WAEC - Mathematics (1996 - No. 28) | ExamPlay

WAEC - Mathematics (1996 - No. 28)

In the diagram above, TRQ is a straight line. Find p, if p = \(\frac{1}{3}\)(a + b + c)

45^o

60^o

90^o

120sup>o

150^o

Explanation

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

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