[" 2.In Fig."6.35,Delta ODC sim Delta OBA,/_BOC=125^(@)],[" and "/_CDO=70^(@)." Find "/_DOC,/_DCO" and "],[/_OAB" ."]

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

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

In the figure beside , DeltaODC~DeltaOBA , /_BOC=125^(@) and /_CDO=70^(@) then /_DOC=

If Fig. ODC ~=Delta OBA , |__ul(BOC)= 125^(@) and |__ul(CDO)= 70^(@) . Find |__ul(DOC), |__ul(OAB) .

In Fig. 6.35, triangleODC~triangleOBA,angleBOC=125^@ and angle CDO=70^@ Find angle DOC,angle DCO,and angle OAB .

In Fig.6.24 Delta…….. approx Delta PQR

In Delta ABC,AC=BC and /_ BAC =70^(@) . Find /_BCA

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

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