a² = 4²+4²-2*4*4cos150
a = 7.72740661031 cm.
Where a is BE.
b = angle ABE = ½(180-150)
b = 15°
c = angle AFE = 180-45-15
c = 120°
(4/sin120) = (d/sin45)
d = 3.26598632371 cm.
Where d is EF.
Noti...
tan15 = a/4
a = 1.07179676972 cm.
Where a is AD.
b² = 1.07179676972²+4²
b = 4.14110472164 cm.
Where b is DF.
c = 4-1.07179676972
c = 2.92820323028 cm.
Where c is CD.
sin60 = d/2.92820323028
d =...
a = acos(3/5)
Where a is angle GCA.
tan(acos(3/5)) = b/3
b = 4 units.
Where b is AG.
c = atan(⅔)
Where c is angle GBE.
d = 180-acos(3/5)-atan(⅔)
d = 93.17983011986°
Where d is angle BEC.
(6/sin9...
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*...
