What is the area of triangle ABC if AB = BC = 12 and angle ABC = 128?
Drop a perpendicular from the apex down to side AC. Call the intersection point "D".
Since this is an isoceles triangle, the perpendicular bisects both AC and angle ABC.
DC / BC = sin (64o)
DC / 12 = 0.89879
DC = 12 • 0.89879
DC = 10.78548 this is half the base
BD / BC = cos (64o)
BD / 12 = 0.43837
BD = 12 • 0.43837
BD = 5.26044 this is the height
Area = (1/2) • base • height
Area = 10.78546 • 5.26044
Area = 56.736 round to 57
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