Mathematics Question and Solution
Let AE = CE = BE = b.
Let OE = a.
4² = a²+b²-2abcos45
16 = a²+b²-√(2)ab --- (1).
6² = a²+b²-2abcos135
36 = a²+b²+√(2)ab --- (2).
Solving (1) and (2) simultaneously.
Subtracting (1) from (2).
(36 = a²+b²+√(2)ab)
-(16 = a²+b²-√(2)ab)
20 = 2√(2)ab --- (3).
From (3).
a = 20/(2√(2)b)
a = 10/(√(2)b)
a = (5√(2)/b) --- (4).
Therefore, Substituting (4) in (1) to get b.
16 = ((5√(2)/b))²+b²-√(2)((5√(2)/b))b
16 = (50/b²)+b²-10
26-b² = (50/b²)
26b²-b⁴ = 50
b⁴-26b²-50 = 0
Let p = b²
p²-26p-50 = 0
Resolving the above quadratic equation.
(p-13)² = 50+(-13)²
(p-13)² = 50+169
p-13 = ±√(219)
p = 13±√(219)
It implies;
p ≠ 13-√(219)
p = 13+√(219)
And p = b²
It implies;
b² = 13+√(219)
b = √(13+√(219))
b = 5.272442374 units.
Calculating a, using (4).
a = (5√(2)/b)
And b = 5.272442374
a= 5√(2)/5.272442374
a = 1.34113705 units.
(1.34113705/sinc) = (4/sin45)
sinc = 0.2370817756
c = asin(0.2370817756)
c = 13.7143681601°
c is angle OAE.
AC = 2b
AC = 2*5.272442374
AC = 10.544884748 units.
Calculating length CD.
Let it be d.
sin13.7143681601 = d/(AC)
sin13.7143681601 = d/10.544884748
d = 10.544884748sin13.7143681601
d = 2.5 units.
CD = 2.5 units.
