Question

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°

Answer 1

`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}`