Question

given figure, BO and Co are the bisectors of the exterior angles meeting each other at O. If Angle A-70 degree
find angle BOC
​

Answer 1

Step-by-step explanation:

A

B C

m 0 l

the figure should be like that. but what a difficult question. but I ll solve it.

given: ∆ abc, bo is the bisector of external angle mbc and co is the bisector external angle lcb. (<=angle)

<mbc + <cba= 180 (linear pair/ straight angle)

<mbc= 180°-<cba -------1

similarly,

<lcb= 180°- <bca ---------2

adding 1 and 2

<mbc +<lcb= 180°-<cba + 180°-<bca

<mbc+<lcb= 360°-(<cba+<bca)

<mbc+<lcb=360° -( 180°- <a)

(explaination:- angle sum property, in ∆abc

<a+ <cba+<bca= 180°

<cba+bca= 180°-<a)

now,

<mbc+<lcb= 360°-180°+<a

<mbc+<lcb= 180°+<a

multiplying 1/2 on both side

1/2<mbc+1/2 <lcb= 1/2*180° +1/2 <a

<obc+<ocb=90 +1/2<a--------3

(explaination:- ob and oc are bisectors so <obc and <ocb are half of <mbc and < lcb)

now in ∆obc

<obc+<ocb+ <boc= 180°

<boc= 180°-(<obc+<ocb)

<boc= 180°-( 90+1/2<a)from equation 3

<boc= 180°-90-1/2<a

<boc= 90-1/2*70°

<boc=90-35°

<boc=55°

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