\fdO'^ - Old Forge Coal Mines

GEOMETRY AND TRIGONOMETRY. 149
A £ and EB. Then,
A£: CEv.CE'.EB,
or 3^: CEy.CE: 10-^,
4 4
or CW' = si X lo\ = 34.9375.
4 4
Extracting the square root, we have
CE=6.%\.
2 X CE = rZ> = 2 X 5.91, or 11.82 inches. Ans.
(299) In 19° 19' 19' there are 69,559 seconds, and
in 360°, or a circle, there are 1,296,000 seconds. There-
69 559
fore, 69,559 seconds equal ., one nnn y '^^ .053672 part of
1,296,000'
a circle. Ans.
(300) In an angle measuring 19° 19' 19' there are
69,559 seconds, and in a quadrant, which is — of 360°, or
90", there are 324,000 seconds. Therefore, 69,559 seconds
69 559
^^"^^ onl Ann * ^^ .214688 part of a quadrant. Ans.
324,000'
(301) Given, OB = OA — — , or 11— inches, and
angle AOB=^ of 360°, or 36°. (See Fig. 8.) In the
right-angled triangle C O B^ we have
sin COB = ^, or CB= OB X sin COB.
Substituting the values oi O B and sin
CO B^ we have
(:r^=ll^X sinl8%
„ „ OT CB=i llL X .30902 = 3.55.
Fig. a 2
Since AB=2 CB, AB = 2x 3.55 = 7.1 inches.
The perimeter then equals 10 X 7. 1 = 71 inches, nearly.
Ans.