In the given figure `Delta ODC ~ Delta OAB, /_BOC=100^@, /_ODC=60^@` then `/_OAB=`

In the given figure Delta ODC ~ Delta OAB , /_BOC= 125^@, /_ODC = 70^@ , then /_OAB is equal to :

In the given figure DeltaODC ~ DeltaOBA, /_BOC= 125^(@)" and "/_CDO= 70^(@) . Find /_DOC, /_DCO" and "/_OAB .

In the given figure, Delta ODC~ Delta OBA, angle BOC =115^(@) and angle COD =70^(@) Find (i) angle DOC (ii) angle DCO (iii) angle OAB (iv) angle OBA

In the figure given below, Delta ODC - Delta OBA , /_ BOC = 125^(@) and /_ CDO = 70^(@) . Find /_ DOC , /_ DCO and /_ OAB

In the given figure, Delta OQP~ Delta OAB, angle OPQ =56^(@) and angle BOQ=132^(@) . Find angleOAB .

In the given figure, Delta OAB~ Delta OCD . If AB=8 cm, BO= 6.4 cm, OC=3.5 cm and CD=5 cm find (i) OA (ii) DO.

In the given figure, AB||CD . If angleAOC=30^(@) and /_OAB=100^(@) then angleOCD=?

In the given figure,O is the centre of the circle /_OAB=30^(@) and /_OCB=40^(@). Calculate /_AOC