Notice, triangle ABC is equilateral.
BD = √(3)+2-1
BD = (√(3)+1) units.
BD = 2.73205080757 units.
BC = 3.73205080757 units.
(CD)² = 2.73205080757²+3.73205080757²-2*2.73205080757*3.73205080757cos6...
Let the radius of the three congruent inscribed circles be r.
Calculating r.
(1+r)² = (1-r)²+(1-3r)²
1+2r+r² = 1-2r+r²+1-6r+9r²
1+2r+r² = 10r²-8r+2
9r²-10r+1 = 0
Therefore, solving the quadratic...
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(AC)² = 2a²
AC = √(2)a units.
b² = 3²+4²
b = 5 units.
c = √(16-a²) units.
d = a-c
d = (a-√(16-a²)) units.
Therefore, considering similar triangle ratios.
a/4 = d/3. And d is (a-√(16-a²)).
Ther...
4 = 18-18cosa
a = acos(7/9)
a = 38.94244126898°
a is angle BOC.
b = 90-38.94244126898
b = 51.05755873102°
b is angle COD.
d = 0.5(180-38.94244126898)
d = 70.52877936551°
d is angle OCB...
Calculating r, radius of the circle.
(r/tan40)+(r/tan30) = 10
r = 3.42020143326 cm.
(a/sin60) = (10/sin40)
a = 13.47296355334 cm.
tan40 = 3.42020143326/b
b = 4.07603734548
(4.07603734548/sin50)...
Let the three congruent lengths be 1 unit.
a² = 2-2cos96
a = 1.48628965095 units.
(1/sin12) = (1.48628965095/sinb)
b = 18°
It implies;
Alpha = b = 18°
AC = √(4²+4²)
AC = √(32)
AC = 4√(2) Units.
(4√(2))² = (4+r)²+r²
32 = 16+8r+r²+r²
2r²+8r-16 = 0
r²+4r-8 = 0
(r+2)² = 12
r = -2±2√(3)
r ≠ -2-2√(3) units.
r = (2√(3)-2) units.
a = r+4
a = (2+2√(3)) u...
r, radius of the inscribed circle is;
2(r/tan22.5) = 12√(2)
r = 6√(12)tan22.5
r = 3.51471862576 cm.
a = 12-3.51471862576
a = 8.48528137424 cm.
b = (8.48528137424-c) units.
c is the radius of the...
a² = 5²+2²
a = √(29) cm.
b² = 3²+2²
b = √(13) cm
c = atan(2/5)°
d = atan(1/3)°
e = atan(2/3)°
f = 90-c-d
f = 90-atan(2/5)-atan(1/3)
f = 49.76364169073°
g = atan(3)°
h = g-e
h = atan(3)-atan(2...
