Let the single side length of the inscribed blue square area be x.
Therefore;
x+(x÷(tan(tan–¹(39/52))))+(x÷(tan(90-tan–¹(39/52))))=65
x=65÷(1+(1÷(tan(tan–¹(39/52))))+(1÷(tan(90-tan–¹(39/52...
Sir Mike Ambrose is the author of the question.
a = ⅛*180(8-2)
a = 135°
a is the single interior angle of the regular octagon.
2² = 2b²
b = √(2) units.
c = 2+b
c = (2+√(2)) units.
tan...
Sir Mike Ambrose is the author of the question.
2² = 2(3²)-2(3²)cosa
18cosa = 18-4
a = acos(7/9)°
a is angle ABC.
Let b be the radius of the inscribed circle.
Calculating b.
c²+1² = 3²...
Sir Mike Ambrose is the author of the question.
Let CD = 1 units.
It implies;
DE = 1 units.
AD = 2 units.
AE = 2 units.
1² = 2²+2²-2*2*2cosa
1 = 8-8cosa
cosa = 7/8
a = acos(⅞)°
b...
Sir Mike Ambrose is the author of the question.
Let the square side length be 2 units.
Calculating Area C and D.
Notice.
Area C = Area D.
C = ½(2²-¼(2*2π))
C = ½(4-π) square units.
T...
Area Triangle is;
16tan(sin–¹(1/3))
= 5.65685424949 square units.
Or
sina = ⅓
a = asin(⅓)°
tana = b/4
b = 4tan(asin(⅓)) units.
c = 2b
c = 8tan(asin(⅓)) units.
Therefore, area triangle is;...
Calculating angle a.
Let the length of the ascribed rectangle be 2 units.
Therefore it's width is 1 unit.
b² = 2²+1²
b = √(5) units.
b is the diagonal of the ascribed rectangle.
tanc...
Calculating x.
Notice.
The composite plane shape is drawn to scale.
2(60+x)+x = 180
120+2x+x = 180
3x = 60
x = 20°
Or
(60-x)+2(60+10) = 180
60-x+140 = 18
200-x = 180...
It will be;
Let the single side length of the regular heptagon be 5 units.
Therefore;
(1/x) + (1/y) will be;
(1÷(√(50-50cos129))) + (1÷(5+10cos51))
= 0.19934168501≈ 0.2 units.
Sir Mike Ambrose is the author of the question.
Let the side length of the introduced equilateral triangle inscribing the circle be 2 units.
Calculating r, radius of the circle.
cos30 = 1/r...
