[" 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 "/_DAE=theta" then "tan theta=]

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=3tan B(B)3tan 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

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

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

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

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

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