ABCD is a parallelogram in which ∠DAB = 70° and ∠CBD=55°. Find ∠CDB and ∠ADB.
Answers
Question :
ABCD is a parallelogram in which ∠DAB = 70° and ∠CBD=55°. Find ∠CDB and ∠ADB.
Solution :
▪Given : ABCD is a parallelogram ∠DAB = 70° and ∠CBD=55°.
▪To Find : ∠CDB and ∠ADB.
▪Solution :-
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Properties of parallelogram :
1 .Opposite angles of parallelogram Are Equal .
2. Alternate Interiors Angle of parallelogram are Equal .
3. Sum of All angles of parallelogram is 360°
4. Two opposite lines are parallel and the Never intersect each other .
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● So , Angle opposite to ∠DAB is ∠CBD
so According to Property 1 ,
=》∠DAB = ∠DCB = 70°
Now , We know that ,
☆ Sum of all Angles of a triangle is 180°
So , In Triangle DBC ,
=》 ∠DCB + ∠CBD + ∠CDB = 180°
=》 70° + 55° + ∠CDB = 180°
=》 125° + ∠CDB = 180 °
=》 ∠CDB = 180° - 125°
=》 ∠CDB = 55°
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Now For ∠ADB ,
Using property 2 ,
∠CBD = ∠ADB = 55°
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Note :
♢ Check the attachment for the diagram .
Answer :
The value of ∠CDB is 55° and the value of ∠ADB is 55° .
