Let us consider a triangle ABC having BC=5 cm, CA=4cm, AB=3cm, D,E are points on BC such BD = DE= EC, `angleCAE=theta`, then:
`AE^(2)` is equal to

Let us consider a triangle ABC having BC=5 cm, CA=4cm, AB=3cm, D,E are points on BC such BD = DE= EC, angleCAE=theta , then: AD^(2) is equal to

Let us consider a triangle ABC having BC=5 cm, CA=4cm, AB=3cm, D,E are points on BC such BD = DE= EC, angleCAE=theta , then: tan theta is equal to

In a triangle ABC, if a = 5, b=4 and c = 3, D and E are the points on BC such that BD = DE = EC. If DeltaDAE =theta, then tan theta =

Can we drawn a triangle, ABC with AB=3 cm, BC= 3.5 cm and Ca=65 cm ?

In a Delta ABC, AB=AC=17cm D is a point on BC such that AD=15, CD=4 cm , find BD=?

In triangle ABC , AB=7 cm, BC = 10 cm, and AC = 8 cm. If AD is the angle bisector of angle BAC , where D is a point on BC, then BD is equal to:

In figure, ABC is a triangle with angle B = 90^@, BC = 3cm and AB = 4cm. D is a point on AC such that AD = 1cm and E is the mid- point of AB. Join D and E and extend DE to meet CB extended at F. Then BF is equal to

ABC is a triangle and D is a point on the side BC. If BC = 12 cm, BD = 9 cm and angle ADC = angle BAC, then the length of AC is equal to