If cos2A+cos2B+cos2C is minimum then A= B=C=60∘
So △ABC is equilateral
Now in-radias r=3
So in △MBD we have
Tan30∘=BDMD=a/2r=a6
1/3=a1=a=63
Perimeter of △ABC=183
Area of △ABC=43a2=273
If \( \cos 2A + \cos 2B + \cos 2C \) is minimum, then \[ A = 60^\circ, \quad B = C = 60^\circ \] \[ \Rightarrow \triangle ABC \text{ is equilateral} \] \[ \text{In-radius} = r = 3 \]
\[ \text{Area} = \frac{\sqrt{3}}{4} a^2 = 27\sqrt{3} \] \[ \text{Thus, area is incorrectly given as } \frac{27\sqrt{3}}{2} \]
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
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