The sides AB and AC of triangle ABC have been produced to D and E respectively The bisectors of - Maths - Lines and Angles

Question

The sides AB and AC of triangle ABC have been produced to D and
E respectively The bisectors of angle CBD and angle BCE meet at O
If angle A = 40o, find angle BOC

Ajanta Trivedi · Accepted Answer

given: ABC is a triangle , such that $∠ B A C = 40 \deg$

let ∠ABC = x deg and ∠ACB = y deg.

sum of interior angles of a triangle is 180 deg.

therefore

$∠ A B C + ∠ A C B + ∠ B A C = 180 x + y + 40 = 180 x + y = 180 - 40 x + y = 140............ (1)$

therefore

∠DBC = 180-∠ABC

∠DBC = 180- x [angles made on the same side of a straight line are 180 deg]

∠ CBO = $\frac{1}{2} ∠ D B C$ [since BO is the angle bisector of ∠DBC]

therefore ∠ CBO = $\frac{1}{2} (180 - x) = 90 - \frac{x}{2}$ .............(2)

similarly ∠BCO = $90 - \frac{y}{2} .............. (3)$

now in the triangle BCO,

sum of interior angles are 180 deg

∠BOC = $180 - (∠ C B O + ∠ B C O)$

from eq(2) and eq(3):

$∠ B O C = 180 - (90 - \frac{x}{2} + 90 - \frac{y}{2}) = 180 - (180 - \frac{x}{2} - \frac{y}{2}) = 180 - 180 + \frac{x}{2} + \frac{y}{2} = \frac{x + y}{2} = \frac{140}{2} [f r o m e q (1)] = 70$

thus ∠BOC = 70 deg

hope this helps you.

cheers!!

The sides AB and AC of triangle ABC have been produced to D and E respectively. The bisectors of angle CBD and angle BCE meet at O. If angle A = 40o, find angle BOC.

The sides AB and AC of triangle ABC have been produced to D and E respectively. The bisectors of angle CBD and angle BCE meet at O. If angle A = 40^o, find angle BOC.