SOLUTION: am stuck with this question on proving in geometry. Given a triangle ABC prove that {{{cot(C) = expr(a/c)cosec(B)- cot(B)}}} the substitution has been slightly tricky

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Question 436542: am stuck with this question on proving in geometry.
Given a triangle ABC prove that
cot%28C%29+=+expr%28a%2Fc%29cosec%28B%29-+cot%28B%29
the substitution has been slightly tricky

Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
am stuck with this question on proving in geometry.
Given a triangle ABC prove that
cot%28C%29+=+expr%28a%2Fc%29cosec%28B%29-+cot%28B%29
the substitution has been slightly tricky




Next draw an altitude (perpendicular) from A to BC, 
and label it h.

That green altitude divides the base BC of length "a" 
into two parts, we'll let "x" represent the right part 
of "a", and the left part will be "a-x"


 
 

Using the right triangle on the right:

cot%28C%29+=+adjacent%2Fopposite+=+x%2Fh

Using the right triangle on the left:

cosec%28B%29+=+hypotenuse%2Fopposite+=+c%2Fh

cot%28B%29+=+adjacent%2Fopposite+=+%28a-x%29%2Fh

Substituting into 

cot%28C%29+=+expr%28a%2Fc%29cosec%28B%29-+cot%28B%29

x%2Fh+=+expr%28a%2Fc%29%2Aexpr%28c%2Fh%29-%28a-x%29%2Fh

Cancel the c's in the first term on the right:

x%2Fh+=+expr%28a%2Fcross%28c%29%29%2Aexpr%28cross%28c%29%2Fh%29-%28a-x%29%2Fh

x%2Fh+=+expr%28a%2Fh%29-%28a-x%29%2Fh

The fractions on the right have a common denominator so
we can combine the numerators:

x%2Fh+=+%28a-%28a-x%29%29%2Fh

x%2Fh+=+%28a-a%2Bx%29%2Fh

x%2Fh+=+x%2Fh

Edwin