Question 1198040
<font color=black size=3>
Here's a slight alternative approach to what the tutor Edwin McCravy has written. 
I'll be using his diagram and the notation he set up of u = tan(38), v = tan(22)


The slight different approach is to let
BG = x
BA = 46
AG = x-46


tan(angle) = opposite/adjacent
tan(angle TBG) = TG/BG
tan(22) = y/x
y = x*tan(22)


tan(38) = y/(x-46)
tan(38) = x*tan(22)/(x-46) .... plug in y = x*tan(22)
u = x*v/(x-46) .... make replacements for u and v
u(x-46) = xv
ux - 46u = xv
ux-xv = 46u
x(u-v) = 46u
x = 46u/(u-v)
x = 46*tan(38)/(tan(38)-tan(22))
x = 95.263733005284, which is the approximate length of segment BG.


y = x*tan(22)
y = 95.263733005284*tan(22)
y = 38.489046505093
y = 38.5, which is the approximate length of segment TG.


Answer: <font color=red>E) 38.5 feet</font>


---------------------------------------------------------------------------


Yet another approach


Refer to the diagram Edwin McCravy has drawn.


Angle TAG = 38 degrees.
angle TAB = 180-angle TAG = 180-38 = 142 degrees
This is angle A of triangle TAB.


Focus on triangle TAB
The interior angles T, A, B must add to 180 degrees.
T + A + B = 180
T + 142 + 22 = 180
T + 164 = 180
T = 180 - 164
T = 16


Use the law of sines to find side 'a' which is opposite angle A.


sin(A)/a = sin(T)/t
sin(142)/a = sin(16)/46
46*sin(142) = a*sin(16)
a = 46*sin(142)/sin(16)
a = 102.74524576336
This is the approximate length of segment TB.


Then focus on triangle TBG to say the following:
sin(angle) = opposite/hypotenuse
sin(angle TBG) = TG/TB
sin(22) = y/102.74524576336
y = 102.74524576336*sin(22)
y = 38.489046505093
y = 38.5



Answer: <font color=red>E) 38.5 feet</font>
</font>