find the value of each variable help fast

Answer:[tex]a=10\sqrt{3}\ units[/tex],[tex]b=5\sqrt{3}\ units[/tex],[tex]c=15\ units[/tex],[tex]d=5\ units[/tex]Step-by-step explanation:see the attached figure with letters to better understand the problemstep 1Find the value of bIn the right triangle BCD[tex]sin(60\°)=\frac{b}{10}[/tex][tex]b=sin(60\°)(10)[/tex]Remember that[tex]sin(60\°)=\frac{\sqrt{3}}{2}[/tex]so[tex]b=\frac{\sqrt{3}}{2}(10)[/tex][tex]b=5\sqrt{3}\ units[/tex]step 2Find the value of dIn the right triangle BCD[tex]cos(60\°)=\frac{d}{10}[/tex][tex]d=cos(60\°)(10)[/tex]Remember that[tex]cos(60\°)=\frac{1}{2}[/tex]so[tex]d=\frac{1}{2}(10)[/tex][tex]d=5\ units[/tex]step 3Find the value of aIn the right triangle ABD[tex]sin(30\°)=\frac{b}{a}[/tex][tex]a=\frac{b}{sin(30\°)}[/tex]Remember that[tex]sin(30\°)=\frac{1}{2}[/tex][tex]b=5\sqrt{3}\ units[/tex]so[tex]a=\frac{5\sqrt{3}}{(1/2)}[/tex][tex]a=10\sqrt{3}\ units[/tex]step 4Find the value of cIn the right triangle ABD[tex]cos(30\°)=\frac{c}{a}[/tex][tex]c=(a)cos(30\°)[/tex]Remember that[tex]cos(30\°)=\frac{\sqrt{3}}{2}[/tex][tex]a=10\sqrt{3}\ units[/tex]substitute[tex]c=(10\sqrt{3})\frac{\sqrt{3}}{2}[/tex][tex]c=15\ units[/tex]therefore[tex]a=10\sqrt{3}\ units[/tex][tex]b=5\sqrt{3}\ units[/tex][tex]c=15\ units[/tex][tex]d=5\ units[/tex]