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 problem
step 1
Find the value of b
In 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 2
Find the value of d
In 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 3
Find the value of a
In 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 4
Find the value of c
In 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]
