JAMB 1997 · UME · Q26

Angles, triangles & polygons

In the figure, $FK \parallel GR$ and $FH = GH$ , $\angle RFK = 34^\circ$ and $\angle FGH = 47^\circ$ . Calculate the angle marked $x$ ( $\angle HFR$ ).

Worked solution (try it first)

$FH = GH$ , so triangle $FGH$ is isosceles with equal angles at $G$ and $F$ : $\angle GFH = 47^\circ$ .
$FK \parallel GR$ , so $\angle GFK$ and $\angle FGH$ are co-interior: $\angle GFK = 180^\circ - 47^\circ$
$= 133^\circ$ .
$\angle GFK$ is made up of $47^\circ + x + 34^\circ$ , so $x = 133^\circ - 81^\circ = 52^\circ$ , option B.

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