<PRE><B>Question 197854</B>
Would you please help me with this question.. 
In triangle ABC, AC = 18, BC =10, and cosC = 1/2. Find the area of triangle ABC to the nearest tenth of a square unit. 
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Assuming the triangle is a right triangle, since cos(C) = 1/2,
sin(C) = [sqrt(3)]/2
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Formula:
Area = (1/2)ab*sin(C)
Area = (1/2)*10*18*[sqrt(3)/2]
= 5*9*sqrt(3)
= 77.94 sq. units
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Cheers,
Stan H.
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