Mathematics Question and Solution
Calculating x, the side length of the square ABCD.
3² = x²+4²-2*4xcosy
9 = x²+16-8xcosy
8xcosy = x²+7
cosy = (x²+7)/(8x) --- (1).
5² = x²+4²-2*x*4cos(90-y)
25 = x²+16-8xcos(90-y)
8x(cos90cosy-sin90siny) = x²-9
-8xsiny = x²-9
8xsiny = 9-x²
siny = (9-x²)/(8x) --- (2).
At (2).
(9-x²) is opposite.
(8x) is hypotenuse.
Calculating adjacent. Let it be a.
a²+(9-x²)² = (8x)²
a² = 64x²-(81-18x²+x⁴)
a² = 64x²-81+18x²-x⁴
a = √(82x²-81-x⁴) units.
Again, a is adjacent.
It implies;
cosy = √(82x²-81-x⁴)/(8x) --- (3).
Equating (1) and (3).
(x²+7)/(8x) = √(82x²-81-x⁴)/(8x)
(x²+7)² = 82x²-81-x⁴
x⁴+14x²+49 = 82x²-81-x⁴
2x⁴-68x²+130 = 0
x⁴-34x²+65 = 0
(x²-17)² = -65+(-17)²
(x²-17)² = 224
x² = 17±√(224)
x² = 17±4√(14)
It implies;
x ≠ √(17-4√(14)) units.
x = √(17+4√(14)) units.
x = 5.65390392093 units.
x is the side length of the square ABCD.
