A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that QR = 15 m. If from a point A on the ground the angle of elevation of R is 60° and the part PR of the tower subtends an angle of 15° at A, then the height of the tower is :
5(2\(\sqrt3\)+3)m
5(\(\sqrt3\)+3)m
10(\(\sqrt3\)+1)m
10(2\(\sqrt3\)+1)m

From △APQ
\(\frac{x+15}{y}\)=tan75º⋯(i)
From △RQA
\(\frac{15}{y}\)=tan60º⋯(ii)
From (i) and (ii)
\(\frac{x+15}{15}\)=tan75ºtan60º=tan(45º+30º)tan60º=3+1(3−1)⋅3
On simplification,
x=10\(\sqrt3\) m
Hence the height of the tower
=(15+10\(\sqrt3\)) m
=5(2\(\sqrt3\)+3) m
If $ \theta \in [-2\pi,\ 2\pi] $, then the number of solutions of $$ 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 $$ is:
For \( \alpha, \beta, \gamma \in \mathbb{R} \), if \[ \lim_{x \to 0} \frac{x^2 \sin(\alpha x) + (\gamma - 1)e^{x^2}}{\sin(2x - \beta x)} = 3, \] then \( \beta + \gamma - \alpha \) is equal to:

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
The relationship between the sides and angles of a right-angle triangle is described by trigonometry functions, sometimes known as circular functions. These trigonometric functions derive the relationship between the angles and sides of a triangle. In trigonometry, there are three primary functions of sine (sin), cosine (cos), tangent (tan). The other three main functions can be derived from the primary functions as cotangent (cot), secant (sec), and cosecant (cosec).
sin x = a/h
cos x = b/h
tan x = a/b
Tan x can also be represented as sin x/cos x
sec x = 1/cosx = h/b
cosec x = 1/sinx = h/a
cot x = 1/tan x = b/a
