Question

In the right triangle shown, \angle B = 60^\circ∠B=60

Answer 1

Answer:

Part 1) $AC=6\sqrt{3}\ units$

Part 2) $AC=12\sqrt{3}\ units$

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

$sin(B)=\frac{AC}{AB}$ ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values

$sin(60^o)=\frac{AC}{12}$

Remember that

$sin(60^o)=\frac{\sqrt{3}}{2}$

substitute

$\frac{\sqrt{3}}{2}=\frac{AC}{12}$

$AC=6\sqrt{3}\ units$

Problem 2

The hypotenuse is the segment BC  and the right angle is A

we know that

In the right triangle ABC

$tan(B)=\frac{AC}{AB}$ ---> by TOA (opposite side divided by the adjacent side)

substitute the given values

$tan(60^o)=\frac{AC}{12}$

Remember that

$tan(60^o)=\sqrt{3}$

substitute

$\sqrt{3}=\frac{AC}{12}$

$AC=12\sqrt{3}\ units$