Question

plzzz solve it fast.......

Answer 1

Answer:

Step-by-step explanation:

mark the intersections 'O'

angle BOC + angle BOA + angle BOD = 180° ( linear pair axiom)

$\\ \frac {5x}{10} - 5 + \frac {x}{2} + x + 9 = 180° \\ \\ \frac {5x - 50}{10} + \frac {x}{2} + x = 171° \\ \frac {5(x-10)}{10} + \frac {x}{2} + x = 171° \\ \frac {2x - 10}{2} + x = 171° \\ x - 5 + x = 171° \\ x = \frac {176}{2} = 86 \\ \\ angle BOC = 5( \frac {86}{2} - 1) = \frac {84}{2} = 42° \\ \\ angle BOA = \frac {86}{2} = 43° \\ \\ angle BOD = 86 + 9 = 95° \\ \\ Since AOC \: is \: exterior \: angle \: of \: traingle \: AO' \: in \: which \: angle \: a \: and \: b \: are \: included \: , so \\ angle AOC = a + b \: \: \: \: (Exterior \: angle \: a \: triangle \: is \: equal \: the \: sum \: of \: two \: opposite \: interior \: angles ) \\ \\ angle BOC + angle BOA = a + b \\ \\ 42 + 43 = a + b \\ \\ a + b = 85°$