(a) see figure a above, In the diagram, MN || ST, NP || QT and < STQ = 70^\circ. Find x.
(b) see figure b above, In the diagram above, AC is a straight line, |BC| = |BD|, \(\stackrel\frown{BCD} = 50^\circ\) and \(\stackrel\frown{BAD} = 55^\circ\). Find \(\stackrel\frown{BDA}\).
Attached Figure / Diagram
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a) see figure above a \(< QTR = < TRP = 70^\circ\) (alternate angle)
\(< MNR = < TRP = 70^\circ\) (alternate angle; MN || ST)
\(< MNR + < PNM = 180^\circ\)
\(70^\circ + < PNM = 180^\circ \implies < PNM = 110^\circ\)
\(\therefore x = 110^\circ\)
(b)see figure b above \(< BDC = < BCD = 50^\circ\) (BD = BC)
\(\therefore \Delta BDC = isosceles\)
\(< ABD = 2(50^\circ) = 100^\circ\) (exterior angle)
\(55^\circ + < BDA + 100^\circ = 180^\circ\)
\(< BDA = 180^\circ - 155^\circ = 25^\circ\)
Source: Myschool Past Questions Archive (Record #53387)