If in a triangle ABC, (s - a)(s - b) = s(s - | vedclass

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Question

(A) Given the relation $(s - a)(s - b) = s(s - c)$.Dividing both sides by $s(s - c)$,we get $\frac{(s - a)(s - b)}{s(s - c)} = 1$.We know the formula

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(A) Given the relation $(s - a)(s - b) = s(s - c)$.Dividing both sides by $s(s - c)$,we get $\frac{(s - a)(s - b)}{s(s - c)} = 1$.We know the formula for the half-angle tangent: $	an \frac{C}{2} = \sqrt{\frac{(s - a)(s - b)}{s(s - c)}}$.Substituting the value,we get $	an \frac{C}{2} = \sqrt{1} = 1$.Since $	an \frac{C}{2} = 1$,we have $\frac{C}{2} = 45^o$.Therefore,$C = 90^o$.

If in a triangle $ABC,$ $(s - a)(s - b) = s(s - c)$,then angle $C$ is equal to.....$^o$

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