[" In a triangle "ABC" ,the length of the bisector of angle "A" is "],[(2bc sin(A/2))/(b+c)],[(2bc cos(A/2))/(b+c)],[(4Delta)/(b+c)cosec(A)/(2)]

In a triangle ABC the length of the bisector of angle A is : (i) 2(bc)/(b+c) sin.(A)/(2) (ii) 2(bc)/(b+c) cos.(A)/(2) (iii) (abc)/(2R(b+c)) cosec.(A)/(2) (iv) (4A)/(b+c) cosec.(A)/(2)

" In a triangle "ABC" ,if "2cos A=sin B" cosec "C" then "

Statement I In any triangle ABC, the square of the length of the bisector AD is bc(1-(a^(2))/((b+c)^(2))). Statement II In any triangle ABC length of bisector AD is (2bc)/((b+c))cos ((A)/(2)).

Statement I In any triangle ABC, the square of the length of the bisector AD is bc(1-(a^(2))/((b+c)^(2))). Statement II In any triangle ABC length of bisector AD is (2bc)/((b+c))cos ((A)/(2)).

In a triangle ABC with fixed base BC,the verte moves such that cos B+cos C=4sin^(2)((A)/(2)) then

In a triangle ABC,bc cos^(2)(A)/(2)+ca cos^(2)(B)/(2)+ab cos^(2)(C)/(2)=

In triangle ABC,a,b,c are the lengths of its sides and A,B,C are the angles of triangle ABC .The correct relation is given by (a) (b-c)sin((B-C)/(2))=a(cos A)/(2) (b) (b-c)cos((A)/(2))=as in(B-C)/(2)(c)(b+c)sin((B+C)/(2))=a(cos A)/(2)(d)(b-c)cos((A)/(2))=2a(sin(B+C))/(2)

For any triangle ABC, find the value of (bc cos^2(A/2)+cacos^2(B/2)+abcos^2(C/2))/(a+b+c .