\fdO'^ - Old Forge Coal Mines

GEOMETRY AND TRIGONOMETRY. 153
Tan A==~= .55176 ; therefore, A = 28° 53' 18'. Ans.
Angle ^ = 90" - 28° 63' 18', or 61° 6' 42'. Ans.
(315) In Fig. 9, angle A OB = \ ot ^0O(°, or 45°. Angle
w(9^ = =-of 45°, or 22^. Side^^ =
— of 56 feet, or 7 feet. Now, in the
8
triangle m O B, we have the angle mOB
1° 7 1
= 22 ^, and mB = -, or d~ feet, given,
to find O B and the angle mB O.
feet.
niB
^\nmO B OB,
ox OB--^ m B
sinw OB'
Fig. 9.
Substituting their values, 0B= ^^^^ = -J^ = 9.146
BFy the diameter of the circle, equals ;
BF= 2 X 9.146 = 18.292 feet = 18 feet 3^- inches.
Z
Therefore,
Angle B O m = 22° 30'.
B O m-\-0 B m = 90°.
2 X B O therefore,
O B m = 90° - B O in = 90° - 22° 30'.
= 67° 30'.
A B C=2 O B m-2 (67° 30') = 135°.
Ans,
By Art. 703, the sum of the interior angles of an octagon
is 2 (8 — 2) = 12 right angles. Since the octagon is regular,
the interior anglies are equal, and since there are eight of
them, each one is — = 1— right angles. 1- X 90° = 135°.
o Z Z
(316) Lay off with a protractor the angle A O C equal
to 67° 8' 49', Fig. 10. Tangent to the circle at A, draw the
line A T. Through the point C, draw the line O C, and
continue it until it intersects the line A 7" at T. From C