helpp plz...trigo?

Question

SOLVE
1. Given the difference x(tan45*-tan30*)=6



2. Solve the following right triangles:

im taking so USE a/sinA=b/sinB=c/sinC



3. In a right triangle, given a= 32.8 ft. and A=28° 30’ 15’’,solve for B and C
taking now use a/sinA=b/sinB=c/sinC

Answer 1

1.) Tan 45 = 1
tan 30 = 1/√3
So:
x(tan 45 - tan30) = 6
x(1 - 1/√3) = 6
x((√3 - 1)/3) = 6
x = 6 * 3 / (√3 - 1)
x = 18 / (√3 - 1) = 24.588

2.) You can't find the lengths of the sides, because you have to have at least one side's length given to use Law of Sines.

3.)
Since you're given a = 32.8 ft, you can now use the law of sines:
First, find c, since sin of c is easy.
a/sin a = 32.8 / sin (28°30'15") = c/sin c = c/1
sin (28°30'15") = sin 28 1815/3600 = 0.47722
c = 32.8/0.47722 = 68.7 ft.

Now, find b:
b/sin b = b/ sin 61°29'45" = 68.7 ft.
b = 68.7 ft. * sin 61°29'45" (can be rewritten as: sin 61 1785/3600)
b = 68.7 ft. * 0.87878 = 60.4 ft.