In a `Delta ABC`, the median AD is perpendicular to AC. If b = 5 and c = 11, then a =

In a Delta ABC , if D is the mid point of BC and AD is perpendicular to AC , then cos B =

In a Delta ABC, if D is the mid point of BC and AD is perpendicular to AC, then cos B=

In Delta ABC, if the median AD drawn through A is perpendicular to the side AC, then 3ca cos A cos C + 2a^(2) =

In Delta ABC , D is the mid point of BC and AD is perpendicular to AC , then show that cos A cos C = ( 2 ( c^(2) - a^(2))/( 3ac)

Let ABC be a triangle with b = 5 , c = 11 if the medium AD is perpendicular to AC , then a =

In Delta ABC, median AD is perpendicular to median BE, if BC=6,AC=8 and AB=x sqrt(5) ,then find x.

In a triangle ABC ,if the median AD through A is perpendicular to AB ,then AD is equal to (A) c tan B B) -ctanA (c) 2c tan A (D) -(c/2)tan A

In triangle ABC,D is the midpoint of BC and AD is perpendicular to AC. Then cos A.cos C=

In Delta ABC, AD and BE are perpendiculars from A and B to the sides BC and AC, then