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TRIGONOMETRY. 1 79
Thus the solution
:
c _ 12.75 X sin. 62° 11’ 55"
sin. 20° 38’ 12"
r _ 12.75 X 0.88456 e ,
6
~ ’
035243 ’
= 32 feet lon^-
Example 2.
Find the length of the side B (see Fig. 20) when angle b
is 41° 33’ 43", side C is 32 feet and side A is 12.75 feet.
In this example two sides and their included angle are
given and the third side is required ; therefore the formula
B = \/A2
+ C2 — 2A C cos. b must be used.
Solution
:
B = \/l2.752
+ 322 — 2 X 12.75 X 32 X 0.748238
B = Vl186.562 — 610.562 = \^hlQ = 24 feet long.
Example 3.
Find the length of the side B when side A is 12.75 feet
long, angle b is 41° 33’ 43" and angle c is 117° 48’ 5’ . ( See
Fig. 20).
In this problem one side and its two adjacent angles are
given; therefore it can not be solved directly by any of the
preceding formulas, but the first thing to do is to find the angle
opposite to side A.
Thus: Angle a = 180° — (41° 33’ 43" + 117° 48’ 5") =
20° 38’ 12". The side B may be found by the formula
A sin, b
h — sin. a
Solution
:
B _ A sin. 41° 33’ 43" _ 12.75 X 0.66343
sin. 20° 38’ 12" 0.35242 = 24 ieet l°ng-
Example 4.
Find length of the side C when B is 24 feet long, angle c is
117° 48’ 5" and the side A is 12.75 feet long. ( See Fig. 20).
Solution
:
C = VA2
+ B1 — 2 A B cos. c
C= Vl2.752
+ 242 — 2 X 12.75 X 24 X (— 0.4664)
C = Vl62.56 +1>76 + 285.44
C = Vl024 = 32 feet long.
Note.—In this example the cos. of 117° 48’ 5" is used,
which, in numerical value, is equal to cos. of 62° 11’ 55" =
0.4664, but cos. in the second quadrant is negative (see page 154);
therefore cos. 117° 48;
5" = ( — 0.46645) and the essential sign
of the last product after it is multiplied by this negative cos.
must change from — to +. ( See Algebra, page 63).
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