6. The width of the gable of a house is 34 ft.; the height of the house above the eaves is 15 ft. Find the length of the rafters and the angle Find the pitch of the roof. (Ex. 10, p. 11.)

of inclination of the roof.

7. A kite string is 250 ft. long and makes an angle of 40° with the level ground. Find approximately the height of the kite above the ground, neglecting the sag in the string.

8. One bank of a river is a bluff rising 75 ft. vertically above the water. The angle of elevation of the top of the bluff from the water's edge on the opposite bank is 20° 27'; find the width of the river.

9. A taut rope 100 ft. long is attached to the top of a building. The free end reaches the ground 24 ft. 7 in. away from the base of the building. Find the height of the building and the angle which the rope makes with the ground.

10. Find the angles which the diagonal of a rectangle 12 ft. wide and 17 ft. long makes with the sides.

11. A chord of a circle is 100 ft. long and subtends an angle of 40° 42! at the center. Find the radius of the circle.

Find

12. A hill rises 8 ft. vertically in a horizontal distance of 40 ft. the angle of inclination of the hill with the horizontal. What is the difference in elevation of two points that are 500 ft. apart measured up the hill ?

13. Find the length of a side of an equilateral triangle circumscribed about a circle of radius 15 inches.

14. Devise a formula for solving an isosceles triangle when the base and the base angles are given; when the base and one of the equal sides

18. From the same figure show that sin 0 = 2 sin (0/2) cos (0/2). [HINT. Find the area of ▲ ABC, using first AB, then BC, as base.] 19. Show that tan (0/2) = (1 - cos 0)/sin 0.

*

CHAPTER III

SOLUTION OF OBLIQUE TRIANGLES

PART I. FUNCTIONS OF OBTUSE ANGLES

18. Obtuse Angles. The solution of oblique triangles involves obtuse as well as acute angles. Let an obtuse angle a be placed on the coördinate axes with the vertex at the origin and one side along the x-axis to the right; then the other side will fall in the second quadrant. The ratios sin a, cos a, etc., are defined in terms of x, y, and r=√x2+ y2 precisely as for acute angles. (See § 6.) It should be noticed, however, that since x is negative while y and rare positive, every ratio which involves is negative for an obtuse angle; thus x/r cosa, y/x= tan a, and their reciprocals, sec a and ctn a, are all negative for obtuse angles.

FIG. 19

We have seen how it is possible to find the ratios of angles greater than 45° from a table extending no farther than 45°, by means of the relations sin (90° — α) : = cos a, etc., which were proved on p. 13. By means of similar relations it is possible to find the ratios of obtuse angles from the same table.

Fa,b)

a

FIG. 20

(a,b)

Let be placed on coördinate axes as described above, and let the supplement of a be denoted by B (which is an acute angle). Lay off B from Ox so that the other side of B falls in the first quadrant. From a point P in the side of a (in second quadrant) and a point P' in the side of B (in first quadrant) at the same distance r from the origin, draw the

* An obtuse angle is an angle which is greater than 90° and less than 180°.

perpendiculars PM, P'M', as in Fig. 20. The value of x for the point P will be negative since P is in the second quadrant. Let its coördinates be (a, b); then, since the triangles OPM, OP'M' are symmetric, the coördinates of P' are (a, b). As in § 6, we have

It follows that if a is an obtuse angle we find its sine by looking for the sine of its supplement, which is an acute angle, and similarly for the other functions, always having regard for the proper sign. The relations just found, together with those of §. 9, enable us to find the values of the functions for any angle which can occur in a triangle, from a table which gives them from 0° to 45°.

EXERCISES IX.-FUNCTIONS OF OBTUSE ANGLES

1. From the accompanying figure prove the following relations :

3. Find the values of the remaining functions of the angles of Ex. 2. 4. Express the following as functions of an angle less than 45°, and look up their values in a table.

PART II.

FUNDAMENTAL PRINCIPLES OF SOLUTION

19. General Method for Oblique Triangles. In the solution of oblique triangles the following cases arise:

Case I.

Case II.

Case III.

Given two angles and a side.

Given two sides and the included angle.

Given the three sides.

Case IV. Given two sides and an angle opposite one of them. A general method for solving oblique triangles in all of these cases consists in dividing the triangle into two right triangles by a perpendicular from a vertex to the opposite side; these right triangles are then solved by the methods of the previous chapter. In all cases except the three side case the perpendicular can be drawn in such a manner that one of the resulting right triangles contains two of the given parts.

This method applied in the various cases leads to formulas for the solution if letters are employed for the sides and angles.

20. Case I: Given Two Angles and a Side. In this case it is immaterial which side is given, since the third angle can be found immediately from the fact that the sum of the three angles is 180°. Drop the perpendicular from either extremity of the given side.

Example 1. An oblique triangle has one angle equal to 43°, another equal to 67°, and the side opposite the unknown angle equal to 51. termine the remaining parts.

B

De

70°

51

It is immediately seen that the third angle is 180° — (43° + 67°) = 70°. To solve this triangle draw the figure approximately to scale and drop the perpendicular CD = p from one extremity C of the known side to AB, the side opposite C. Denote the unknown side CB by a. In the right triangle ACD, the hypotenuse and one angle are known; hence by (13), § 6, p = 51 sin 67° = 46.95. An angle and the side opposite, in the right triangle BCD, are now known; hence by (15), § 6, a = p/sin 70° 46.95/.93969 = 49.958.

FIG. 21

The side AB may be found in the same manner. Check as in § 3, p. 2.

21. Case II: Given Two Sides and the Included Angle. The triangle can be divided into two right triangles, one of which contains two of the known parts, by a perpendicular drawn from either extremity of the unknown side to the side opposite. Two sides of a triangle are 26.5 and 32.8; the included Find the remaining parts.

Example 1. angle is 58° 18'.

=

In the figure let AB 32.8, AC 26.5, and the angle at A = 52° 18'. Drop a perpendicular p from B to the opposite side. Denote the unknown side by a and the segments of AC by x and y as in Fig. 22; then p, x, y, a, can be computed in the following order:

32.8

26.5 FIG. 22

The student may determine the angles at B and C.

22. Case III: Given the Three Sides. In this case it is not possible to divide the triangle into two right triangles in such a way that one of them contains two of the given parts; however, if a perpendicular is dropped to the longest side from the vertex of the angle opposite, the segments into which this side is divided by the perpendicular are easily computed, as in the example below. There is one and only one solution, provided no side is greater than the sum of the other two.

Example 1. The sides of a triangle are a = 36.4, b = 50.8, and c = 72.5. Determine the angles.

Draw a figure and drop a perpendicular from B upon AC. Denote the segments of the base by x and y as in Fig. 23; then

Since we now know x and y, the angles at A and C are easily found. The student may complete the solution. See also § 25.