20² = 12²+a²
a² = 400-144
a = √(256)
a = 16 cm.
b = 16-9
b = 7 cm.
c = atan(3/4)°
Let Alpha be d.
d = ½(c)
d = 0.5atan(¾)°
tan(0.5atan(¾)) = e/16
e = ⅓(16) cm.
tan(0.5atan(¾))...
Sir Mike Ambrose is the author of the question.
a² = 32-32cos120
a = 4√(3) units.
b = 2a
b = 8√(3) units.
c² =32-32cos150
c = 7.72740661031 units.
sin30 = d/4
d = 2 units.
sin60 = e/...
Calculating x.
½(x+2)(x-2) = 30
x²-4 = 60
x² = 64
x = 8 m.
Therefore;
AB = 8-2 = 6 m.
BC = 8+2 = 10 m.
It implies;
L² = 6²+10²
L² = 36+100
L² = 136
L = √(136)
L = √(2*2*34)
L = 2√(34) m.
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Sir Mike Ambrose is the author of the question.
a = atan(¼)°
b² = 12²+3²
b = 3√(17) cm.
c = ½(atan (4/3))°
Let the radius of the inscribed circle be r.
Calculating r.
tan(½(atan(4/3))) = (r/(9-r)...
Let the side length of the regular pentagon be a.
b² = a²+(0.5a)²-2*a*(0.5a)cos108
b = 1.24860602048a units.
Calculating angle AGE.
Let it be c.
(1.24860602048a/sin108) = (a/sinc)
c = 49.6138224...
a² = 5-4cos120
a = √(7) units.
(√(7)/sin120) = (2/sinb)
b = 40.89339464913°
c = 180-2*40.89339464913
c = 98.21321070174°
Calculating r.
0.5*r(r-1)sin98.21321070174 +(3/2) = (98.21321070174π*r²)...
Let the side length of the regular pentagon be 1 unit.
c² = 2-2cos108
c = 1.61803398875 units.
(d/sin36) = (1/sin108)
d = 0.61803398875 units.
e = c-d
e = 1 unit.
Calculating a.
sin72 = a/1
a =...
Sir Mike Ambrose is the author of the question.
Let the two equal length be 2 units each.
2 = 2/a
a = 1 unit.
sin(atan(½)) = b/3
b = ⅕(3√(5)) units.
c = 2b
c = ⅕(6√(5)) units.
Where c is the squa...
Calculating equation of the curve.
2a² = 4²
a = 2√(2)
b = (4+2√(2))
b is the maximum turning point.
Therefore considering points;
(0, 0) and ((4+2√(2)), (4+2√(2)))
y-(4+2√(2)) = c(x-(4+2√(2)))²...
