Mathematics Question and Solution
Let a be the blue length.
b² = 4²+3²
b = √(25)
b = 5 units.
Let alpha be c.
cos(2c) = 4/5
cos(c+c) =4/5
(cosc)²-(sinc)² = 4/5 --- (1).
cosc = 3/a --- (2).
a² = 3²+d²
d = √(a²-9) units.
sinc = d/a
sinc = √(a²-9)/a --- (3).
Substituting (2) and (3) in (1).
(3/a)²-(√(a²-9)/a)² = 4/5
(9/a²)-((a²-9)/a²) = 4/5
9-(a²-9) = ⅕(4a²)
18-a² = ⅕(4a²)
90-5a² = 4a²
9a² = 90
a² = 10
a = √(10) units.
a = 3.16227766017 units.
Again, a is the required blue length.
Or
tan(2c) = (3/4)
2c = atan(3/4)
c = ½(atan(3/4))
c = 18.4349488229°
c is alpha.
It implies, the required length, a is;
cos(18.4349488229) = 3/a
a = 3/cos(18.4349488229)
a = 3.16227766017 units.
a = √(10) units.
