Question

Pls help with this Geometry Maths question, thanks

Answer 1

Hi,

Explanation:

`triangles\ isoceles\\\angle{ADO}=\angle{DAO}=28^o\\\angle{OCA}=\angle{OAC}=24^o\\\\\angle{AOB}=2*\angle{ADB}=2*28^o=56^o\\\\\angle{AOC}=180^o-2*\angle{OAC}=180^o- 48^o=132^o\\\angle{BOC}=132^o-56^o=76^o\\\angle{OBC}=(180^o- 76^o)/(2)=52^o\\\angle{ABC}=62^o+52^o= 114^o\\\angle{ACB}=52^o -24^o=28^o\angle{CAB}=62^o -24^o=38^o`