Notice!
x+90 = 120
x = 120-90
x = 30°
Where x is the required angle.
a = ½(180-90-30)
a = ½(60)
a = 30°
b = ½(180-30)
b = ½(150)
b = 75°
c = b-a
c = 75-30
c = 45°
Therefore, the required angle, x is;
x = 180-45-45
x = 180-90
x = 90°
Or
x =...
r = √(7²+6²)
r = √(85) cm.
r is the radius of the semi circle.
x² = √(85)²+6²
x² = 85+36
x = √(121)
x = 11 cm.
r² = x²+(6+8)² --- (1).
r² = (10+x)²+6² --- (2).
Where r is the radius of the ascribed quarter circle.
Equating (1) and (2) to get x.
r² = r²
x²+(6+8)² = (10+x)²+6²
x²+196 = 100+20x+x²+36
196-136...
Calculating angle x.
Let the side length of the regular pentagon be 1 unit.
Calculating r, radius of the inscribed two equal circles.
a = ⅕*180(5-2)
a = 108°
b = ½(a)
b = ½*108
b = 54...
a = 2+2+R+R
a = (4+2R)
a is the diameter of the ascribed circle.
b = ½(a)
b = ½(4+2R)
b = (2+R) units.
b is the radius of the ascribed circle.
c = b-R
c = (2+R)-R
c = 2 units.
d = b...
a² = 13²-5²
a² = 169-25
a = √(144)
a = 12 units.
a is AE.
tanb = 5/12
b = atan(5/12)°
b is angle BAE.
c = x+a
c = (x+12) units.
c is AG.
Notice!
Triangle ABG is isosceles.
AB = AG
Therefore;
13...
Calculating R, radius of the inscribed circle.
Observing similar plane shape (right-angled triangle) side length ratios.
(12+x) - 6
6 - x
Cross Multiply.
36 = 12x+x²
x²+12x-36 = 0
(x+6)² =...
a² = 16+9-2*3*4cos150
a = 6.7664325675 cm.
a is the side length of the regular triangle.
(6.7664325675/sin150) = (4/sinb)
b = 17.192123734°
c = 60-b
c = 60-17.192123734
c = 42.807876266...
