Mathematics Question and Solution
Calculating Area Triangle A.
Let the two equal inscribed angles be a each.
(0.5*14*12sina)÷(0.5*12*35sina)
= 14/35
= 2/5
= 2:5
Where 2:5 is the ratio of the bases of the both triangles with equal angle opposite their base.
Let the total length be x.
It implies;
(2x)² = 14²+12²-2*14*12cosa
4x² = 340-336cosa
cosa = (340-4x²)/336 --- (1).
(5x)² = 12²+35²-2*12*35cosa
25x² = 1369-840cosa
cosa = (1369-25x²)/840 --- (2).
Calculating x.
Equating (1) and (2).
(340-4x²)/336 = (1369-25x²)/840
840(340-4x²) = 336(1369-25x²)
210(340-4x²) = 84(1369-25x²)
35(340-4x²) = 14(1369-25x²)
11900-140x² = 19166-350x²
350x²-140x² = 19166-11900
210x² = 7266
x = √(7266/210)
x = 5.88217646794 units.
Calculating a, using (1).
cosa = (340-4(5.88217646794)²)/336
cosa = 0.6
a = acos(0.6)
a = 53.1301023542°
Therefore, Area A is;
½*12*35sin53.1301023542
= 210sin53.1301023542
= 168 square units.
