a = atan(2)°
b = 180-2a
b = (180-2atan(2))°
Where b is angle FBP.
Notice;
CQ = 5 cm.
Calculating FP using sine rule.
Let FP be c.
(c/sin(180-2atan(2))) = (5/sin(atan(2)))
c = 4.472135955 cm.
T...
a² = 10²+10²-2*10*10cos150
a = 19.31851652578 cm.
Where a is AF which is equal AE.
(19.31851652578/sin150) = (10/sinb)
b = 15°
Where b is angle BAF.
tan15 = c/10
c = 2.67949192431 cm.
It implies;...
Let r be the radius of the ascribed semi circle.
Therefore;
½(r) is the radius of the inscribed circle.
(2r)² = 2a²
a² = 2r²
a = √(2)r cm.
b = ½(a)
b = ½(√(2)r) cm.
c² = a²+b²
c² = (√(2)r)²+(½(√...
Notice!
Since we are resolving to get;
Area Blue/Area Yellow
Let x be 1 unit.
It implies;
Area Yellow is;
1²
= 1 square unit.
Ascribed or bigger square side length is;
1+3(1)
= 1+3...
Let the single side length of the bigger equilateral triangle be 2 units.
Therefore;
R = ⅓(2√(3)) unit
It implies;
L will be;
(2√(3)/3)² = (⅓(√(3)))² + L² - 2*(⅓(√(3)))*Lcos150
Therefore;
L =...
Let the small inscribed square side be a.
Let b be the side of the big inscribed square.
It implies;
b = 10-a
Therefore;
Calculating a.
tan(90-75) = a/(10-a)
a(1+tan15) = 10tan15
a = (10tan15)/(...
Sir Mike Ambrose is the author of the question.
Area R exactly in its simplest decimal form is;
Area triangle with height 4.54406194214 units and base 2sin(atan(2)) units.
= ½*4.54406194214*2sin(a...
Let the radius of the inscribed circle be r.
The square side is 8 cm.
b = 8-r
c = 2+r
Therefore;
(2+r)² = 2²+(8-r)²
4+4r+r² = 4+64-16r+r²
4r = 64-16r
20r = 64
5r = 16
r = (16/5) cm.
Area circle...
Sir Mike Ambrose is the author of the question.
Let the side let of the square be 2 units.
Therefore;
Area square is;
2²
= 4 square units.
It implies;
Area R is;
Area trapezoid with parallel sid...
a = ½(4√(6)-2√(6))
a = √(6) cm.
tan60 = b/√(6)
b = 3√(2).
b is the height of the trapezoid.
c² = (√(6))²+(3√(2))²
c² = 6+18
c = √(24)
c = 2√(6) cm.
Or
cos60 = √(6)/c
c = √(6)/(½)
c = 2√(6) cm.
c i...
