Question

15.Two lines AB and CD intersect at point
O, such that angleBOC + angleAOD = 280° then
angleAOC + angleBOD=?*
O 180°
O 40°
O 80°
O 140°

hey guys plz answer fast....​

Answer 1

${\bigstar}$GIVEN${\bigstar}$

∠BOC + ∠AOD = 280° ➜(1)

${\bigstar}$TO FIND${\bigstar}$

∠AOC + ∠BOD = ?

${\bigstar}$ SOLUTION${\bigstar}$

➠sum of all the angles at O = 360°

➠∠AOC + ∠BOC + ∠BOD + ∠AOD = 360°

➠280° + ∠BOD + ∠AOD = 360°

➠∠BOD + ∠AOD = 360° - 280°

➠∠BOD + ∠AOD = 80° ➜(2)

EXTRA :

VERTICALLY OPPOSITE ANGLES :

$when \: two \: lines \: each other \: \\ they \: form \: two \: sets \: of \: equal \: opposite angles \: \\ which \: is \: said \: to be \: vertically \: opposite \: angles </p><p>$

here ,in this problem ,

vertically opposite angles are

• ∠BOC = ∠AOD (3)
• ∠AOC = ∠BOD (4)

by applying this condition in (1) and (2)

(1)➜∠BOC + ∠AOD = 280°

➠∠BOC +∠BOC = 280°

➠2∠BOC = 280°

➠∠BOC = 140°

also

➠∠AOD = 140° [from (3)]

➠(2)➜ ∠BOD + ∠AOD = 80°

➠∠BOD +∠ AOD = 80°

➠∠BOD + ∠BOD = 80°

➠2∠BOD = 80°

➠∠BOD = 40°

also

➠∠AOD = 40° [from (4)]

therefore

angles are

• ∠BOC = 140°
• ∠AOD = 140°
• ∠AOC = 40°
• ∠BOD = 40°