Trigonometry

Using the properties of medians and the given angles, the triangle geometry can be solved using the sine rule and median length formulas. The circumradius R is calculated as (3+sqrt(5)) * cosec(C).

Evaluate the terms: sin(30)=1/2, cos(60)=1/2, cos(45)=1/sqrt(2), sin(90)=1, tan(45)=1, cot(45)=1. The expression becomes 4((1/16 + 1/16))(1/2 - 1) + 1 - 1 = 4(1/8)(-1/2) = -1/4.

The tangent at P(at^2, 2at) has slope 1/t. The normal has slope -t. The angle between the tangent and the normal is 90 degrees. The circle through P, T, G has the normal as a diameter. The angle between the tangent at P to the parabola and the tangent at P to the circle is tan^-1(t).

Initial ladder length L = sqrt(x^2 + (2x)^2) = sqrt(5x^2) = x*sqrt(5). After sliding, the base is at x+y. The new height h is sqrt(L^2 - (x+y)^2) = sqrt(5x^2 - (x+y)^2). The slide of the upper end is the original height minus the new height: 2x - sqrt(5x^2 - (x+y)^2).

tan A = 5/12. In a right triangle, opposite=5, adjacent=12. Hypotenuse = sqrt(5^2 + 12^2) = 13. sin A = opposite/hypotenuse = 5/13.

The angle between a line and the four diagonals of a cube has a cosine value of 1/sqrt(3). The sine of that angle is sqrt(1 - cos^2) = sqrt(1 - 1/3) = sqrt(2/3).

sin 30 = 1/2, cos 60 = 1/2, cos 45 = 1/sqrt(2), sin 90 = 1, tan 45 = 1, cot 45 = 1. Expression: 4((1/16) + (1/16)) * (1/2 - 1) + 1 - 1 = 4(2/16) * (-1/2) = 4(1/8) * (-1/2) = -1/4.

Let pillar height be h, flagstaff 2h, distance x. Angle at ground: tan(theta) = h/x. Also tan(2*theta) = (h+2h)/x = 3h/x. Using tan(2*theta) = 2tan(theta) / (1-tan^2(theta)), we get 3h/x = 2(h/x) / (1 - (h/x)^2). Simplifying gives 3 = 2 / (1 - (h/x)^2), so 1 - (h/x)^2 = 2/3, (h/x)^2 = 1/3, h/x = 1/sqrt(3).

First, find the tangent of the difference of the two angles using the identity tan(A - B) = (tan A - tan B) / (1 + tan A tan B), which yields 1/7. Then, use the double-angle formula sin(2 theta) = 2 tan(theta) / (1 + tan^2(theta)) with theta = A - B to get 2(1/7) / (1 + 1/49) = 7/25.

This is a known property for a point O inside a triangle satisfying the given condition. The sum cot A + cot B + cot C for such a point O where the angles are 15 degrees is 2 + sqrt(3).

Given sin(alpha) = sin(beta) = sin(gamma) = sin(delta) = K, and alpha, beta, gamma, delta are in ascending order, we have alpha = arcsin(K), beta = pi - arcsin(K), gamma = 2pi + arcsin(K), delta = 3pi - arcsin(K). Substituting these into the expression and using trigonometric identities simplifies the result to 2*sqrt(1+K).