Question

geometry help

Answer 1

@dr0zier99

Answer 2

wouldnt it be 40?

Answer 3

explain?

Answer 4

You have two different 30-60-90 triangles.
Find AC and BC using the triangles.
The answer is AC - BC

Answer 5

i think that the answer is b

AC: \(\normalsize\tan(60) = \frac{AC}{10 \sqrt{3}} \implies \sqrt{3} = \frac{AC}{10 \sqrt{3}}\) \(\normalsize\ AC = 10 \sqrt{3} \times \sqrt{3} = 30\) BC: \(\normalsize\ tan(30) = \frac{BC}{10 \sqrt{3}} \implies \frac{1}{\sqrt{3}} = \frac{BC}{10 \sqrt{3}} \implies BC = 10\) \(\ AB = 30 - 10 = 20\)

Answer 6

is that write

Answer 7

wow.. i'm so confused right now

Answer 8

i say that the answer is b

Answer 9

In a 30-60-90 triangle, the long leg is \(\sqrt 3\) times longer than the short leg.

Answer 10

so 20 feet?

Answer 11

i would think so

Answer 12

me also