WAEC 2019 · Paper 1 · Q43

In the diagram, PQPQ is parallel to RSRS, ∠QFG=105∘\angle QFG = 105^\circ and ∠FEG=50∘\angle FEG = 50^\circ. Find the value of mm.

50°105°mnEFGPQRS
Worked solution (try it first)
  1. Angles on the straight line PQPQ: ∠EFG=180∘−105∘\angle EFG = 180^\circ - 105^\circ
    =75∘= 75^\circ.
  2. The angles of triangle EFGEFG add up to 180∘180^\circ: m=180∘−50∘−75∘m = 180^\circ - 50^\circ - 75^\circ
    =55∘= 55^\circ, option D.

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