The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32°, ∠AOB = 70°, then ∠DBC is equal to:
Answer:
<DBC = 38°
Step-by-step explanation:
Given:
ABCD is a parallelogram .
AC,BD diagonals intersecting at "O".
<DAC = 32°, <AOB = 70°
Proof:
i) AB//DC , AC is a transversal.
<DAC = <BCA = 32°
/* Alternate angles */
ii) In ∆BCO , CO extended to A.
Sum of interior opposite angles = Exterior angle at O
=> <BCA + <OBC = <AOB
=> <BCA + <DBC = <AOB
=> 32° + <DBC = 70°
=> <DBC = 70°-32°
=> <DBC = 38°
The opposite angles of a parallelogram are (3x – 2)° and (50 – x)° the measure of these angles is ______.
In parallelogram ABCD, if ∠A = 2x + 15°, ∠B = 3x – 25°, then value of x is:
2x+15+3x-25=180
[co interior angles]
5x-10=180
5x=190
x=38
option d is right answer
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