Question

ABCD is a kite in which angle BDC =35 BAC= 25 Find BAC,BCD,ABC​

Answer 1

Answer:

< BDC =35 ,< BAC= 25

DC= BC  ,  AB =AD

let the intersection point between diagonals be O

the diagonals of a kite intersects into 90* . so <DOC=90* and ,BOC=90*

now  in  ΔBOC and  ΔDOC

DC= BC

<DOC =<BOC=90*

CO is common

ΔBOC ≅ΔDOC  

so < BDC = <DBC=35

so <BCD = 180-< BDC+<DBC

<BCD=110

in  ΔAOD

<AOD = 90 ,< BAC= 25

so <DAC = 180-<AOD +< BAC

    <DAC=65

in ΔAOD and ΔAOB

AB =AD

<AOD = 90= <AOB

AO is common

so ΔAOD  ≅ ΔAOB

<DAC=65= <BAC

< ABC=25 +35= 60

Answer 2

Answer:

Given

BDC - 35

BAC - 25

Find

BAC , BCD, ABC

Step-by-step explanation:

BAC-25

IN triangle BCD

side BC and CD are equal (given)

angle CDB = angle CBD ( if two sides of triangle are equal then the base angles will also be equal.

In triangle BCO ( I have marked the point of intersection of the two diagonals as O)

angle OBC -35

angle BOC - 90 ( the bigger diagonal bisects the smaller one)

angle BCO -55(angle sum property) ----------1

Similarly

in triangle OCD

angle OCD -55 ---------2.

add 1 and 2

angle BCD - 110

in triangle BAC

angle BAC - 25

angle ACB - 55

By angle sum property

angle ABC - 180 -(angle BAC + angle ACB)

angle ABC - 100.

DONE.