(a) In the diagram, MN || ST, NP || QT and < STQ = 70°. Find x.
(b) In the diagram above, AC is a straight line, |BC| = |BD|, (stackrelfrown{BCD} = 50°) and (stackrelfrown{BAD} = 55°). Find (stackrelfrown{BDA}).
Explanation
(a) (< QTR = < TRP = 70°) (alternate angle)
(< MNR = < TRP = 70°) (alternate angle; MN || ST)
(< MNR + < PNM = 180°)
(70° + < PNM = 180° implies < PNM = 110°)
(therefore x = 110°)
(b) (< BDC = < BCD = 50°) (BD = BC)
(therefore Delta BDC = isosceles)
(< ABD = 2(50°) = 100°) (exterior angle)
(55° + < BDA + 100° = 180°)
(< BDA = 180° – 155° = 25°)