please solve my doubt of quadrilateral
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abo= (180-25+75)=80°
odc=80°
acb=35
odc=80°
acb=35
JoydipGhosh:
i am in metro.
can you please tell me your qualifications
Class 12 studying
ok thanks ☺
And your qualification??
class 8th studying but i m curious to enchance my knowledge of class 10 math and science
Which school do you read in?
kishore gk English school
are you Ahmadabad?
yess
Answered by
1
angle DOC=angle AOB(vertically oppo.)
So,angle AOB =75°
In triangle AOB
angle AOB+angle OAB+angle ABO=180°
75°+25°+angle ABO=180°
angle ABO +100°=180°
angle ABO=80°
so, angle ABD =80°. 1
ABCD is a parallelogram
So,AD parallel to BC
BD intersect them
So,angle ABD=angle BDC(alternate angles)
from 1
angle BDC =80°
So, angle ODC=80°
angle AOD+angle COD=180°(linear pair)
So, angle AOD =105°
So, by applying angle sum property in triangle AOD we will get
angle ODA =40°
So,angle BDA=40°
also, angle BDA =angle CBD(alternate angles)
So,angle CBD =40°
So,angle AOB =75°
In triangle AOB
angle AOB+angle OAB+angle ABO=180°
75°+25°+angle ABO=180°
angle ABO +100°=180°
angle ABO=80°
so, angle ABD =80°. 1
ABCD is a parallelogram
So,AD parallel to BC
BD intersect them
So,angle ABD=angle BDC(alternate angles)
from 1
angle BDC =80°
So, angle ODC=80°
angle AOD+angle COD=180°(linear pair)
So, angle AOD =105°
So, by applying angle sum property in triangle AOD we will get
angle ODA =40°
So,angle BDA=40°
also, angle BDA =angle CBD(alternate angles)
So,angle CBD =40°
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