Answer:
Step-by-step explanation:
[tex]BD=\sqrt{10^2-2^2}\\ BD=\sqrt{100-4} \\BD=\sqrt{96}\\ BD=4\sqrt{6}[/tex]
Similar triangles
[tex]tan(\alpha )=tan(\alpha )\\\\ \\\frac{BC}{BA} =\frac{BD}{AD} \\\\\frac{BC}{10} =\frac{4\sqrt{6} }{2}\\\\BC=2\sqrt{6} *10\\BC=20\sqrt{6}[/tex]
Pythagorean theorem
[tex]CD=\sqrt{BC^2-BD^2} \\CD=\sqrt{(20\sqrt{6} )^2-(4\sqrt{6} )^2} \\CD=\sqrt{400(6)-16(6)} \\CD=\sqrt{2304} \\CD=48[/tex]
