been obtained from A+B+C = 180°. But then it would have been necessary to use the law of sines as a check, and little would have been gained.

CASE IV. Given the three sides, the three angles are found by the law of cosines, or one may be found by the law of cosines and then the others by the law of sines.

EXAMPLE 4. The lengths of a triangular lot are found by pacing to be 80 feet, 50 feet, and 100 feet. Find the angles at the corners and the

area.

80,

b =

A triangle is determined by a = 50, c = 100. Substituting these values in the three forms of the law of cosines we get

To find the area, draw the altitude from C and denote it by h. Then

computations, and the angles to four significant figures for the purpose of obtaining drill in interpolation.

In the next chapter, a labor saving device to assist in the computations will be considered, and further exercises given.

EXERCISES

1. In Example 4, find B and C by the law of sines. What point in the solution might be overlooked, which one is more likely to notice in using the law of cosines?

2. Is a triangle always determined if values are given for a side and two angles? For two sides and the included angle? For the three sides? 3. If two sides, say a and b, and the angle opposite one of them, say A, are given, show that

(a) If A is obtuse, there is one solution if a>b, and there is no solution if ab.

(b) If A is right, there is one solution if a > b, and there is no solution if ab.

(c) If A is acute, there is one solution if ab, there are two solutions if b sin A <a <b, there is one solution if a = b sin A, and there is no solution if a <b sin A.

4. Determine the number of solutions, by Exercise 3, if,

5. Solve the following triangles, and check the solution:

(a) B = 62°.74, C = 87°.20, a =

10.

6. A tunnel is to be built through a hill from a point A to another point B. A point C, at the same level as A and B, is 1000 feet from A and 800 feet from B and ACB = 42°. Find the length of the tunnel.

7. A man starts from camp and walks N.E. for 5 miles, and then 22°.5 east of south until he reaches a point from which the camp is visible in a direction due west. How far has he walked, and how far is he from camp? 8. Two batteries of artillery. A and B, are four miles apart. An enemies' battery is located at a point C such that ZBAC = 64°.22 and LABC

==

43°.17. Find the range for each battery.

9. A field is bounded by two roads intersecting at right angles, and by two other straight lines. Find its area if the lengths of the sides, beginning at the corner at the crossing of the roads and measured in order around the field, are 30, 60, 70 and 40 rods.

10. A schooner sails 10° west of north at the rate of 7 knots an hour across the Gulf Stream at a place where it flows N.E. with a velocity of 4 knots an hour. Find the actual velocity of the schooner in direction and magnitude.

11. The current in a river flows at the rate of 2 miles an hour, and a man rows at the rate of 4 miles an hour. If he desires to cross the river at an angle of 70° with the bank, in the direction of the current, in what direction should he row?

12. Solve the preceding exercise if the man desires to cross at the same angle with the bank but in the upstream direction.

13. Resolve a velocity of 50 feet per second into two components inclined at 10° and 40° respectively to the direction of the given velocity.

14. A boy in an automobile moving 40 feet per second throws a ball in a horizontal direction inclined at 50° to the road with a speed of 30 feet per second. At what angle to the road will the ball move?

15. A road runs up a hill at an angle of 20°. At a point on it, 500 feet from the foot, the angle of depression of a horseman on the road leading to the hill is 5°. How far is he from the foot of the hill?

16. To find the width of a river, two points A and B are taken on one bank 100 feet apart. If C is a point on the opposite bank such that LBAC 62°.34 and ZABC = 49°.82, find the width of the river.

=

17. At a certain point the angle of elevation of the top of a mountain is 45°, and at a point 1000 feet nearer the mountain and at the same level as the first, the angle of elevation is 54°.13. How high is the mountain?

18. A ship steams due east at the rate of 25 miles an hour, and the smoke from its funnel is blown in a direction 20° south of west. The wind gauge shows an apparent velocity of 35 miles an hour for the wind. Find the actual velocity of the wind in direction and magnitude.

74. Inverse Trigonometric Functions. To find the inverse of sin x (pages 40 and 114) we set y sin x, and interchange x and y, obtaining xsin y. The solution of this equation for У in terms of x requires the introduction of a new function which is called the angle whose sine is x, and which is denoted by arc sin x. Hence,

A table of sines may be regarded as a table of angles whose sines are given (see "finding 0 if sin 0 is given " page 178). Thus Example 2, page 179, might have been stated: find arc sin 0.4332, the result being

0 =

=

arc sin 0.4332

=

25°.76 + n360° or 154°.33 + n360°.

This example illustrates the fact that for a given value of x arc sin x has not only one but a boundless number of values. This is apparent from the graph of arc sin x, which is sym

DEFINITION. The principal value of any one of the inverse. trigonometric functions for a given value of x is that one of the boundless number of values of the function which is smallest. numerically. If two values of the function are equal numerically, but opposite in sign, the positive value is the principal value. Unless the contrary is indicated,

the symbols arc sin x, arc cos x, etc., will be used in this work to denote the principal values only.

2

+

arc sin

1

The part of the graph which represents the principal values of arc sin x is given in the figure.

Inverse trigonometric functions are of much importance, although

we shall use them but little in this course. venient in stating a general result, as in this

FIG. 124.

They are con