A cyclist rode from town A to town B, 14km apart on a bearing of N 40 degree E. At B he rode due East to another town C, 19km away from B. Calculate, correct to the nearest whole number a) the distance between A and C.

Question

b) how far North of A is B.

Answer 1

All angles are measured CW from +y-axis.
AC = AB + BC = 14km[40o] + 19km[90o].

X = 14*sin40 + 19*sin90 = 28 km.
Y = 14*Cos40 + 19*Cos90 = 10.7 km.

a. AC = sqrt(X^2+Y^2).

b. TanA = X/Y.

Answer 2

b. Cos40 = d/14.
d = 14*Cos40 = 10.7 km = Vertical distance from A.