Mathematics Question and Solution
Calculating Area Triangle CBA.
Let the two equal lengths of the ascribed isosceles right-angled triangle be x.
It implies;
4² = 2²+x²-2*2xcosa
16 = 4+x²-4xcosa
12 = x²-4xcosa --- (1).
b = (90-a)°
(2√(2))² = 2²+x²-2*2xcosb
8 = 4+x²-4xcos(90-a)
4 = x²-4x(cos90scosa+sin90sina)
4 = x²-4xsina
4xsina = x²-4
sina = (x²-4)/(4x) --- (2).
Calculating cosa from (2).
(x²-4) = Opposite.
4x = Hypotenuse.
Calculating Adjacent, let it be c.
c²+(x²-4)² = (4x)²
c² = 16x²-x⁴+8x²-16
c = √(-x⁴+24x²-16) units.
cosa = √(-x⁴+24x²-16)/(4x) --- (3).
Therefore, substituting (3) in (1).
12 = x²-4x(√(-x⁴+24x²-16)/(4x))
12 = x²-√(-x⁴+24x²-16)
√(-x⁴+24x²-16) = x²-12
-x⁴+24x²-16 = (x²-12)²
-x⁴+24x²-16 = x⁴-24x²+144
2x⁴-48x²+160 = 0
x⁴-24x²+80 = 0 --- (4).
From (4), calculating x.
x⁴-20x²-4x²+80 = 0
x²(x²-20)-4(x²-20) = 0
(x²-4)(x²-20) = 0
It implies;
x² ≠ 5 square units.
x² = 20 square units.
x = √(20)
x = 2√(5) units.
Again, x is the two equal side lengths of the ascribed isosceles right-angled triangle, AC = AB.
Therefore, area triangle CBA is;
½*x²
= ½(2√(5))²
=½(20)
= 10 square units.
