Mathematics Question and Solution
Calculating area of the square.
Let x be the side length of the square.
7² = x²+3²-2*x*3cosy
49 = x²+9-6xcosy
6xcosy = x²-40
cosy = (x²-40)/(6x) --- (1).
5² = x²+3²-2*x*3cos(90-y)
25 = x²+9-6x(cos90cosy+sin90siny)
6xsiny = x²-16
siny = (x²-16)/(6x) --- (2).
At (2).
(x²-16) is opposite.
(6x) is hypotenuse.
Calculating Adjacent.
Adjacent = √((6x)²-(x²-16))
= √(36x²-(x⁴-32x²+256))
Adjacent = √(68x²-x⁴-256) units
It implies;
cosy = √(68x²-x⁴-256)/(6x) --- (3).
Calculating x², the area of the square.
Equating (1) and (3).
(x²-40)/(6x) = √(68x²-x⁴-256)/(6x)
(x²-40)² = 68x²-x⁴-256
x⁴-80x²+1600 = 68x²-x⁴-256
2x⁴-148x²+1856 = 0
x⁴-74x²+928 = 0
(x²-37)² = -928+(-37)²
(x²-37)² = 441
x² = 37±√(441)
x² = 37±21
It implies;
x² ≠ 37-21
x² ≠ 16 square units.
x² = 37+21
x² = 58 square units.
Again, x² is the area of the square.
