Mathematics Question and Solution
a = x°
b = (90-x)°
R² = 6²+16²-2*6*16cosx
R² = 292-192cosx
cosx = (292-R²)/192 --- (1).
Notice.
At (1).
(292-R²) is adjacent.
192 is hypotenuse.
Let c be opposite.
c² = 192²-(292-R²)²
c² = 36864-85264+584R²-R⁴
c = √(-R⁴+584R²-48400) units.
It implies;
sinx = √(-R⁴+584R²-48400)/192 --- (3).
Again.
R² = 10²+6²-2*10*6cos(90-x)
R² = 136-120(cos90cosx+sin90sinx)
R² = 136+120sinx --- (4).
Substituting (3) in (4).
R² = 136+120(√(-R⁴+584R²-48400)/192)
R²-136 = 5√(-R⁴+584R²-48400)/8
(8R²-1088)² = 25(-R⁴+584R²-48400)
64R⁴-17408R²+1183744 = -25R⁴+14600R²-1210000
89R⁴-32008R²+2393744 = 0
(R²-(16004/89))² = (-16004/89)-(2393744
/89)
(R²-(16004/89))² = (256128016-213043216)/7921
(R²-(16004/89))² = (43084800/7921)
R² = (16004/89)±√(43084800/7921)
R² = (16004±480√(187))/89
Therefore;
R = √(16004-480√(187))÷89)
R = 10.2989574693 units.
Or
a = (10-x) units.
b = a-x
b = (10-2x) units.
c² = 6²-x²
c = √(36-x²) units.
d = 16-c
d = (16-√(36-x²)) units.
e = d-c
e = (16-√(36-x²))-√(36-x²)
e = (16-2√(36-x²)) units.
It implies;
Calculating x.
10(10-2x) = 16(16-2√(36-x²))
50-10x = 128-16√(36-x²)
16√(36-x²) = 78+10x
8√(36-x²) = 39+5x
64(36-x²) = 1521+390x+25x²
2304-64x² = 1521+390x+25x²
89x²+390x-783 = 0
Therefore;
x ≠ -5.8786 units.
x = 1.49657 units.
Recall.
c = √(36-x²)
And x = 1.49657 units.
c = √(36-1.49657²)
c = 5.8103595616 units.
Recall Again.
d = 16-c
And c = 5.8103595616 units.
d = 16-5.8103595616
d = 10.1896404384 units.
Therefore R, radius of the ascribed circle is;
R = √(d²+x²)
R = √(10.1896404384²+1.49657²)
R = 10.2989559679 units.
