The points X, Y and Z are located such that Y is 15 km south of X, Z is 20 km from X on a bearing of 270″.
Calculate, correct: (a) two significant figures, |YZ|
(b) The nearest degree, the bearing of Y from Z
Explanation
|YZ|\(^2\) = 15\(^2\) + 20\(^2\)
= 225 + 400 = 625
YZ = √625
YZ = 25
(b) tan θ = \(\frac{15}{20}\)
tan θ = 0.75
θ = tan -1 (0.75)
= 36.87°
Bearing of Y from Z = (90 + 36.87)° = 126.87°
= 127°