Please solve que no. 9
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<A+<B+<C+<D =360<A+<B +100+60 =360
<A+<B +160=360
<A+<B =360-160
<A+ <B =200
DIVIDE EACH TERM WITH 2
<A/2 +<B/2 =200/2
<PAB +<PBA =100-----(1)
<PAB +<PBA+<APB =180 [ SUM OF the angles in a triangle PAB=180
100 +<APB = 180 [FROM (1)]
<APB = 180-100=80