SOLUTION: In triangle ABC, (bsinC)(bcosC + c(cosB)) = 42. What is the area of the triangle?

Algebra ->  Trigonometry-basics -> SOLUTION: In triangle ABC, (bsinC)(bcosC + c(cosB)) = 42. What is the area of the triangle?      Log On


   

Question 1177730: In triangle ABC, (bsinC)(bcosC + c(cosB)) = 42. What is the area of the triangle?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
D = projection of A on BC.
Then:
b%2Asin%28C%29+=+AD => height of the triangle ABC with respect to base+BC
b%2Acos%28C%29+=+DC =>projection of side b+=+AC onto base BC
c%2Acos%28B%29+=+BD=>projection of side c+=+AB onto base BC
Thus
b%2Acos%28C%29%2Bc%2Acos%28B%29+=+BD+%2B+DC+=+BC+=+base

So
Therefore the area of ABC+=+42%2F2+=+21+ square units.