Calculating x.
7² = x²+5²+2*5*xcosa
49 = x²+25+10xcosa
24 = x²+10xcosa
10xcosa = 24-x²
cosa = (24-x²)/(10x) --- (1).
b = (x+5) units.
7² = x²+(x+5)²-2x(x+5)cosa
49 = x²+x²+10x+25-(2x²+10x)cosa
2...
The required angle, x is;
a = ½(90)
a = 45°
it implies;
x = ½(180-a)
x = ½(180-45)
x = ½(135)
x = 67.5°
Notice!
The ascribed polygon is a regular nonagon.
The inscribed polygon is a regular hexagon.
Calculating the required angle, x.
a = ⅙*180(6-2)
a = 120°
a is the single interior angle of the re...
Sir Mike Ambrose is the author of the question.
Let the side of the regular nonagon be 1 unit.
Therefore;
Area nonagon is;
½(9)*(1/(2tan(180/9)))
= ¼(9)*(1/tan(20))
= 6.18182419377 square units....
Sir Mike Ambrose is the author of the question.
Let the side of the regular nonagon be 1 unit.
Therefore;
Length Green is;
((1+2sin50)sin20/sin80) + √(1.08506357513²+1²)
= 0.87938524157 + 1.4755...
a = 8+9
a = 17 units.
a is the radius of the half circle.
b²+8² = 17²
b = √(225)
b = 15 units.
c²+d² = 17²
c = √(289-d²) units.
e = (17-d) units.
f = ½(c)
f = ½√(289-d²) unit.
g = 17+17-9
g = 3...
Let the equal side lengths be a.
14² = a²+16²-2*16acosb
196 = a²+256-32acosb --- (1).
21² = 2a²+2a²cosb
cosb = (441-2a²)/2a² --- (2).
Substituting (2) in (1).
196 = a²+256-32a(441-2a²...
Notice!
r, radius of the inscribed quarter circle is also the side length of the ascribed square.
Let alpha be a.
Notice again.
2a = 60°
a = 30°
Alpha = 30°
It implies;
sin30 =...
Let the square side length be a.
Observing similar plane shape (right-angled) side length ratios.
8 - ½(a)
b - a
Cross Multiply.
½(ab) = 8a
b = 16 units.
c² = 8²+16²
c = √(64+256)
c = 8√(5) units...
let theta be a.
tana = k/3k
a = atan(⅓)°
Notice!
3k = 4.
b = 2a
b = 2atan(⅓)°
It implies;
tan(2atan(⅓)) = c/4
c = 3 units.
Therefore, ? the required length is;
Let if be d.
d = 4-c
d = 4-3
d...
