Mathematics Question and Solution
Calculating length AD.
Let AD be x.
Let alpha be y.
Notice.
AB = CD = 6 cm.
a = (x+6) cm.
tany = x/6 --- (1).
tan(2y) = 6/(x+6)
(2tany)/(1-tan²y) = 6/(x+6) --- (3).
Calculating x.
Substituting (1) in (3).
(2tany)/(1-tan²y) = (6/(x+6)
(2(x/6))/(1-(x/6)²) = 6/(x+6)
(x/3)/(1-(x²/36)) = (6/(x+6))
(x/3)/((36-x²)/36) = (6/(x+6))
36x/(108-3x²) = (6/(x+6))
36x(x+6) = 6(108-3x²)
6x(x+6) = 108-3x²
6x²+36x = 108-3x²
9x²+36x-108 = 0
x²+4x-12 = 0
(x²+6x)-(2x-12) = 0
x(x+6)-2(x+6) = 0
(x+6)(x-2) = 0
It implies;
x ≠ -6 cm.
x = 2 cm.
Where x is length AD.
