Answer:
Part A) The length of the hypotenuse of triangle ABC is
[tex]AB=\frac{40\sqrt{3}}{3}\ units[/tex]
Part B) The length of the shorter leg of triangle ABC is
[tex]AC=\frac{20\sqrt{3}}{3}\ units[/tex]
Part C) The length of the longer leg of triangle ABC is
[tex]BC=20\ units[/tex]
Step-by-step explanation:
see the attached figure to better understand the problem
Step 1
Find the length of the longer leg of triangle ABC (side BC)
we know that
In the right triangle DBC
[tex]sin(30\°)=\frac{DC}{BC}[/tex]
substitute the values and solve for BC
[tex]sin(30\°)=\frac{10}{BC}[/tex]
[tex]BC=10/sin(30\°)=20\ units[/tex]
Step 2
Find the length of the shorter leg of triangle ABC (side AC)
we know that
In the right triangle ACD
[tex]sin(60\°)=\frac{DC}{AC}[/tex]
substitute the values and solve for AC
[tex]sin(60\°)=\frac{10}{AC}[/tex]
[tex]AC=10/sin(60\°)=\frac{20\sqrt{3}}{3}\ units[/tex]
Step 3
Find the length of the hypotenuse of triangle ABC (side AB)
Applying the Pythagoras theorem
[tex]AB^{2}=BC^{2}+AC^{2}[/tex]
we have
[tex]AC=\frac{20\sqrt{3}}{3}\ units[/tex]
[tex]BC=20\ units[/tex]
substitutes
[tex]AB^{2}=20^{2}+(\frac{20\sqrt{3}}{3})^{2}[/tex]
[tex]AB^{2}=400+(\frac{400}{3})[/tex]
[tex]AB^{2}=(\frac{1600}{3})[/tex]
[tex]AB=\frac{40\sqrt{3}}{3}\ units[/tex]