JAMB 1986 · UME · Q35

Elevation, depression & bearings

A girl walks 45 m in the direction $050^\circ$ from a point $Q$ to a point $X$ . She then walks 24 m in the direction $140^\circ$ from $X$ to a point $Y$ . How far is she then from $Q$ ?

Worked solution (try it first)

At $X$ the direction back to $Q$ is $050^\circ + 180^\circ = 230^\circ$ , and she sets off on $140^\circ$ .
These differ by $90^\circ$ , so the angle at $X$ is a right angle.
By Pythagoras, $QY^2 = 45^2 + 24^2$ , which is $2025 + 576 = 2601$ .
So $QY = \sqrt{2601} = 51$ m, option C.

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