Asked by Olayinka
3 town abc are such that the distance A and B is 50km and distance between A and C is 90km if the bearing of B from A is 075 degree and the bearing C from A is 310 degree find the distance between B and C and bearing of C from B
Answers
Answered by
mathhelper
after making my sketch, I had triangle ABC, with angle A = (50+75)° or 125°
BC^2 = 50^2 + 90^2 - 2(50)(90)cos 125°
= ...
BC = √...... = .....
You can then use the sine law to find angle B, and with your bearing
marked on your sketch, you can find the bearing you want.
BC^2 = 50^2 + 90^2 - 2(50)(90)cos 125°
= ...
BC = √...... = .....
You can then use the sine law to find angle B, and with your bearing
marked on your sketch, you can find the bearing you want.
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