Mathematics Question and Solution
Calculating x, length AD.
Let AB = AC = a.
cosb = x/32 --- (1).
a² = 32²+x²-2*32*xcosb
a² = 1024+x²-64xcosb --- (2).
a² = 23²+x²+2*23*xcosb
a² = 529+x²+46xcosb --- (3).
Equating (2) and (3).
1024+x²-64xcosb = 529+x²+46xcosb
1024-64xcosb = 529+46xcosb
495 = 110xcosb
cosb = 495/(110x) --- (4).
Therefore, x, length AD is;
Equating (1) and (4).
x/32 = 495/(110x)
110x² = 32*495
x² = (32*495)/110
x² = 144
x = 12 units.
