WAEC - Mathematics (1996 - No. 28)
In the diagram above, TRQ is a straight line. Find p, if p = \(\frac{1}{3}\)(a + b + c)
45o
60o
90o
120sup>o
150o
ব্যাখ্যা
a + b + c + p = 180 (sum of angles on a straight line)
But p = \(\frac{1}{3}\)(a + b + c), Then, (a + b + c) = 3p
3p + p = 180
4p = 180
p = \(\frac{180}{4}\) = 45º