Question

Please solve que no. 9

Answer 1

<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