Find cable lengths AB and BC the angle C and the angle at Ba

Find cable lengths AB and BC, the angle C, and the angle at B(angle ABC).

22\' 7\" = 271 \" ; 26\' 8\" = 320 \"

In triangle ABD, apply sine rule:

65deg50\'= 65.83 deg ;

320/sin65.83 = 271/sinB

sinABD = (sin65.83*271)/320 = 0.772

ABD = 50.59 deg

Angle BDA = 180 - 65.83 -50.59 = 63.58 deg

Again applying sine rule : AB/sin63.58 = 320/sin65.83

AB = 314.11 inch = 26\'2\"

Triangle:BCD

Angle BDC = 180 - 63.58 = 116.42 deg

BD = 320\" ; CD = 342\"

Applying cosine rule : BC^2 = BD^2 + CD^2 -2BD*CDcos116.42

BC^2 = 320^2 + 342^2 -2*342*320cos116.42

BC = 562.80 \" = 46\' 11\"

Applying sine rule : BD/sinC = BC/sinD

sinC = BD*sinD/BC = 320*sin116.42/562.80

C = 30.61 deg

Angle DBC = 180 - 30.61 - 116.42 = 32.97 deg

AngleB = 50.59 +32.97 = 83.56 deg

$Find cable lengths AB and BC, the angle C, and the angle at B(angle ABC). Solution22\' 7\$