Answer:
see below
Step-by-step explanation:
[tex]3(i)\\\\AD=?\\\\\cos (\angle BDA)=\frac{AD}{BD}\\\\\cos (27)=\frac{AD}{3}\\\\AD=3\cos 27\\\\AD\approx 2.67[/tex]
[tex]3(ii)\\\\\sin (\angle BCD) = \frac{BD}{CD}\\\\\sin(41)=\frac{3}{CD}\\\\CD=\frac{3}{\sin 41}\\\\CD\approx 4.57[/tex]
I'm sorry but i have to go I'll come back in half an hour and finish answering
