SOLUTION: If sides of a right angled triangle ABC are in A.P.(angle B =90),then sinC=?

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Question 1104855: If sides of a right angled triangle ABC are in A.P.(angle B =90),then sinC=?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The side lengths are in arithmetic progression, so let the side lengths be x-a, a, and x+a. Then

%28x-a%29%5E2%2Bx%5E2+=+%28x%2Ba%29%5E2
x%5E2-2ax%2Ba%5E2%2Bx%5E2+=+x%5E2%2B2ax%2Ba%5E2
x%5E2-4ax+=+0
x%28x-4a%29+=+0


So x = 4a, and the side lengths of the triangle are
x-a = 4a-a = 3a;
x = 4a; and
x+a = 4a+a = 5a

So the side lengths are in the ratio 3:4:5.

That means the sines of the acute angles (A and C) are 3/5 and 4/5.

But we can't finish the problem, because there is nothing in your statement of the problem to tell us which of the acute angles is angle C.