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

The answer is the first one (A)

Answer 2

`AC=(11√(3))/(3)`

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)=(opposite)/(adjacent)`

`\tan\left(B\right)=(AC)/(BC)`

`\tan\left(30^o\right)=(AC)/(11)`

`(1)/(√(3))=(AC)/(11)`

`(11)/(√(3))=AC`

`AC=(11)/(√(3))`

or

`AC=(11)/(√(3))\cdot(√(3))/(√(3))`

or

`AC=(11√(3))/(3)`

Hence final answer is
`AC=(11√(3))/(3)`.