## Trigonometry (10th Edition)

Note that the angle at point $C$ is $110^{\circ}$. We can use the law of cosines to find the length $AB$, which is the length of the tunnel: $AB^2 = AC^2+BC^2-2(AC)(BC)~cos~C$ $AB = \sqrt{AC^2+BC^2-2(AC)(BC)~cos~C}$ $AB = \sqrt{(3800~m)^2+(2900~m)^2-(2)(3800~m)(2900~m)~cos~110^{\circ}}$ $AB = \sqrt{30388123.96~m^2}$ $AB = 5512.5~m$ The length of the tunnel is 5512.5 m