`DeltaABC` में, `angleABC=5 angleACB` तथा `angleBAC=3 angleACB` हे, तब `angleABC=?`

angleBAC:angleABC:angleACB = What?

In fig. angleABC=angleACB and angle3=angle4 . Prove that angle1=angle2

In the figure above, BC || DE and angleABC=angleCED. angleA=angleACB-30^(@) . Find angleA .

In a DeltaABC, angleBAC gt 90^(@) , then angleABC and angleACB must be :

In given figure angleABC = angleACB and angle3 =angle4 Show that angle1 = angle2

In triangle ABC, angleABC=angleACB . The bisectors of angleABCandangleACB intersect AC and AB at E and F respectively. Prove that FE||BC.

In the adjacent figure, it is given that AB AC, angleBAC = 36^@ , angleADB = 45^@ and angleAEC = 40^@ . Find : angleABC .

In DeltaABC, angleBAC=75^(@), angleABC=45^(@) , BC is extended to D. If angleACD=x^(@) then 60^(@) of (x)/(3)% is?

In the fig side BC of triangleABC is produced to angleACD=128^@ and angleABC=43^@ find angleBAC and angleACB .