Plane trigonometry with practical applications, by Leonard E. Dickson.

Ch. XI] GRAPHS, RADIANS, INVERSE FUNCTIONS 163
Solution. The required angles x are the abscissas of the points of intersection of y = sin x and y = 2(x -. The latter represents a straight
line, two of whose points are
A-(, )0andB= (, j. B
The abscissa xi of the point /
of intersection is just less than
5w/12 or 75 ~./
A more exact value of xl can
be found by Table II and a I
table (Art. 110) of the radians
in 1~,.., 9~. We shall compute 0 /A7
the values of
7r. TFIG. 94
F = x — sinx -
for x = 72~ and x = 73~, and find that these values have opposite signs;
hence F is zero for a value xl between 72~ and 73~.
sin 72~ = 0.9511 sin 73~ = 0.9563
~ sin 72~ = 0.4755 2 sin 73~ = 0.4781
= 0.7854 = 0.7854
_
1.2609 1.2635
72~ = 1.2566 radians 73~ = 1.2741 radians
F = - 0.0043 F = + 0.0106
The difference of the two F's is 0.0149. Hence xl exceeds 72~ by
43
49 X 60' = 17'; xl = 72~17', approximately.
EXAMPLE 2. The area of a segment of a circle of radius 5 feet is 28.56
square feet. Find the length of the chord.
Solution. Let x be the number of radians in the angle at the center subtended by the chord. By Ex. 9, Art. 110,
25
2- (x - sin x) = 28.56, x - sin x = 2.2848.
Thus x is the abscissa of the point of intersection of y = sin x and y = x
2.2848, whose graphs are shown in Fig. 94, the straight line being determined
by the points having x = 0 and x = r. As in Ex. 1, we get x = 15504',
nearly. The length of the chord is 10 sin i x = 9.764 ft.