JAMB 1987 · UME · Q36✱✱

In the figure, two parallel lines are cut by a transversal, and the two interior angles on the same side of the transversal are bisected. The bisectors meet at the angle marked xx. Find the value of xx.

aabbx
Worked solution (try it first)
  1. Co-interior angles between parallel lines add up to 180∘180^\circ.
  2. The lower angle is 2a2a and the upper is 2b2b, so 2a+2b=180∘2a + 2b = 180^\circ.
  3. Halve it: a+b=90∘a + b = 90^\circ.
  4. The two bisectors and the transversal form a triangle with angles aa, bb and xx, so x=180∘−(a+b)=90∘x = 180^\circ - (a + b) = 90^\circ, option C.

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