tana = 4/7
a = atan(7/4)°
b = 180-a
b = (90+atan(4/7))°
c² = 7²+4²
c = √(49+16)
c = √(65) units.
Therefore, Blue Area is;
Area triangle with height √(65) units and base 4sin(90+atan...
a = 180-44
a = 136°
b = ½(180+44)
b = ½(224)
b = 112=
c = ½(a)
c = 68°
d = ½(b)
d = 56
e = 180-c-d
e = 180-68-56
e = 180-124
e = 56°
Where e is the required angle, angle CBO....
7² = 5²+6²-2*5*6cosa
60cosa = 25+36-49
cosa = (61-49)/60
a = acos(12/60)
a = 78.4630409672°
Calculating r, radius of the inscribed half circle.
5r+6r = 5*6sin78.4630409672
11r = 30sin78....
tan30 = a/h
a = htan30 units.
tan60 = b/h
b = htan60 units.
It implies;
a+b = 100
htan30+htan60 = 100
⅓(√(3)h)+√(3)h = 100
⅓(4√(3)h) = 100
4√(3)h = 300
h = 300/(4√(3))
h = ¼*⅓*300√...
a²+R² = 7²
a = √(49-R²) units.
b = (3+R) units.
c²+R² = 3²
c = √(9-R²) units.
d = a+c
d = √(49-R²)+√(9-R²) units.
R is;
(√(49-R²)+√(9-R²))² = (3+R)²+√(49-R²)²
49-R²+2√((49-R²)(9-...
0.5*a*6sin75 = 9
a = 9/(3sin75)
a = 3/sin75
a = 3.1058285412 units.
a is AB.
b² = 3.1058285412²+6²-2*3.1058285412*6cos75
b = 6 units.
b is AC.
It implies, triangle ABC is isosceles....
Calculating length CD.
Notice!
The composite plane shape is not drawn to scale.
tan60 = 8/a
a = 8/√(3)
a = ⅓(8√(3)) units.
a is OE.
(⅓(8√(3)))²+8² = b²
b² = ⅓(64)+64
b² = ⅓(64+192)
b²...
Let the square side length be 2 units.
Therefore, r, radius of the circle is;
r = 1 unit.
Calculating the required angle, let it be a.
cosb = 1/2
b = acos(1/2)
b = 60°
c² = 2²+2²
c...
Notice!
The side length of the inscribed biggest square is 5 units.
Calculating inscribed red area.
tana = 2.5/5
a = atan(½)°
b = 2atan(½)°
tanc = 5/2.5
c = atan(2)°
d = 2atan(2)°...
Let the side length of the three equal regular triangles be 2a units each.
b² = 2a²-2a²cos120
b = √(3)a unit.
(6√(7))² = (2a)²+(√(3)a)²-2*2a*√(3)acos90
252 = 4a²+3a²
252 = 7a²
a² = 36
a...
