11.180. On the edge of a dihedral angle of $120^{\circ}$, a segment of length $c$ is taken, and from its ends, perpendiculars to it are drawn, lying in different faces of the given dihedral angle and having lengths $a$ and $b$. Find the length of the segment of the line connecting the ends of these perpendiculars.
## Solution.

Let the line $A_{1} B_{1}$ be the edge of the given dihedral angle (Fig. 11.72), $A_{1} B_{1}=c, A A_{1} \perp A_{1} B_{1}, B B_{1} \perp A_{1} B_{1}, A A_{1}=a, B \dot{B}_{1}=b$. In the plane $A A_{1} B_{1}$, draw the perpendicular $B_{1} C$ to $A_{1} B_{1}, B_{1} C=A_{1} A$. Then $\angle B B_{1} C$ is the linear angle of the given dihedral angle, $\angle B B_{1} C=120^{\circ}, A C \| A_{1} B_{1}$, $A C=A_{1} B_{1}=c$. Since $A_{1} B_{1}$ is perpendicular to the plane $B B_{1} C_{1}$, then $A_{1} B_{1} \perp B C \cdot A_{1} B_{1} \| A C$. Therefore, $A C \perp B C$. From $\triangle B B_{1} C: B C^{2}=B B_{1}^{2}+B_{1} C^{2}-$ $-2 B B_{1} \cdot B_{1} C \cdot \cos \angle B B_{1} C=b^{2}+a^{2}-2 a b \cos 120^{\circ}=a^{2}+a b+b^{2}$.

From $\triangle B C A\left(\angle B C A=90^{\circ}\right): A B=\sqrt{B C^{2}+A C^{2}}=\sqrt{a^{2}+b^{2}+c^{2}+a b}$.

Answer: $\sqrt{a^{2}+b^{2}+c^{2}+a b}$.