math questions help!!

(a) cos A - sinA
= AB/AC - BC/AC
= (AB - BC)/AC
= (BC + 1 - BC) / (root 2)
= 1/root 2

(b) (cos A - sinA)^2 = (1/root 2)^2
(cos A)^2 - 2(cos A)(sin A) + (sin A)^2 = 1/2
(sin A)^2 + (cos A)^2 - 2(cos A)(sin A) = 1/2
1 - 2(cos A)(sin A) = 1/2
- sin(2A) = - 1/2
sin(2A) = 1/2
2A = 30(degree) or [150(degree)]rejected
so, A = 15(degree)

(c) AC^2 = AB^2 + BC^2
(root 2)^2 = (BC + 1)^2 + BC^2
2 = BC^2 + 2BC + 1 + BC^2
2(BC^2) + 2BC - 1 = 0
BC = { -2 ± root [2^2 - 4(2)(-1)] } / [2(2)]
BC = ( -2 ± root 12 )/4
BC = ( -2 ± 2 * root 3 )/4
BC = ( -1 + root 3 )/2 or [( -1 - root 3 )/2]rejected
(as BC(length) cannot be negative)
so, BC = ( -1 + root 3 )/2

sin A = BC/AC
sin A = [ ( -1 + root 3 )/2 ] / (root 2)
sin A = ( -1 + root 3 ) / (2 * root 2)
sin A = [ ( -1 + root 3 )(root 2) ] / [ (2 * root 2)(root 2) ]
sin A = [ - root 2 + (root 3)(root 2) ] / (2 * 2)
sin A = [ - root 2 + root (3 * 2) ] / 4
sin A = [ - root 2 + root 6 ] / 4
sin A = (root 6 - root 2) / 4

(d) let the length of the perpendicular from B to AC be h
you may reference
http://www.analyzemath.com/Geometry_calculators/area_triangle_sine.html
(1/2)(AC)(AB)(sin A) = (1/2)(h)(AC)
(AC)(AB)(sin A) = (h)(AC)
(AB)(sin A) = h
h = (AB)(sin A)
h = (BC+1)(sin A)
h = [ ( -1 + root 3 )/2 + 1 ] [ (root 6 - root 2) / 4 ]
(using results from BC and sin A)
h = { [ ( -1 + root 3 ) + 2 ] / 2 } [ (root 6 - root 2) / 4 ]
h = (1 + root 3) (root 6 - root 2) / (2 * 4)
h = [ root 6 - root 2+ (root 3)(root 6) - (root 3)(root 2) ] / 8
h = (root 6 - root 2+ root 18 - root 6) / 8
h = ( - root 2+ 3 * root 2 ) / 8
h = 2 * root 2 / 8
h = (root 2) / 4
so, the length of the perpendicular from B to AC is (root 2) / 4