Mathematics Question and Solution
Calculating x, length AD.
Let z be alpha.
Let CD = y.
a = (x+y) units.
a is AC.
b²+y² = 10²
b = √(100-y²) units.
b is BC.
(x+y)²+√(100-y²) = 12²
x²+2xy+y²+100-y² = 144
x²+2xy = 44
2xy = 44-x²
y = (44-x²)/(2x) --- (1).
sinz = y/10 --- (2).
cosz = √(100-y²)/10 --- (3).
x² = 12²+10²-2*12*10cos(2z)
x² = 244-240(cos²z-sin²z)
x² = 244-240(1-2sin²z) --- (4).
Substituting (2) in (4).
x² = 244-240(1-2(y/10)²)
x² = 244-240(1-(2y²/100)
x² = 244-240+⅕(24y²)
x² = 4+⅕(24y²)
x² = ⅕(20+24y²)
5x² = 20+24y² --- (5).
Calculating x.
Substituting (1) in (5).
5x² = 20+24((44-x²)/(2x))²
5x² = 20+24((1936-88x²+x⁴)/(4x²))
5x² = 20+6((1936-88x²+x⁴)/(x²))
5x⁴ = 20x²+6(1936-88x²+x⁴)
5x⁴ = 20x²+11616-528x²+6x⁴
x⁴-508x²+11616 = 0
(x²-254)² = -11616+(-254)²
(x²-254)² = 52900
x² = 254±√(52900)
x² = 254±230
It implies;
x ≠ √(254+230)
x = √(254-230)
x = √(24)
x = 2√(6) units.
x = 4.89897948557 units.
x is length AD.
