WAEC - Mathematics (1990 - No. 29)

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

72^o

64^o

52^o

32^o

26^o

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°