Make your sketch.
label the start I, the point of turning K, and the junction as J
In triangle KIJ, you have angle JKI = 120°, KI = 500, KJ = 300
using the cosine law,
IJ^2 = 500^2 + 300^2 - 2(500)(300)cos120°
IJ = .... , (careful with the cos120°, it is negative)
Once you have IJ, use the sine law to find angle I
An aeroplane flies due north from Ikeja airport for 500km.if their flies on a bearing of 060 for a further distance of 300km before overflying a road junction . calculate the distance of the aeroplane from Ikeja airport when it was directly above the junction (b).the bearing of the aeroplane from Ikeja airport at this instance
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2 answers
All angles are measured CW from +y-axis,
a. d = 500kmn[0o] + 300km[60o],
X = 500*sin0 + 300*sin60 = 259.8 km,
Y = 500*Cos0 + 300*Cos60 = 650 km,
d = sqrt(X^2 + Y^2) =
b. Tan A = X/Y,
A = ?.
a. d = 500kmn[0o] + 300km[60o],
X = 500*sin0 + 300*sin60 = 259.8 km,
Y = 500*Cos0 + 300*Cos60 = 650 km,
d = sqrt(X^2 + Y^2) =
b. Tan A = X/Y,
A = ?.
10 answers