. 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
20
Step-by-step explanation:
ABC_64; ADC_100; BAD_98; BDC_98
bhai can you explain how BCD is 98. i have done all but my BCD is 90
can you explain the last one
Answered by
98
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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