Math, asked by rockaditya45, 11 months ago

. In the figure given alongside ABCD is a kite whose diagonals AC and BD intersect each other at O. if ‹ ABO= 32º and < OCD = 40°, find 1.<ABC (ii) <ADC (iii) <BAD (iv) <BCD
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rockaditya45: pls provide solution . i will mark as brain list

Answers

Answered by AshwanikumarRai

Step-by-step explanation:

ABC_64; ADC_100; BAD_98; BDC_98

rockaditya45: bhai can you explain how BCD is 98. i have done all but my BCD is 90

rockaditya45: can you explain the last one

Answered by tarunbhoi65

In triangle ADC

COD=AOD

:the sum of all angles in a triangle is 180 degree .

therefore , 40°+40°+x= 180°

80°+x=180°

x= 100°

In triangle ABO

32°+90°+x=180°

122°+x=180°

x= 58°

In triangle COB

32°+68°+x= 180°

100°+x= 180°

x= 80°

angle ABC = 32°+32°= 68°

angle ADC =100° ( given in above solution )

angle BAD= 40°+ 58°= 98°

angle BCD = 40°+58°= 98°

I hope its help you , plz mark me as brilliant

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