Question

# In the figure, in ∆ABC, the angle bisectors of ∠B and ∠C meet at a point O. Find the measure of ∠BOC. Figure

Open in App
Solution

## $\mathrm{In}∆ABC,\mathrm{using}\mathrm{angle}\mathrm{sum}\mathrm{property},\mathrm{we}\mathrm{have}:\phantom{\rule{0ex}{0ex}}\angle CAB+\angle ABC+\angle BCA=180°\phantom{\rule{0ex}{0ex}}⇒70°+\angle ABC+\angle BCA=180°\phantom{\rule{0ex}{0ex}}\therefore \angle ABC+\angle BCA=110°-----\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}BO\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle ABC.\phantom{\rule{0ex}{0ex}}⇒\angle OBC=\frac{1}{2}\angle ABC\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}CO\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle BCA.\phantom{\rule{0ex}{0ex}}⇒\angle OCB=\frac{1}{2}\angle BCA\phantom{\rule{0ex}{0ex}}\therefore \angle OBC+\angle OCB=\frac{1}{2}\left(\angle ABC+\angle BCA\right)-----\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{In}∆OBC,\mathrm{using}\mathrm{angle}\mathrm{sum}\mathrm{property},\mathrm{we}\text{get}:\phantom{\rule{0ex}{0ex}}\angle BOC+\angle OBC+\angle OCB=180°\phantom{\rule{0ex}{0ex}}\mathrm{Using}\left(2\right)\mathrm{we}\text{get}:\phantom{\rule{0ex}{0ex}}\angle BOC+\frac{1}{2}\left(\angle ABC+\angle BCA\right)=180°\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{using}\left(1\right)\mathrm{we}\text{get}:\phantom{\rule{0ex}{0ex}}\angle BOC+\frac{1}{2}\left(110°\right)=180°\phantom{\rule{0ex}{0ex}}⇒\angle BOC+55°=180°\phantom{\rule{0ex}{0ex}}⇒\angle BOC=180°-55°=125°\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\angle BOC=125°\phantom{\rule{0ex}{0ex}}$

Suggest Corrections
2
Related Videos
Basic Properties of a Triangle
MATHEMATICS
Watch in App