Question

Trying to find AB and BC.

Answer 1

I tried a few methods but none seem to work.

Answer 2

??

Answer 3

Just need some assistance on the right method. I used tan(30) = BC/5sqrt(3) to find BC,

Then I found DB to use Tan(45) = AB/DB to find AB, but when I subtract BC from AB, it seems to be wrong. I tried working it out again and got something different.

Answer 4

are u able to find BC ?

Answer 5

the fastest way to do this is using the special triangles :
http://www.mathwarehouse.com/geometry/triangles/right-triangles/images/picture-30-60-90-special-right-triangle.png

Answer 6

Just by looking at the diagram, you can say that \(BC = 5\)

Answer 7

\[\tan(30)= \frac{ BC }{ 5 \sqrt3 } = 5\]

\[Sin(30) = \frac{ 5 }{ BD } = \frac{ 5 }{0.5 } = 10; BD = 10\]

\[\tan(45) = \frac{ AB }{ 10 } = 1 * 10 = AB; AB = 10\]

So then I subtract BC from AB because I am trying to find the measurements of JUST AB, and I get 5. However, 5 isn't a solution. So where did I go wrong?

Answer 8

mistake !