Question

In the figure, ABCD is a rectangle in which diagonal AC is produced to E. If $\angle ECD={146}^{\circ }$, find $\angle AOB$.

Answer 1

In rectangle ABCD, Diagonals AC and BD bisect each other at O

AC is produced to E and $\angle ECD={146}^{\circ }$

$\angle ECD+\angle DCA={180}^{\circ }\left(Linearpair\right)$

$\Rightarrow {146}^{\circ }+\angle DCA={180}^{\circ }$

$\Rightarrow \angle DCA={180}^{\circ }-{146}^{\circ }$

$\Rightarrow \angle DCA={34}^{\circ }$

$\therefore \angle CAB=\angle DCA={34}^{\circ }\left(Alternateangles\right)$

We know that diagonals of rectangle are equal and bisect each other.

Diagonals AC and BD bisect each other at O and are also equal.

So, $AC=BD$

$\Rightarrow \frac{1}{2}AC=\frac{1}{2}BD$

$\Rightarrow OA=OB$

$\therefore \angle OAB=\angle OBA={34}^{\circ }$ [Angles opposite to equal sides]

Now in $\Delta AOB,$

$\angle AOB={180}^{\circ }-(\angle OAB+\angle OBA)$ [Angle Sum property]

$={180}^{\circ }-({34}^{\circ }+{34}^{\circ })$

$={180}^{\circ }-{68}^{\circ }={112}^{\circ }$