Mathematics Question and Solution
Let AB = a
Notice.
AB = CD
a² = 2²+1²+2*2*1cosx
a² = 5+4cosx --- (1).
a² = 3²+4²-2*3*4cosx
a² = 25-24cosx --- (2).
Calculating x.
Equating (1) and (2).
5+4cosx = 25-24cosx
28cosx = 20
cosx = (20/28)
x = acos(5/7)
x = 44.4153085972°
y = 180-x
y = 180-44.4153085972
y = 135.5846914028°
Let BC = b.
AD = 3b.
(3b)² = 1²+4²+2*4*1cosc
9b² = 17+8cosc
b² = ⅑(17+8cosc) --- (3).
b² = 2²+3²-2*3*2cosc
b² = 13-12cosc --- (4).
Calculating b.
Equating (3) and (4).
⅑(17+8cosc) = 13-12cosc
17+8cosc = 117-108cosc
116cosc = 100
cosc = 100/116
cosc = 25/29
c = acos(25/29)
c = 30.4503140214°
d = 180-c
d = 180-30.4503140214
d = 149.5496859786°
It implies;
At (4).
b² = 13-12cosc
And c = 30.4503140214°
b² = 13-12cos30.4503140214
b = 1.6294699794 units.
b is BC.
AD = 3b
AD = 3*1.6294699794
AD = 4.8884099382 units.
Or
AD = √(1²+4²-2*1*4cosd)
d = 149.5496859786°
AD = √(17-8cos149.5496859786)
AD = 4.8884099382 units.
Therefore;
BC ÷ AD is;
= 1.6294699794÷4.8884099382
= ⅓
