CPhill

I think that you want  n(A U B)  in (5)

If so, we have

n(A U B)  =  n(A) + n(B)  -  n(A ∩ B)

n(A U B)  =   7    +   12    -       5        =   14

Here's (4)

Here's (3)  ....I'm assuming that the "16"  in set "F" is supposed to be "15"

Here's (2)

Here's (1)

4cos^2θ = 6 - 9sin θ

4(1 -sin^2 θ) = 6 - 9sinθ

4 - 4sin^2θ  = 6 - 9sin θ     re-arrange as

4sin^θ - 9sinθ + 2  = 0   factor as

(4sinθ - 1) (sin θ  - 2)  = 0

Set each factor to 0.........the second one will have no solution since  sinθ  = 2  is impossible

4sinθ  = 1

sinθ  = 1/4    

And this happens at about 14.48° + n*360°   and at about 165.52° + n360°.....where n is an integer

sin(a/2)  =  ± √ [ (1 - cos(a))  / 2 ]     (1)

sin^2a + cos^2a = 1

(25/169) + cos^2a  = 1

cos^2a  =  1 - (25/169)

cos^2a = 144/169

cosa = ± 12/13

"a" will be an angle in the first or second quadrant since the sine is positive in both....so  "a/2"   will be in the first quadrant..and the sine  and cosine will be positive here....so 

a =  sin-1(5/13)  ≈ 22.62°   or   ≈ 157.38°

So.....a/2  ≈  11.31°    or  a/2   ≈ 78.69°

However,  "a/2"  falls into the first quadrant in both cases, so the positive root  in (1)  is taken

So.....using  (1) and assumng that a ≈ 22.62° :  sin(a/2)  =  + √ [ (1 - (12/13))  / 2 ]  ≈  0.196

[Note that "a" falls in the first quadrant so the positive value of the cosine is used in the identity]

Verify for yourself that the  sin (11.31°) ≈  0.196

And using (1) and assuming that a ≈ 157.38° : sin(a/2)  =  + √ [ (1 - (-12/13))  / 2 ]  ≈ 0.9806

[Note that "a" occurs in the second quadrant here, so the negative value of the cosine is used in the identity ]

And verify that sin (78.69°)  ≈ 0.9806

[30] / [1/2]  + 10  =

[30 / 1 ] * [2/1]  + 10

60/ 1  + 10 =

60 + 10  =

70

(3)    1 acre = 43,560 ft^2 * 144 in^2 /1 ft^2   = 6,272,640 in^2

So 18 acres = 112,907,520 in^2

So the linear scale of the map  =   √1000 : √112,907,520  

So....the actual length of the street will be

15cm * √ [112,907,520 / 1000]  ≈ 5040 cm  ≈ 50.4m