Question

solve this Question plz​

Answer 1

$} question$

In the adjoining figure, ABCD is a parallelogram in which < CAD =40° , <BAC =35° and

<COD = 65°

FIND value of 1) <ABD 2) < BDC 3)< ACB

4)< CBD

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GIVEN : < CAD = 40° <BAC= 35° <COD = 65°

1)

<COD = 65 °

→ <COD =< AOB

( V.O.P angles )

therefore ,< AOD = 65°

In triangle AOB

<BAC + <AOB + <ABO = 180°

35° + 65 ° + < ABO = 180°

100° + < ABO = 180°

<ABO = 180° -100°

<ABO = 80°

therefore <ABD = 80°

2)

AB||CD and BD is transversel

therefore, ABD = BDC

<BDC = 80°

3)

<AOB + <BOC = 180°

65° + <BOC = 180°

<BOC = 110°

<DAO = <OCB

( alternate angles )

therefore,

<OCB = 40 °

In triangle OCB

<BOC + <OCB + <CBO = 180°

115° + 40° +<CBO = 180°

<CBO = 25°

therefore < CBD = 25 °