9. In the adjoining figure, ABCD is a parallelogram and
diagonals intersect at O.Find the
angle ACD
Answers
Answer:
Here, ABCD is a parallelogram.
∠CBD=46
o
[ Given ]
AD∥BC and AC is a transversal.
∴ ∠CBD=∠BDA [ Alternate angles ]
∴ ∠BDA=46
o
.
So, ∠ODA=46
o
In △ODA,
⇒ ∠ODA+∠OAD+AOD=180
o
⇒ 46
o
+∠OAD+68
o
=180
o
⇒ 114
o
+∠OAD=180
o
⇒ ∠OAD=66
o
∴ ∠CAD=66
o
⇒ ∠D+∠A=180
o
[ Sum of adjacent sides are supplementary in parallelogram ]
⇒ (∠CDB+∠BDA)+(∠CAD+∠BAC)=180
o
⇒ (30
o
+46
o
)+(66
o
+∠BAC)=180
o
⇒ 142
o
+∠BAC=180
o
⇒ ∠BAC=38
o
⇒ ∠BAC=∠ACD [ Alternate angles ]
∴ ∠ACD=38
o
Now,
⇒ ∠ADC=∠CDB+∠BDA=30
o
+46
o
=76
o
∴ We get, ∠CAD=66
o
,∠ACD=38
o
and ∠ADC=76
o
Answer:∠ACD = 38°
mark as brainliest
Explanation:
∠AOD +∠DOC = 180° ----> BEING ST. LINE
or, 68° + ∠DOC = 180°
∴∠DOC = 112°
in triange DOC,
∠DOC +∠ODC+ ∠ACD = 180° ---> sum of angles of a triangle
or, 112° + 30° +∠ACD =180°
∴∠ACD = 38°