To solve the problem of finding the number of triangles whose vertices are at the vertices of a regular octagon, but none of whose sides is a side of the octagon, we can follow these steps:
First, calculate the total number of triangles that can be formed by choosing any three vertices from the octagon. A regular octagon has 8 vertices. The number of ways to choose 3 vertices from 8 is given by the combination formula:
\(^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\).
Next, subtract the number of triangles where at least one side is a side of the octagon from the total number of triangles.
To determine these triangles, notice that for a triangle to have a side of the octagon, it must include two adjacent vertices:
There are 8 such pairs of adjacent vertices (one pair for each side of the octagon). Thus, the number of unwanted triangles (where one side is a side of the octagon) is:
\(8 \times 5 = 40\).
Finally, subtract the number of unwanted triangles from the total number possible:
\(56 - 40 = 16\).
Thus, the number of triangles where none of the sides is a side of the octagon is 16.
Therefore, the correct answer is 16.
The number of triangles having no side common with an \( n \)-sided polygon is given by:
\[ \text{no. of triangles having no side common with a } n \text{-sided polygon} = \binom{n}{1} \times \binom{n-4}{2} \div 3 \]
Substitute \( n = 8 \):
\[ = \binom{8}{1} \times \binom{4}{2} \div 3 \]
\[ = 8 \times 6 \div 3 \]
\[ = 16. \]
Thus, the number of such triangles is 16.
If \( \theta \in \left[ -\frac{7\pi}{6}, \frac{4\pi}{3} \right] \), then the number of solutions of \[ \sqrt{3} \csc^2 \theta - 2(\sqrt{3} - 1)\csc \theta - 4 = 0 \] is equal to ______.
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is _____. 