Mathematics Question and Solution
Calculating the required angle, let it be x.
Let a be the side length of the equilateral triangle.
It implies;
6² = a²+10²-2*10*a cosy
36 = a²+100-20acosy
20acosy = a²+64
cosy = (a²+64)/(20a) --- (1).
8² = a²+10²-2*10*acos(60-y)
64 = a²+100-20a(½*cosy+½√(3)siny)
64 = a²+100-10acosy-10√(3)asiny
10acosy+10√(3)asiny = a²+36 --- (2).
At (1).
cosy = (a²+64)/(20a)
Calculating siny.
siny = opp/hyp
opp = √((20a)²-(a²+64)²)
opp = √(400a²-(a⁴+128a²+4096))
opp = √(-a⁴+272a²-4096)
Therefore;
siny = √(-a⁴+272a²-4096)/(20a) --- (3).
Calculating a.
Substituting (1) and (3) in (2).
10a((a²+64)/(20a))+10√(3)a(√(-a⁴+272a²-4096)/(20a)) = a²+36
½(a²+64)+½(√(3)*√(-a⁴+272a²-4096)) = a²+36
(a²+64)+(√(3)*√(-a⁴+272a²-4096)) = 2a²+72
(√(3)*√(-a⁴+272a²-4096)) = a²+8
3(-a⁴+272a²-4096) = (a²+8)²
-3a⁴+816a²-12288 = a⁴+16a²+64
4a⁴-800a²+12352 = 0
It implies;
a² = 16.8616
or
a² = 183.138
Therefore;
a ≠ √(16.8616) units.
a = √(183.138) units.
a = 13.532848924 units.
Again, a is the side length of the equilateral triangle.
Therefore, x the required angle is;
13.532848924² = 6²+8²-2*6*8cosx
96cosx = 100-13.532848924²
96cosx = -83.1379999998
cosx = -83.1379999998/96
x = acos(-83.1379999998/96)
x = 150°
Again, x is the required angle.
