Trigonometric EquationHard
Question
In a triangle ABC with fixed base BC, the vertex A moves such that cosB + cosC = 4sin2 
. If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C, respectively, then
. If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C, respectively, thenOptions
A.b + c = 4a
B.b + c = 2a
C.locus of point A is an ellipse
D.locus of point A is a pair of straight lines
Solution
2 cos
cos 
= 4 sin2 
cos = 2 sin (A/2)
⇒
⇒
⇒ b + c = 2a (constant).
cos = 4 sin2 cos
= 2 sin (A/2)⇒
⇒
⇒ b + c = 2a (constant).
Create a free account to view solution
View Solution FreeMore Trigonometric Equation Questions
General solution of sin3x + cos3x + sin2x = 1...The equation a sin x + b cos x = c, where |c| > has -...Number of solutions of $\sqrt{3}cos2\theta + 8cos\theta + 3\sqrt{3} = 0,\theta \in \lbrack - 3\pi,2\pi\rbrack$ is:...Solution of the differential equation cos x dy = y(sin x - y) dx, 0 is...If tan and tan are the roots of the equation 8x2 - 26x + 15 = 0 then cos (α + β) is equal to-...