a = 8+6
a = 14 units.
tanb = 6/14
b = atan(3/7)°
c = atan(7/3)°
d = 180-45-atan(7/3)
d = 68.1985905136°
tane = 2/(6+4)
e = atan(1/5)°
f = e+45
f = (45+atan(1/5))°
f = 56.309932...
Calculating x.
√(1+√(1+x)) = x^(⅓)
1+√(1+x) = x^(⅔)
√(1+x) = x^(⅔)-1
1+x = (x^(⅔)-1)²
1+x = x^(4/3)-2x^(⅔)+1
x = x^(4/3)-2x^(⅔)
Dividing through by x.
1 = x^(⅓)-2x^(-⅓)
1 = x^(⅓)-(2/...
Calculating d.
0.5*4*d = 24
4d = 48
d = 12 m.
Calculating c.
c² = 5²+d²
c² = 25+144
c = √(169)
c = 13 m.
Calculating b.
b² = (4+5)²+d²
b² = 81+144
b = √(225)
b = 15 m.
Cal...
Let a be the radius of the inscribed half circle.
15² = 13²+14²-2*14*13cosb
364cosb = (169+192)-225
cosb = 140/362
b = 67.380135052°
Calculating a, radius of the inscribed half circle....
a = (2+x) units.
a is an adjacent of the inscribed red triangle.
b = (2+2x) units.
b is the diameter of the half circle.
2²+(2+x)² = (2+2x)²
4+4+4x+x² = 4+8x+4x²
8+4x+x² = 4+8x+4x²
3x²+4...
Shaded area is;
Area triangle with height = base = √(42) cm - area triangle with height 6.15093874956 cm and base 4.39211425448, and angle 53°
= ½(√(42))² - ½(6.15093874956*4.39211425448)sin5...
Let the side length of the big inscribed regular triangle be a.
It implies, the side length of the biggest inscribed regular triangle is 3a.
Calculating a.
2² = a²+(2a)²-2*a*2acos120
4 =...
Let AD = 1 unit.
a = 180-62-56
a = 62°
a is angle ADC.
It implies;
AD = AC = 1 unit.
b = 180-48-88
b = 44°
b is angle ABC.
(1/sin44) = (c/sin88)
c = 1.4386796007 units.
c is AB...
Let the inscribed square side length be a.
b² = 2a²
b = √(2)a units.
b is AC = BD, the diagonal of the inscribed square.
a² = 2c²
Where c is OA = OD.
c² = ½(a²)
c = √(a²/2)
c = ½(√(2)a)...
a = 180-45-15
a = 120°
b = 360-120-90
b = 150°
c = ½(360-2(150))
c = ½(60)
c = 30°
d = 30+45
d = 75°
e = 150-90
e = 60°
(6/sin60) = (f/sin75)
f = 6.6921304299 units.
(6/sin...
