218
[CHAP. 31
OBLIQUE TRIANGLES
For c :
c2 ¼ a2 þ b2 2ab cos C ¼ ð132Þ2 þ ð224Þ2 2ð132Þð224Þ cos 28– 400 ¼ ð132Þ2 þ ð224Þ2 2ð132Þð224Þð0:8774Þ ¼ 15 714 and c ¼ 125
For A :
sin A ¼
a sin C 132 sin 28– 400 132ð0:4797Þ ¼ ¼ 0:5066 ¼ c 125 125
and
A ¼ 30– 300
For B :
sin B ¼
b sin C 224 sin 28– 400 224ð0:4797Þ ¼ ¼ 0:8596 ¼ c 125 125
and
B ¼ 120– 400
(Since b > a; A is acute; since A þ C < 90– ; B > 90– :Þ Check: A þ B þ C ¼ 179– 500 : The required parts are A ¼ 30– 300 ; B ¼ 120– 400 ; c ¼ 125.
31.12
Two forces of 17.5 1b and 22.5 1b act on a body. If their directions make an angle of 50– 100 with each other, find the magnitude of their resultant and the angle which it makes with the larger force. In the parallelogram ABCD (see Fig. 31-13), A þ B ¼ C þ D ¼ 180– and B ¼ 180– 50– 100 ¼ 129– 500 : In the triangle ABC, b2 ¼ c2 þ a2 2ca cos B
½cos129– 500 ¼ cos ð180– 129– 500 Þ ¼ cos 50– 100
¼ ð22:5Þ þ ð17:5Þ 2ð22:5Þð17:5Þð 0:6406Þ ¼ 1317 2
sin A ¼
2
b ¼ 36:3
and
a sin B 17:5 sin 129– 500 17:5ð0:7679Þ ¼ ¼ ¼ 0:3702 36:3 b 36:3
and
A ¼ 21– 400
The resultant is a force of 36.3 1b; the required angle is 21– 400 .
Fig. 31-13
31.13
From A a pilot flies 125 mi in the direction N 38– 200 W and turns back. Through an error, he then flies 125 mi in the direction S 51– 400 E. How far and in what direction must he now fly to reach his intended destination A? Denote the turn back point as B and his final position as C. In the triangle ABC (see Fig. 31-14), b2 ¼ c2 þ a2 2ca cos B ¼ ð125Þ2 þ ð125Þ2 2ð125Þð125Þ cos 13– 200 ¼ 2ð125Þ2 ð1 0:9730Þ ¼ 843:7 sin A ¼
and
b ¼ 29:0
a sin B 125 sin 13– 200 125ð0:2306Þ ¼ ¼ ¼ 0:9940 29:0 b 29:0
and
A ¼ 83– 400
Since —CAN1 ¼ A —N1 AB ¼ 45– 200 ; the pilot must fly a course S 45– 200 W for 29.0 mi in going from C to A.