We are given that the interior angles are in arithmetic progression (A.P.)
with a common difference of 6° and the largest angle is 219°. The sum of the interior angles of an \( n \)-sided polygon is given by: \[ \frac{n}{2} \left( 2a + (n-1) \times 6 \right) = (n-2) \times 180 \] where \( a \) is the first angle. Simplifying: \[ an + 3n^2 - 3n = (n-2) \times 180 \] Now, using the condition that the largest interior angle is 219°, we have: \[ a + (n-1) \times 6 = 219 \] which simplifies to: \[ a = 225 - 6n \] Substitute this value of \( a \) into the sum equation: \[ (225 - 6n) + 3n^2 - 3n = (n-2) \times 180 \] Solving the resulting quadratic equation gives \( n = 20 \).
A point particle of charge \( Q \) is located at \( P \) along the axis of an electric dipole 1 at a distance \( r \) as shown in the figure. The point \( P \) is also on the equatorial plane of a second electric dipole 2 at a distance \( r \). The dipoles are made of opposite charge \( q \) separated by a distance \( 2a \). For the charge particle at \( P \) not to experience any net force, which of the following correctly describes the situation?

If $10 \sin^4 \theta + 15 \cos^4 \theta = 6$, then the value of $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is: