Asked by fredrick
three towns A,B,C are situated such that AB is equal to 60km and AC is 100km.The bearing of C from A is 290degree and the bearing of B from A is 060 degree calculate
a)the distance BC
b)the bearing of B from C
a)the distance BC
b)the bearing of B from C
Answers
Answered by
Steve
use the law of cosines to get BC, knowing that angle a is 130 degrees.
then use the law of sines to get angle B, and then you can figure the bearing of C.
then use the law of sines to get angle B, and then you can figure the bearing of C.
Answered by
henry2,
All angles are measured CW from +y-axis.
Given: AB = 60 km[60o], AC = 100 km[290o], BC = ?
BC = BA + AC = 60[60+180] + 100[290o]
BC = ( 60*sin240+100*sin290) + (60*cos240+100*cos290)I
BC = -146 + 4.2i = 146.1km[-88] = 146.1km[272o].
CB = 146.1km[272-180] = 146.1km[92o].
Given: AB = 60 km[60o], AC = 100 km[290o], BC = ?
BC = BA + AC = 60[60+180] + 100[290o]
BC = ( 60*sin240+100*sin290) + (60*cos240+100*cos290)I
BC = -146 + 4.2i = 146.1km[-88] = 146.1km[272o].
CB = 146.1km[272-180] = 146.1km[92o].
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