Mathematics Question and Solution
Let AD be x.
a² = x²+4²-2*4*xcos30
a = √(x²+16-4√(3)x) units.
a is BD.
b² = x²+3²-2*3*xcos30
b = √(x²+9-3√(3)x) units.
b is CD.
c² = 4²+3²-2*3*4cos60
c² = 25-12
c = √(13) units.
c is BC.
Calculating length AD (x).
It implies;
a+b = c
√(x²+16-4√(3)x)+√(x²+9-3√(3)x) = √(13)
x²+9-3√(3)x = (√(13)-√(x²+16-4√(3)x))²
x²+9-3√(3)x = 13-2√(13)√(x²+16-4√(3)x)+x²+16-4√(3)x
20-√(3)x = 2√(13)√(x²+16-4√(3)x)
(20-√(3)x)² = 52(x²+16-4√(3)x)
400-40√(3)x+3x² = 52x²-208√(3)x+832
49x²-168√(3)x+432 = 0
49x²-290.984535672x+432 = 0
Therefore;
x = 2.96924 units.
Again, x is AD.
