If `alpha, beta, gamma` are lengths of internal bisectors of angles `A,B,C` respectively of `DeltaABC`, then `sum 1/alpha cos(A/2)` is

If alpha, beta, gamma are lenghts of internal bisectors of angles A,B,C respectively of DeltaABC , then sum (1)/(alpha) "cos" (A)/(2) is

If alpha,beta,gamma are lengths of internal bisectors of angles A,B,C respectively of Delta ABC, then sum(1)/(alpha)cos((A)/(2)) is

If alpha,beta,gamma are the lengths of the altitudes of Delta ABC, then (cos A)/(alpha)+(cos B)/(beta)+(cos C)/(gamma)=

IF alpha,beta,gamma are the lengths of the altitudes of Delta ABC , then 1/alpha+1/beta-1/gamma-(2ab)/((a+b+c)Delta)cos^2""C/2=

If alpha,beta ,gamma are direction angles of a line , then cos 2alpha + cos 2beta + cos 2 gamma =

If alpha,beta, gamma are the direction angles of a vector and cos alpha=(14)/(15), cos beta=(1)/(3) , then cos gamma=

If alpha,beta, gamma are the direction angles of a vector and cos alpha=(14)/(15), cos beta=(1)/(3) , then cos gamma=

If alpha,beta,gamma are the angles of a triangle and system of equations cos(alpha-beta)x+cos(beta-gamma)y+cos(gamma-alpha)z=0 cos(alpha+beta)x+cos(beta+gamma)y+cos(gamma+alpha)z=0 sin(alpha+beta)x+sin(beta+gamma)y+sin(gamma+alpha)z=0 has non-trivial solutions, then triangle is necessarily a. equilateral b. isosceles c. right angled "" d. acute angled

If alpha,beta,gamma are the angles of a triangle and system of equations cos(alpha-beta)x+cos(beta-gamma)y+cos(gamma-alpha)z=0 cos(alpha+beta)x+cos(beta+gamma)y+cos(gamma+alpha)z=0 sin(alpha+beta)x+sin(beta+gamma)y+sin(gamma+alpha)z=0 has non-trivial solutions, then triangle is necessarily a. equilateral b. isosceles c. right angled "" d. acute angled