Start by using trig to find the length of the line LJ
The triangle KJL (big right angled triangle) has been given the following dimensions
Hypotenuse =

The adjacent angle is 30 degrees
Since we have the hypotenuse and the angle we must use the equation
opposite = Sin(angle) x Hypotenuse
Opposite= sin30 x

Opposite=

Therefore line LJ is

Now look at the smaller right angled triangle (LMJ)
Hypotenuse is the line LJ which is

The adjacent angle is 45
Since we have hypotenuse and angle we must use the equation opposite = sin(angle) * h
therefore
x=

* sin45= 4
6 ft for each side, bc area is side times side, which 6x6=36
D is the answer to this question
1)
I:y=3x-4
II:9x-3y=14
substitute y into II:
9x-3*(3x-4)=14
9x-9x+12=14
12=14
this is obviously not equal so there is no solution, the lines are parallel
2)
I:y=4x+6
II:5x-y=6
substitute y into II:
5x-(4x+6)=6
5x-4x-6=6
x=12
substiute x into II:
5*12-y=6
-y=6-60
-y=-54
y=54
the solution is (12,54)
A) <DAE = 180-126 = 54
b) <EBC = 90 - 48 = 42
c) <BAE = 180-48-54=78