CPhill

tan A =  2 sin (A/2)cos (A/2)  = 2 * 1/sqrt 17 * 4/sqrt 17 =  8/17  =  24 / 51

tan B = 2 sin (B/2) cos (B/2) = 2 * 1/sqrt 10 * 3/sqrt 10 =  6/10 = 3/5  = 24/ 40

AB = ( 51 + 40) =  91

AC =  sqrt [ 24^2 + 51^2] = 3sqrt (353)

sin B =  24/ [8 sqrt (34)] = 3 /sqrt 34

Law of Sines

sin C  / AB =  sin B / AC

sin C /  91 = [3/sqrt 34] / [ 3sqrt (353)]

sin C  = 91 / sqrt [ 34* 353] =  91 / sqrt [12002]

cos C  =  sqrt [ 12002 - 91^2  ] / sqrt [ 12002[  =  61/ sqrt [12002]

tan C =  91/61

arctan (91/61)  = C =  56.16°   or   180 - 56.16  = 123.84°

tan (C/2)  =  tan (123.84 / 2) ≈  1.8744

Call the original price = P

He pays   ( P -.20P) = .80P for the first pair of shoes  and .50P  for the second pair

So

.80P + .50P  =  156

1.3P =156

P = 156 / 1.3  =  120

The first pair cost  120 (.80)  =  $96

And the second piar cost  120 (.50) = $60

4/5 - 1/2 - (1/4)(4/5)  = 

4/5 - 1/2 - 4/20

4/5 - 1/2 - 1/5

3/5 - 1/2

6/10 - 5/10

1/10 kg was left

Call the total number of  coins = C

Originally, (2/7) of the coins were 50 cent  coins and (5/7)were dollar coins

When his mother gives him 10 one dollar coins he has  10 + (5/7)C dollar coins

When his mother gives him 5 dollars of 50 cent coins he  has 10 of them in addition to the (2/7)C worth of  50 cent coins

So

In the end, he had twice the number of  one dollar coins as  the number of  fifty cent coins 

2 [ (2/7C + 10 ] =  [ 5/7C + 10]     simplify

(4/7)C + 20  = (5/7)C + 10  

20 -10  =  (5/7)C - (4/7)C

10 = (1/7)C

70 = C = the number he had at first

tan (theta)  =  sqrt [2^2 - 1^2] / 1  =  sqrt 3

cos α =  -2 / sqrt [ 2^2 + 1^2]   =   -2 / sqrt 5

cos A = 2/5     →  tan A =  sqrt 21 / 2 →  sqrt (63) / sqrt (12)

cos B = 1/2    →   tan B =  sqrt 3 / 1  →   sqrt (63) / sqrt (21)

AC =  sqrt (63 +12)  = sqrt (75)

Law of Sines

sin C / AB =  sin B / AC

sin C / ( sqrt (12) +sqrt (21))  =  (sqrt (3) / 2) / sqrt (75)

sin C / ( sqrt (12) + sqrt (21))  =  1 / [ 2 sqrt 25 ]

sin C / (sqrt (12) + sqrt (21)) = 1/10

sin C =  [ sqrt (12) + sqrt (21) ] /10

cos C  =   sqrt [ 10^2 - ( sqrt 12 + sqrt 21)^2] /10   =

sqrt [ 100 - ( 12 + 21 + 2sqrt (12)sqrt (21) ] /10   =

sqrt [ 67 - 12sqrt 7] /10

AB  = sqrt [ (4K)^2 - (3K)^2 ] =  K sqrt  7

sin C =   K sqrt 7  / 4K  =  sqrt (7) / 4 =  0.66

Any line of the form y = ax  will go through the origin and bisect the area of the square

So, b = 0

Slope of diagonal AC =  [ 6-3] / [ 3 -7]  = -3/4

Equation of line through AC is

y =(-3/4)(x -3) + 6

y = (-3/4)x + 9/4 + 6

y = (-3/4)x + 33/4       (1)

Slope of diagonal  BD = [ 6-1] / [ 9-1] = 5/8

Equation of line through BD is

y = (5/8)(x -1) + 1

y = (5/8)x + 3/8         (2)

Set (1)  = (2)   to find the x intersection pt

(-3/4)x + 33/4  = (5/8)x + 3/8

33/4 - 3/8 = (5/8 + 3/4)x

63/8 = (11/8)x

x = 63/11  

y = (5/8) ( 63/11 ) + 3.8

y = 87/22

Intersection pt =  ( 63/11, 87/22 )