Question

1. In ABC, C is a right angle and BC = 11. If B = 30°, find AC. (1 point)

PLEASE HELP I HAVE ONE DAY TO COMPLETE THE CONNECTION PRECALCULUS B UNIT 8: SEMESTER EXAM! I would do ANYTHING PLEASE?!

Answer 1

Answer:

$AC=\frac{11\sqrt{3}}{3}$

Step-by-step explanation:

Given that triangle ABC is a right angle triangle. Where angle C is a right angle. Also we have been given that BC = 11, B = 30°. Now we need to find the value of AC.

Apply formula:

$\tan\left(\theta\right)=\frac{opposite}{adjacent}$

$\tan\left(B\right)=\frac{AC}{BC}$

$\tan\left(30^o\right)=\frac{AC}{11}$

$\frac{1}{\sqrt{3}}=\frac{AC}{11}$

$\frac{11}{\sqrt{3}}=AC$

$AC=\frac{11}{\sqrt{3}}$

or

$AC=\frac{11}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}$

or

$AC=\frac{11\sqrt{3}}{3}$

Hence final answer is $AC=\frac{11\sqrt{3}}{3}$.

Answer 2

The answer is the first one (A)