3²+4²+2*3*4cosa = 5²+6²-2*5*6cosa
25+24cosa = 61-60cosa
84cosa = 36
cosa = (36/84)
cosa = (3/7)
a = acos(3/7)
a = 64.6230664748°
b = 2a
b = 129.2461329497°
c² = 5²+6²-2*5*6cos64.6230664748
c = 5.9...
Calculating the shaded area.
a² = 4²+6²-2*4*6cos30
a = 3.2296719057 units.
sin15 = b/6
b = 1.5529142706 units.
c = 90-15
c = 75°
Or
cosc = 1.5529142706/6
c = acos(1.5529142706/6)
c = 75°
(3.22...
2r² = 2²
r = √(2) units.
r is the radius of the circle.
Area circle is;
πr²
= π√(2)²
= 2π square units.
Sir Mike Ambrose is the author of the question.
Let the radius of the ascribed semicircle be 2 units.
Therefore;
Radius of the inscribed circle is 1 unit.
It implies;
Area R is;
Area sector with...
Length DG is;
cos15 = 0.5(12)/(DG)
DG = 6/cos15
DG = 6.21165708246 cm.
x = 12.416666 - (1).
10x = 124.16666 - (2).
100x = 1241.6666 - (3).
1000x = 12416.6666 - (4).
Subtracting (3) from (4).
900x = 11175
It implies x (p/q) is;
x = 11175/900
x = 2235/180
x = 447/36
x...
a² = 4²+2²
a² = 20
a = √(20)
a = 2√(5) units.
tanb = 2/4
b = atan(½)°
b is angle theta.
c = 180-2b
c = (180-2atan(½))°
d = 360-90-c-b
d = 360-90-(180-2atan(½))-atan(½)
d = (90+atan(½))°
e² = √(...
Let r be the radius of the inscribed circle.
Calculating r.
√(5)² = (2r)²+r²
5 = 5r²
r = 1 unit.
Notice!
The square side is 4r.
Let a be the square side.
a = 4r
And r = 1 unit.
a = 4(1)
a = 4...
sin30 = a/4
Where a is the red length.
a = 4sin30
a = 2 units.
Calculating PQ.
Let r be the radius of the smaller circle.
7 units is the radius of the bigger circle.
sin60 = a/7
√(3)/2 = a/7
a = ½(7√(3)) units.
a = 6.0621778265 units.
a is the radius, r of t...
