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

Mathematics
WAEC 1996

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

  • A. 45o
  • B. 60o
  • C. 90o
  • D. 120sup>o
  • E. 150o
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Correct Answer: Option A
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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