In the diagram, O is the centre of the circle. TAQ, PAB and PQR are straight lines. Given that ∡APO = 14°, ∡OBA = 40°, calculate
(a) ∡ARB,
(b) ∡AOP
(c) ∡TAB.
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Answer:
(a) ∡BAO = ∡ABO = 40° (isosceles triangle)
∡AOB = 180° - 40° - 40° = 100° (angles in a triangle add up to 180°)
∡ARB = 100° / 2 = 50° (angle at centre = 2 * angle at circumference)
=> Hint: Draw a line from R to B to see it clearer
(b) ∡OAP = 180° - 40° = 140° (angles on a straight line)
∡AOP = 180° - 140° - 14° = 26° (angles on a triangle add up to 180°)
(c) ∡OAQ = (180° - 26°) / 2 = 77° (isosceles triangle)
∡TAB = 180° - 77° - 40° = 63° (angles on a straight line)
