Question

please answerrrrr only one questionnnnnnnn​

Answer 1

Answer:

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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