Question

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

• A. $90$
• B. $45$
• C. $30$
• D. $60$

Answer 1

The correct option is A $90$

$\Rightarrow \frac{1}{2}ab.\sin C=\sqrt{s(s-a)(s-b)(s-c)}$
$(s-a)(s-b)=s(s-c)$
$\Rightarrow \frac{1}{2}ab.\sin C=\sqrt{s^2{(s-c)}^2}$
$\Rightarrow \frac{1}{2}ab.\sin C=s(s-c)$
$\Rightarrow \frac{\sin C}{2}=\frac{s(s-c)}{ab}$
$\Rightarrow \frac{\sin C}{2}={\left(\sqrt{\frac{s(s-c)}{ab}}\right)}^2$
$\sqrt{\frac{s(s-c)}{ab}}=\cos \frac{C}{2}$
$\Rightarrow \frac{\sin C}{2}={\left(\cos \frac{C}{2}\right)}^2$
$\Rightarrow \sin C=2{\cos }^2\frac{C}{2}$
$2\sin \frac{C}{2}\cos \frac{C}{2}=2{\cos }^2\frac{C}{2}\Rightarrow \sin \frac{C}{2}=\cos \frac{C}{2}$
$\Rightarrow \frac{\sin \frac{C}{2}}{\cos \frac{C}{2}}=1$
$\Rightarrow \tan \frac{C}{2}=1$
$\Rightarrow \frac{C}{2}=45$
$\Rightarrow C=90$