WAEC - Mathematics (1990 - No. 29)
In the diagram above, |PQ| = |PR| = |RS| and ∠RPS = 32°. Find the value of ∠QPR
72o
64o
52o
32o
26o
ব্যাখ্যা
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°