Question 1104855
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The side lengths are in arithmetic progression, so let the side lengths be x-a, a, and x+a.  Then<br>
{{{(x-a)^2+x^2 = (x+a)^2}}}
{{{x^2-2ax+a^2+x^2 = x^2+2ax+a^2}}}
{{{x^2-4ax = 0}}}
{{{x(x-4a) = 0}}}
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So x = 4a, and the side lengths of the triangle are
x-a = 4a-a = 3a;
x = 4a; and
x+a = 4a+a = 5a<br>
So the side lengths are in the ratio 3:4:5.<br>
That means the sines of the acute angles (A and C) are 3/5 and 4/5.<br>
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.