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Given :-
ABCD is a rhombus.
/_CAB = 44°.
To find :- /_BCD ; /_ABC.
Solution :-
ABCD is a rhombus .............(Given)
Therefore, AC _|_ BD..........(Property of rhombus)
In ∆AOB ,
/_AOB + /_OBA + /_BAO = 180°
..........(Angle Sum Property of Triangle)
90°. + /_OBA + 44°. = 180°
/_OBA = 46° ............(1)
/_OBA = 1/2 /_ABC .......(BD is bisector of /_ABC)
/_ ABC = 2/_OBA
/_ABC = 2 x 46°
/_ABC = 92°
Similarly /_BCO = 44°...............(2)
/_BCO = 1/2/_BCD .......(AC is bisector of /_BCD)
/_BCD = 2/_BCO
/_BCD = 2 x 44°
/_BCD = 88°
Answer :- /_BCD = 88° ; /_ABC = 92°.
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