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 =`

If in a triangle ABC, BC = 5, AC = 4, AB = 3. D, E are the points on BC such that BD = DE = EC, angle CAE = theta . Then

In a triangle ABC, points D and E are taken on side BC such that BD= DE= EC. If angle ADE = angle AED = theta , then: (A) tan theta = 3 tan B (B) 3 tan theta = tan C

In a triangle ABC, points D and E are taken on side BC such that BD= DE= EC. If angle ADE = angle AED = theta , then: (A) tan theta = 3 tan B (B) 3 tan theta = tan C (C) (6tan theta )/( tan^2 theta -9)= tanA (D) angleB = angle C

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, points D and E are taken on side BC such that BD=DE=EC. If angle ADE = angle AED=theta, then: (A)tan theta=3tan B(B)3tan theta=tan C

In Delta ABC , /_ B = /_C , D and E are the points on AB and AC such that BD = CE , prove that DE || BC .

If in a DeltaABC, BC =5, CA=4, AB=3 and D, E are the point on BC such that BD=DE=EC, then 8tan (angleCAE) must be

In Fig.30D and E are to points on BC such that BD=DE=EC. Show that ar(ABD)=ar(ADE)=ar(AEC)

In Delta ABC, if c=(a-b) sec theta, then tan theta=

In triangleABC , D and E are two points on AB and AC such that DE || BC and AD : BD = 3 : 2, then DE : BC =