Answer:
Part 1) [tex]AC=6\sqrt{3}\ units[/tex]
Part 2) [tex]AC=12\sqrt{3}\ units[/tex]
Step-by-step explanation:
I will analyze two problems
see the attached figure to better understand the problem
Problem 1
The hypotenuse is the segment AB and the right angle is C
we know that
In the right triangle ABC
[tex]sin(B)=\frac{AC}{AB}[/tex] ---> by SOH (opposite side divided by the hypotenuse)
substitute the given values
[tex]sin(60^o)=\frac{AC}{12}[/tex]
Remember that
[tex]sin(60^o)=\frac{\sqrt{3}}{2}[/tex]
substitute
[tex]\frac{\sqrt{3}}{2}=\frac{AC}{12}[/tex]
[tex]AC=6\sqrt{3}\ units[/tex]
Problem 2
The hypotenuse is the segment BC and the right angle is A
we know that
In the right triangle ABC
[tex]tan(B)=\frac{AC}{AB}[/tex] ---> by TOA (opposite side divided by the adjacent side)
substitute the given values
[tex]tan(60^o)=\frac{AC}{12}[/tex]
Remember that
[tex]tan(60^o)=\sqrt{3}[/tex]
substitute
[tex]\sqrt{3}=\frac{AC}{12}[/tex]
[tex]AC=12\sqrt{3}\ units[/tex]
