Asked by Kim
Village Q is 7km from village P on a bearing of 060. VIllage R is 5 Km from village P on a bearing of 315. Using scale drawing with a scale of 1 cm : 1 km.
find:
a. the direct distance from village Q to village R
b. the bearing of village Q from village R.
find:
a. the direct distance from village Q to village R
b. the bearing of village Q from village R.
Answers
Answered by
Steve
angle QPR is 105°
so, using the law of cosines, the distance
QR^2 = 7^2+5^2-2*7*5*cos105°
QR = 9.6 km
Expressing the vectors PQ and PR as complex numbers relative to P,
RQ = 7cis60°-5cis105° = 4.8+1.2i
so the angle θ of RQ has
tanθ = 1.2/4.8
θ = 14°
Thus, the bearing of Q from R is 90-14 = 76°
so, using the law of cosines, the distance
QR^2 = 7^2+5^2-2*7*5*cos105°
QR = 9.6 km
Expressing the vectors PQ and PR as complex numbers relative to P,
RQ = 7cis60°-5cis105° = 4.8+1.2i
so the angle θ of RQ has
tanθ = 1.2/4.8
θ = 14°
Thus, the bearing of Q from R is 90-14 = 76°
Answered by
Disha agarwal
From where 105 come?
Answered by
keara
I think 105 came from 315-180=135 and then 135-(90-60)
Answered by
Ntokozo
Where does the 1.2 and 4.8 come from
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