Mathematics - Properties of Triangle Question with Solution | TestHub

First, we find $\cos C$ using the cosine rule: $\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{3^2+4^2-5^2}{2(3)(4)}=\frac{9+16-25}{24}=\frac{0}{24}=0$. This implies $C=90^{\circ }$. The length of the angle bisector $l_c$ of angle C is given by $l_c=\frac{2ab}{a+b}\cos (C/2)$. Since $C=90^{\circ }$, $C/2=45^{\circ }$, and $\cos (C/2)=\cos 45^{\circ }=\frac{1}{\sqrt{2}}$. Substituting the values, $l_c=\frac{2(3)(4)}{3+4}\cdot \frac{1}{\sqrt{2}}=\frac{24}{7}\cdot \frac{1}{\sqrt{2}}=\frac{24\sqrt{2}}{14}=\frac{12\sqrt{2}}{7}$.