a = asin(1/6)
b = 2a
b = 2asin(1/6)
Where b is angle BAC
cos(2asin(1/6)) = c/6
c = ⅓(17) units.
Where c is AD.
d = 6-c
d = 6-⅓(17)
d = ⅓ units
Where d is CD.
sin(2asin(1/6)) = e/6
e = 1.97202659...
a = 15-11.25
a = 3.75 m.
b = atan(3/3.75)°
c = 90-atan(3/3.75)
c = 51.34019174591°
Calculating Length CD.
tan(51.34019174591) = 15/CD
CD = 15/tan(51.34019174591)
CD = 12 m....
Let the square side be a.
tan60 = a/b
b = a/tan60 cm.
Where b is BS.
Calculating a.
2b+a = 6
2(a/tan60)+a = 6
a((2/tan60)+1) = 6
a = 6/((2/tan60)+1)
a = 6(2√(3)-3) cm.
Area Shaded...
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....
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...
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...
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*...
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)...
Sir Mike Ambrose is the author of the question.
Let length AB = 2 units.
Therefore;
Side length of the square, x is;
2x² = 2²
x = √(2) units.
It implies;
Area square is;
x²...
Sir Mike Ambrose is the author of the question.
Inscribed square single side length is;
7.80540954246 cm
Area green in cm² to 1 decimal place is;
Area trapezoid with parallel sides 12 cm...
