ABCD is a rectangle,with angleBAC=42.

determine angle DBC.

Hi Priyanka!
Here is the answer to your question.
 
 
∠BAC = 42°.
Let the diagonals AC and BD intersect in O
∴ OB = OC       (Diagonals of rectangle are equal and bisect each other)
In DOBC, OB = OC
⇒ ∠OCB = ∠OBC      (Equal sides have equal angles opposite to them)
In DABC,
∠BAC + ∠ABC + ∠ACB = 180°   (angle sum property)
⇒ 42° + 90° + ∠ACB = 180°
⇒ ∠ACB = 180° – 132° = 48°
∴ ∠OBC = 48°    (∠OCB = ∠OBC)
Thus, the measure of ∠DBC is 48°

Hope! You got the answer.
 
Cheers!

  • 20

BAC=42  SO,ACD=42 (alternative interior angles)  ABC= 90-ACD=48  So,DBC=48 (co-interior angles)

  • 4

in a rectangle all angles are equaland each angle is a right angle

  • -5
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