Question
ABCD is a Rectangle, diagonals AC and BD intersect each other at P. If $\angle\text{APD} = 52^\circ,$ find $\angle\text{ACB}$ and $\angle\text{DBA}=\ ?$
✓
Answer
- 64º and 26º
Solution:
In Rectangle, diagonals are equal and bisect each other.
In $\triangle\text{APD, AP = PD}$
$⇒ \angle\text{ADP} = \angle\text{PAD} = \text{x}$ (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)
$64^\circ + \angle\text{DBA} + 90^\circ = 180^\circ$
$\angle\text{DBA} = 180^\circ - 154^\circ = 26^\circ$
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