Mathematics Question and Solution
Calculating The Required Angle.
Let it be x.
a = ½(12)+b
a = (6+b) cm.
b is the radius of the inscribed sector.
Calculating b.
c² = 6²+12²
c² = 36+144
c = √(180)
c = 6√(5) cm.
It implies;
a = c
6+b = 6√(5)
b = (6√(5)-6) cm.
b = 7.416407865 cm.
Again, b is the radius of the inscribed sector.
tand = 12/6
d = 63.4349488229°
e = 180-d
e = 116.565051177°
f = ½(d)
f = 31.7174744115°
cos31.7174744115 = g/12
g = 10.2078097002 cm.
7.416407865² = 12²+12²-2*12*12cosh
55.00310562 = 288-288cosh
288cosh = 288-55.00310562
h = acos((288-55.00310562)/288)
h = 36°
j = 90-h
j = 54°
k = ½(180-36)
k = ½(144)
k = 72°
l = k-(90-d)
l = 72-(90-63.4349488229)
l = 45.4349488229°
m² = 2(12)²-2(12)²cos54
m = 10.8957719937 cm.
n² = 2(7.416407865)²-2(7.416407865)²cos45.4349488229
n = 5.72824632767 cm.
Therefore, x, the required angle is;
10.8957719937² = 5.72824632767²+10.2078097002²-2*10.2078097002*5.72824632767cosx
118.717847339 = 137.012184866-116.945696857cosx
116.945696857cosx = 137.012184866-118.717847339
cosx = (137.012184866-118.717847339)/116.945696857
x = acos(0.15643446504)
x = 81°
Again, x is the required angle.
