Trigonometry

If tan(A/2), tan(B/2), tan(C/2) are in HP, then 1/tan(A/2), 1/tan(B/2), 1/tan(C/2) are in AP, which means cot(A/2), cot(B/2), cot(C/2) are in AP. In any triangle, cot(A/2) + cot(C/2) = 2 * cot(B/2) * (something). Given the properties of triangles, the minimum value of cot(B/2) for this condition is sqrt(3).

Using cot(A+B) = (cotA cotB - 1) / (cotA + cotB), we can manipulate the expression. The expression simplifies to tan(50 degrees).

Simplify alpha: (tan^2 - sin^2)/(tan^2 * sin^2) = (sin^2/cos^2 - sin^2)/(sin^4/cos^2) = (sin^2(1-cos^2)/cos^2)/(sin^4/cos^2) = sin^2 * sin^2 / sin^4 = 1. Similarly, beta = 1. If roots are 1 and 1, the discriminant D = b^2 - 4ac = (-2)^2 - 4(1)(1) = 0.

This is a fundamental trigonometric identity: cosec^2(theta) - cot^2(theta) = 1.

Let h be the height of the posts and x be the distance from the 60-degree post. Then h/x = tan(60) = sqrt(3) and h/(50-x) = tan(30) = 1/sqrt(3). So h = x*sqrt(3) and h = (50-x)/sqrt(3). Equating: x*sqrt(3) = (50-x)/sqrt(3) => 3x = 50-x => 4x = 50 => x = 12.5.

This forms a right triangle with base 12 and height (11-6) = 5. The hypotenuse is sqrt(12^2 + 5^2) = sqrt(144 + 25) = sqrt(169) = 13.

Using the law of cosines: c^2 = a^2 + b^2 - 2ab cos(C). Here, c = AB = 4, a = BC = 5, b = AC = 6. 4^2 = 5^2 + 6^2 - 2(5)(6) cos(C). 16 = 25 + 36 - 60 cos(C). 60 cos(C) = 45. cos(C) = 45/60 = 3/4. C = cos^-1(3/4).

The total frequency in a distribution is the sum of frequencies. Given the cumulative frequencies at the boundaries, the total frequency is the sum of the frequencies of all classes. The data provided (less than CF 35 at 60 and greater than CF 25 at 60) implies the total frequency is 35 + 25 = 60.

For unit vectors A and B, A.B = cos(theta). The expression becomes (1 - cos(theta)) / (1 + cos(theta)). Using trigonometric identities, 1 - cos(theta) = 2*sin^2(theta/2) and 1 + cos(theta) = 2*cos^2(theta/2). The ratio is (2*sin^2(theta/2)) / (2*cos^2(theta/2)) = tan^2(theta/2).

If sides are in A.P., they are a-d, a, a+d. By Pythagorean theorem, (a-d)^2 + a^2 = (a+d)^2. This simplifies to a^2 - 2ad + d^2 + a^2 = a^2 + 2ad + d^2, so a^2 = 4ad, or a = 4d. Sides are 3d, 4d, 5d. Sine of angles are 3/5 and 4/5.

The distances from the circumcenter to the sides of a triangle are given by R cos A, R cos B, and R cos C, where R is the circumradius. Therefore, the ratio of these distances is cos A : cos B : cos C.

The angle theta between tangents from a point (x1, y1) to a circle with radius r and center (h, k) satisfies sin(theta/2) = r/d, where d is the distance from the point to the center. Here, the center is (2.5, -2), the radius is sqrt(2.5^2 + (-2)^2 + 2) = sqrt(6.25 + 4 + 2) = sqrt(12.25) = 3.5, and the distance d from (-1, 0) to (2.5, -2) is sqrt((2.5 - (-1))^2 + (-2 - 0)^2) = sqrt(3.5^2 + 2^2) = sqrt(12.25 + 4) = sqrt(16.25). Using sin(theta/2) = 3.5/sqrt(16.25) leads to tan(theta/2) = 7/4, so theta = 2*tan^-1(7/4).

This is a standard trigonometric identity related to the heights of pillars and angles of elevation from specific points. The given equation is a known result in trigonometry problems involving pillars.

The horizontal separation of the tops is 12 m, while their vertical separation is 11 - 6 = 5 m. The distance between the tops is therefore sqrt(12^2 + 5^2) = 13 m.