Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Calculating Exactly Angle Alpha - Angle Beta.
Let Alpha be x.
Let Beta be y.
(6+6√(2))² = 8²+10²-2*8*10cosa
36+72√(2)+72 = 164-160cosa
160cosa = 164-36-72-72√(2)
160cosa = 56-72√(2)
20cosa = 7-9√(2)
cosa = (7-9√(2))/20
a = acos((7-9√(2))/20)°
a = 106.642318459°
8² = 10²+(6+6√(2))²-2*10(6+6√(2))cosb
64 = 100+36+72√(2)+72-(120+120√(2))cosb
(120+120√(2))cosb = 144+72√(2)
cosb = (144+72√(2))/(120+120√(2))
cosb = ⅕(3√(2))
b = acos(⅕(3√(2)))°
cosb = c/10
(⅕(3√(2))) = c/10
c = ⅕(30√(2))
c = 6√(2) cm.
c is AD.
10² = 8²+(6+6√(2))²-2*8(6+6√(2))cosd
100 = 64+36+72√(2)+72-(96+96√(2))cosd
(96+96√(2))cosd = 72√(2)+72
cosd = (72+72√(2))/(96+96√(2))
cosd = ¾
d = acos(¾)°
cosd = e/8
¾ = e/8
e = 6 cm.
e is AC.
f² = (6√(2))²+12²
f = 6√(6) cm.
f is BD.
g² = 6²+12²
g = 6√(5) cm.
g is BC
It implies;
Angle Alpha (x) is;
x = 180-acos(6√(2)/(6√(6)))-acos(10/(6√(6)))
x = (180-acos(⅓√(3))-acos(5√(6)/18))°
Angle Beta, y is;
y = 180-acos(6/(6√(5)))-acos(8/(6√(5)))
y = (180-acos(⅕√(5))-acos(4√(5)/15))°
It implies;
Alpha - Beta (x-y) exactly is;
(180-acos(⅓√(3))-acos(5√(6)/18))-(180-acos(⅕√(5))-acos(4√(5)/15))
= (acos(⅕√(5))+acos(4√(5)/15)-acos(⅓√(3))-acos(5√(6)/18))°
= 14.9710507296°
