In the diagram above, |PQ| = |PR| = |RS| and ∠RPS = 32°. Find the value of ∠QPR

  • A 72o
  • B 64o
  • C 52o
  • D 32o
  • E 26o
waec 1990mathematics

The correct answer is C. 52o

From the figure, < PSR = 32° (base angles of an isos. triangle)

\(\therefore\) < PRS = 180° - (32° + 32°) = 116° (sum of angles in a triangle)

< QRP = 180° - 116° = 64° (angle on a straight line)

< PQR = 64° (base angles of an isos. triangle)

< QPR = 180° - (64° + 64°) = 52°

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