how to do this question ??
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in triangle AED,
angle AED=90°
thus,
angle ADE=180-(90+60)
=>ADE=180-150
THUS,
ADE=30°
ABCD is a cyclic quadrilateral,
thus,
opp. angles are supplementary
thus,
ABC=180-70
ABC=110°
ACD=90°
In triangle ABC,
ACB=180-(110+30)
ACB=180-140
ACB=40°
Thus, BCD=BCA+ACD
BCD=40+90
BCD=130°
angle AED=90°
thus,
angle ADE=180-(90+60)
=>ADE=180-150
THUS,
ADE=30°
ABCD is a cyclic quadrilateral,
thus,
opp. angles are supplementary
thus,
ABC=180-70
ABC=110°
ACD=90°
In triangle ABC,
ACB=180-(110+30)
ACB=180-140
ACB=40°
Thus, BCD=BCA+ACD
BCD=40+90
BCD=130°
bhumika14:
and if we have to find angle CAD ..
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