In a triangle ABC, `angle A = 30^(2), BC = 2 + sqrt5`, then find the distance of the vertex A from the orthocenter

In a triangle ABC, angle A = 30^(o), BC = 2 + sqrt5 , then find the distance of the vertex A from the orthocenter.

In a triangle ABC, angle A = 30^(@), BC = 2 + sqrt5 , then find the distance of the vertex A from the orthocenter

In a triangle ABC, angelA=30^(@), BC=2+ sqrt(5) , then the distance of the vertex A from the orthocentre of the triangle is

In a triangle ABC, if AB = 20 cm, AC = 21 cm and BC = 29 cm, then find the distance vertex A and mid point of BC.

In triangle ABC , /_B=90^@ and AB = BC. If AC = 2sqrt2 , then find AB and BC.

Distance of the orthocenter H from vertex B is

If triangle ABC , angle B = 90^@ . If AB = 12, angle C = 30^@ . Then find BC.

In a triangle ABC the angle A=60^(@) and b:c = (sqrt(3)+1):2 . Find the other two angles B and C.

In triangle ABC, AB = (2a-1)cm, AC = 2sqrt(2a) cm, BC = (2a+1)cm, find angle BAC .