ABCD is a Rectangle, diagonals AC and BD intersect each other at …

MCQ

$ABCD$ is a Rectangle, diagonals $AC$ and $BD$ intersect each other at $P$. If $\angle\text{APD} = 52^\circ,$ find $\angle\text{ACB}$ and $\angle\text{DBA}=\ ?$

✓
$64^\circ $ and $26^\circ $
B
$20^\circ $ and $120^\circ $
C
$100^\circ $ and $260^\circ $
D
$25^\circ $ and $25^\circ $

✓

Answer

Correct option: A.

$64^\circ $ and $26^\circ $

In Rectangle, diagonals are equal and bisect each other.
In $\triangle\text{APD, AP = PD}$
$\Rightarrow \angle\text{ADP} = \angle\text{PAD} = \text{x}$ (angle opposite to equal sides are equal)
In $\triangle\text{APD}, \angle\text{APD} + \angle\text{PDA} + \angle\text{DAP} = 180^\circ$ (angle sum property)
$52^\circ + \text{x} + \text{x} = 180^\circ$
$2\text{x} = 180^\circ - 52^\circ = 128^\circ$
$\text{x} = 64^\circ$
$\angle\text{DAC} = \angle\text{BCA} = 64^\circ$ (alternate angles)
In $\triangle\text{ADB}, \angle\text{ADB} + \angle\text{DBA} + \angle\text{BAD} = 180^\circ$ (angle sum property)
$64^\circ + \angle\text{DBA} + 90^\circ = 180^\circ$
$\angle\text{DBA} = 180^\circ - 154^\circ = 26^\circ$