Mathematics Question and Solution
100 = 49+64-2*7*8cosA
112cosA = 113-100
A = acos(13/112)
A = 83.3345727438°
(7/cosB) = (10/cos83.3345727438)
B = 44.0486256741°
(8/cosC) = (10/cos83.3345727438)
C = 52.6168015821°
a²= 2.5²+10²-20*2.5cos52.6168015821
a= 8.7116506555 cm.
(8.7116506555/sin52.6168015821) = (2.5/sinb)
b = 13.1808743912°
(b1)² = 2.5²+10²-20*2.5cos44.0486256741
b1 = 8.3852549156 cm.
(8.3852549156/sin44.0486256741) = (2.5/sinc)
c = 11.9635307444°
d = 180-11.9635307444-13.1808743912
d = 154.8555948644°
(10/sin154.8555948644) = (e/sin11.9635307444)
e = 4.878524367 cm.
Area Orange to 2 decimal places is;
0.5*7*8sin83.3345727438 - 0.5*4.878524367*10sin13.1808743912
= 27.8107443266-5.5621488653
= 22.2485954613
≈ 22.25 cm²
Calculating red length.
f = 4r/(3π). r = 3.5 cm.
f = 1.4854461355 cm.
tang = 1.4854461355/3.5
g = 22.9970076718°
sin22.9970076718 = 1.4854461355/h
h = 3.802177037 cm.
(10/sin154.8555948644) = (i/sin13.1808743912)
i = 5.366563146 cm.
j = 52.6168015821-11.9635307444
j = 40.6532708377°
k = 40.6532708377+22.9970076718
k = 63.6502785095°
Red Length to 2 decimal places is;
Let it be l.
l² = 3.802177037²+5.366563146²-2*5.366563146*3.802177037cos63.6502785095
l = 5.0143202057 cm.
Therefore red length to 2 decimal places is;
≈ 5.01 cm.
It implies;
Area Orange ≈
22.25 cm²
Red Length ≈ 5.01 cm.
