37.0k views
4 votes
In ΔABC the angle bisectors drawn from vertices A and B intersect at the point D. Find m∠ADB if:

m∠A = 50°, m∠B = 100°

In ΔABC the angle bisectors drawn from vertices A and B intersect at the point D. Find-example-1
User Mhb
by
7.7k points

1 Answer

4 votes

Answer:


m\angle ADB=105^o

Explanation:


\mathrm{From\ figure,}\\\mathrm{\angle A=2\angle BAD\ \ \ \ \ \mathrm{[AD\ is\ angle\ bisector.]} }\\\mathrm{or,\ 50^o=2\angle BAD}\\\mathrm{\therefore\ \angle BAD=25^0}\\\mathrm{Also,\ \angle B=2\angle ABD\ \ \ \ \ \ [BD\ is\ angle\ bisector.]}\\\mathrm{or,\ 100^o=2\angle ABD}\\\mathrm{or,\ \angle ABD=50^o}\\


\mathrm{Now\ in\ \triangle ABD,}\\\mathrm{\angle ABD+\angle BAD+\angle ADB = 180^o\ \ \ \ [Sum\ of\ \ interior\ angles\ of\ triangle\ is\ 180^o.]}\\\mathrm{or,\ 50^o+25^o+\angle ADB=180^o}\\\mathrm{or,\ 75^o+\angle ADB=180^o}\\\mathrm{\therefore\ m\angle ADB=105^o}

User Drussey
by
8.3k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories