Find the lengths of line segments BC, DE, and FG 
For this picture, I need to find the lengths of the line segments inside triangle FGA.
Given this is a 30-60-90 triangle, I know the ratio of side lengths can be simplified to 1; the square root of 3; and 2.
If every side is multiplied by 1.5, then the sides of triangle FGA are 1.5; 1.5*square root of 3; and 3.
1.5, therefore is the length of line segment FG. This also leads to believe the length of segment DF is 1.5.
How do you find the length of the other two line segments when you you’re only given the length of one side?
In other words, how can I find the lengths of line segments BC and DE, when I only know the lengths of line segments AB and AD?
Thanks for your help.
 A: Rather than relying on rules like the 30-60-90 rule, I recommend thinking of this in terms of the unit circle. You can set it up as 3 different triangles (ABC , ADE , AFG) and use known sides and the angle to solve for the length of each vertical component.
For ABC you have an angle of 30 degrees and a hypotenuse of 1. Sine of 30 degrees is equal to the y value (called line BC here) divided by the hypotenuse (AB = 1):
sin 30 = BC / 1
Solve algebraically from here to get BC.
Apply this same though process to the other two triangles to find the length of the vertical component. Hint: since you don't know the length of the hypotenuse for AFG you will not be able to use Sin 30 to solve.
A: Given
$\quad\angle A =30^\circ\\
   BC=\dfrac{AB}{2}=\dfrac{1}{2}=0.5, \\ 
   DE=\dfrac{AD}{2} = \dfrac{1.5}{2}=0.75$
$$\text{and }\quad 
AC=\sqrt{1^1-0.5^2}=0.5\sqrt{3}\\
\text{and }\quad
AE=\sqrt{1.5^2-0.75^2}=.75 \sqrt(3)
$$
Since  $\space AG=3AC=1.5\sqrt{3},\quad$
then $\space AF=3AB=3$
It then holds
$\quad FG=\dfrac{3}{2}=1.5$
A: Note the lengths given are along the hypotenuse. The legs whose length you want to find are opposite to the thirty degree angle. That means the ratio to their corresponding length of the hypotenuse is 1/2. The longer the hypotenuse, the longer that vertical leg.
AB=1, so BC=1/2. Now we continue along the hypotenuse AD is the new hypotenuse length so AD=1+0.5=1.5 . That means DE=0.75.
FG is a bit more complicated. We are not given the rest of the hypotenuse, but we are given the entirety of the base. Given the base of the right triangle and adjacent leg, the remaining leg divided by the base is the tangent. The tangent of a 30 degree angle is $1/\sqrt{3}$ and the base is $15\sqrt{3}$, so you multiply them to get FG=15.
