Question

Question 2
A diagonal of a rectangle is inclined to one side of the rectangle at ${25}^{\circ }$. The acute angle between the diagonals is
(a) ${55}^{\circ }$
(b) ${50}^{\circ }$
(c) ${40}^{\circ }$
(d) ${25}^{\circ }$

Answer 1

$\therefore AC=BD$

$\Rightarrow \frac{1}{2}AC=\frac{1}{2}BD$ [dividing both sides by 2 ]

$\Rightarrow OA=OB$ [since, O is the mid - point of AC and BD]

$\angle 2=\angle 1$ [angles opposite to equal sides are equal]

$={25}^{\circ }$ $\therefore \angle 3=\angle 1+\angle 2$ [exterior angles is equal to the sum of two opposite interior angles]

$={25}^{\circ }+{25}^{\circ }={50}^{\circ }$

Hence, the acute angle between the diagonals is ${50}^{\circ }$.

Alternate Method

Given, in a rectangle ABCD,

$\angle ACD={25}^{\circ }$

$\therefore \angle CAB={25}^{\circ }$ Now, $\angle BCA={90}^{\circ }-{25}^{\circ }={65}^{\circ }=\angle DAC$ [alternate interior angles]

In rectangle, diagonals bisect each other.

$\therefore$OD = OB and OA = OC

So, $\Delta ODCand\Delta OAB$ are congruent. [by SSS congruence rule]

$\therefore \angle OBA={25}^{\circ }=\angle DCA$ [ by CPCT rule]

Now, $\angle AOD$ is exterior angle of $\Delta AOB.$

$\therefore \angle AOD=\angle OAB+\angle OBA={25}^{\circ }+{25}^{\circ }={50}^{\circ }$