Mathematics Question and Solution
Calculating FB.
Let a be OC = OE = CE.
b = (a+16) units.
b is the radius of the ascribed half circle (OD = OA).
c = b-5
c = (a+16)-5
c = (a+11) units.
c is OB.
d = c+a
d = (a+11)+a
d = (2a+11) units.
d is BC.
Calculating a.
e² = (a+11)²+a²-2a(a+11)cos120
e² = 2a²+22a+121+a²+11a
e² = 3a²+33a+121 --- (1).
e² = (a+11)²+(a+16)²-2(a+11)(a+16)cos60
e² = a²+22a+121+a²+32a+256-((a+11)(a+16))
e² = 2a²+54a+377-(a²+27a+176)
e² = a²+27a+201 --- (2).
Calculating a.
Equating (1) and (2).
3a²+33a+121 = a²+27a+201
2a²+6a-80 = 0
a²+3a-40 = 0
Therefore;
a ≠ -8
a = 5 units.
Again, a is OC = OE = CE.
Recall.
OB = a+11
And a = 5 units.
OB = 5+11
OB = 16 units.
Therefore, the required length, FB is;
(FB)² = 16²+5²-2*16*5cos120
(FB)² = 361
FB = √(361)
FB = 19 units.
