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
Let \(S=\left\{0∈(0,\frac{π}{2}) : \sum^{9}_{m=1} \sec(θ+(m-1)\frac{π}{6})\sec(θ+\frac{mπ}{6}) = -\frac{8}{\sqrt3}\right\}\)
Then,
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
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
