Question 1

In a rectangle, all four corner angles are $90^\circ$. Opposite sides are parallel and equal. The diagonals are equal in length and bisect each other. This means the four segments formed by the intersection of the diagonals are all equal, making the four triangles formed by the diagonals isosceles.

Step 1 — Angles in Rectangle ABCD (i)

Let O be the intersection point of the diagonals AC and BD. The given angle is $\angle CAB = \mathbf{30^\circ}$. We will call this $\angle 1$.

The corner angle $\angle DAB$ of the rectangle is $90^\circ$. We can find $\angle DAC$, which we will call $\angle 2$.

$\angle 2 = \angle DAB - \angle CAB$ $= 90^\circ - 30^\circ$

$\boxed{\angle 2 = 60^\circ}$

Since AD is parallel to BC, $\angle DAC$ and $\angle ACB$ are alternate interior angles. So, $\angle ACB = \angle DAC$. We will call $\angle ACB$ as $\angle 6$.

$\angle 6 = \angle DAC$

$\boxed{\angle 6 = 60^\circ}$

Also, AB is parallel to DC, so $\angle CAB$ and $\angle ACD$ are alternate interior angles. So, $\angle ACD = \angle CAB$. We will call $\angle ACD$ as $\angle 5$.

$\angle 5 = \angle CAB$

$\boxed{\angle 5 = 30^\circ}$

In $\triangle AOB$, $OA = OB$ because diagonals of a rectangle are equal and bisect each other. So, $\triangle AOB$ is an isosceles triangle. This means the base angles are equal: $\angle OAB = \angle OBA$. We know $\angle OAB = \angle CAB = 30^\circ$. So, $\angle OBA = 30^\circ$. We will call this $\angle 8$.

$\angle 8 = \angle OAB$

$\boxed{\angle 8 = 30^\circ}$

The sum of angles in $\triangle AOB$ is $180^\circ$. So, $\angle AOB = 180^\circ - (\angle OAB + \angle OBA)$. We will call $\angle AOB$ as $\angle 9$.

$\angle 9 = 180^\circ - (30^\circ + 30^\circ)$ $= 180^\circ - 60^\circ$

$\boxed{\angle 9 = 120^\circ}$

$\angle AOB$ and $\angle COD$ are vertically opposite angles. So, $\angle COD = \angle AOB$. We will call $\angle COD$ as $\angle 11$.

$\angle 11 = \angle AOB$

$\boxed{\angle 11 = 120^\circ}$

$\angle AOB$ and $\angle BOC$ form a linear pair (angles on a straight line). So, $\angle BOC = 180^\circ - \angle AOB$. We will call $\angle BOC$ as $\angle 10$.

$\angle 10 = 180^\circ - 120^\circ$

$\boxed{\angle 10 = 60^\circ}$

$\angle BOC$ and $\angle DOA$ are vertically opposite angles. So, $\angle DOA = \angle BOC$. We will call $\angle DOA$ as $\angle 12$.

$\angle 12 = \angle BOC$

$\boxed{\angle 12 = 60^\circ}$

Now we find the remaining angles at the corners. In $\triangle BOC$, $OB = OC$, so it is an isosceles triangle. This means $\angle OBC = \angle OCB$. We know $\angle OCB = \angle ACB = 60^\circ$. So, $\angle OBC = 60^\circ$. We will call this $\angle 7$.

$\angle 7 = \angle OCB$

$\boxed{\angle 7 = 60^\circ}$

In $\triangle COD$, $OC = OD$, so it is an isosceles triangle. This means $\angle OCD = \angle ODC$. We know $\angle OCD = \angle ACD = 30^\circ$. So, $\angle ODC = 30^\circ$. We will call this $\angle 4$.

$\angle 4 = \angle OCD$

$\boxed{\angle 4 = 30^\circ}$

In $\triangle DOA$, $OD = OA$, so it is an isosceles triangle. This means $\angle ODA = \angle OAD$. We know $\angle OAD = \angle DAC = 60^\circ$. So, $\angle ODA = 60^\circ$. We will call this $\angle 3$.

$\angle 3 = \angle OAD$

$\boxed{\angle 3 = 60^\circ}$

Step 2 — Angles in Rectangle PSRQ (ii)

Let O be the intersection point of the diagonals PR and QS. The given angle is $\angle QOR = \mathbf{110^\circ}$. We will call this $\angle 9$.

$\angle QOR$ and $\angle POS$ are vertically opposite angles. So, $\angle POS = \angle QOR$. We will call $\angle POS$ as $\angle 11$.

$\angle 11 = \angle QOR$

$\boxed{\angle 11 = 110^\circ}$

$\angle QOR$ and $\angle QOP$ form a linear pair (angles on a straight line). So, $\angle QOP = 180^\circ - \angle QOR$. We will call $\angle QOP$ as $\angle 10$.

$\angle 10 = 180^\circ - 110^\circ$

$\boxed{\angle 10 = 70^\circ}$

$\angle QOP$ and $\angle ROS$ are vertically opposite angles. So, $\angle ROS = \angle QOP$. We will call $\angle ROS$ as $\angle 12$.

$\angle 12 = \angle QOP$

$\boxed{\angle 12 = 70^\circ}$

In $\triangle POS$, $OP = OS$ because diagonals of a rectangle are equal and bisect each other. So, $\triangle POS$ is an isosceles triangle. This means the base angles are equal: $\angle OPS = \angle OSP$. The sum of angles in $\triangle POS$ is $180^\circ$. So, $\angle OPS + \angle OSP + \angle POS = 180^\circ$. $2 \times \angle OPS + 110^\circ = 180^\circ$. $2 \times \angle OPS = 180^\circ - 110^\circ$. $2 \times \angle OPS = 70^\circ$. $\angle OPS = 35^\circ$. We will call $\angle OPS$ as $\angle 1$.

$\angle 1 = 35^\circ$

We will call $\angle OSP$ as $\angle 8$.

$\angle 8 = \angle OPS$

$\boxed{\angle 8 = 35^\circ}$

The corner angle $\angle QPS$ of the rectangle is $90^\circ$. We can find $\angle QPO$, which we will call $\angle 2$.

$\angle 2 = \angle QPS - \angle OPS$ $= 90^\circ - 35^\circ$

$\boxed{\angle 2 = 55^\circ}$

In $\triangle QOP$, $OQ = OP$, so it is an isosceles triangle. This means $\angle OQP = \angle OPQ$. We know $\angle OPQ = \angle QPO = 55^\circ$. So, $\angle OQP = 55^\circ$. We will call this $\angle 3$.

$\angle 3 = \angle OPQ$

$\boxed{\angle 3 = 55^\circ}$

The corner angle $\angle PQR$ of the rectangle is $90^\circ$. We can find $\angle RQS$, which we will call $\angle 4$.

$\angle 4 = \angle PQR - \angle PQS$ $= 90^\circ - 55^\circ$

$\boxed{\angle 4 = 35^\circ}$

In $\triangle QOR$, $OQ = OR$, so it is an isosceles triangle. This means $\angle OQR = \angle ORQ$. We know $\angle OQR = \angle RQS = 35^\circ$. So, $\angle ORQ = 35^\circ$. We will call this $\angle 5$.

$\angle 5 = \angle OQR$

$\boxed{\angle 5 = 35^\circ}$

The corner angle $\angle QRS$ of the rectangle is $90^\circ$. We can find $\angle ORS$, which we will call $\angle 6$.

$\angle 6 = \angle QRS - \angle ORQ$ $= 90^\circ - 35^\circ$

$\boxed{\angle 6 = 55^\circ}$

In $\triangle ROS$, $OR = OS$, so it is an isosceles triangle. This means $\angle ORS = \angle OSR$. We know $\angle ORS = \angle 6 = 55^\circ$. So, $\angle OSR = 55^\circ$. We will call this $\angle 7$.

$\angle 7 = \angle ORS$

$\boxed{\angle 7 = 55^\circ}$

Answer

(i) The angles in rectangle ABCD are: $\angle 1 = 30^\circ$, $\angle 2 = 60^\circ$, $\angle 3 = 60^\circ$, $\angle 4 = 30^\circ$, $\angle 5 = 30^\circ$, $\angle 6 = 60^\circ$, $\angle 7 = 60^\circ$, $\angle 8 = 30^\circ$, $\angle 9 = 120^\circ$, $\angle 10 = 60^\circ$, $\angle 11 = 120^\circ$, and $\angle 12 = 60^\circ$. (ii) The angles in rectangle PSRQ are: $\angle 1 = 35^\circ$, $\angle 2 = 55^\circ$, $\angle 3 = 55^\circ$, $\angle 4 = 35^\circ$, $\angle 5 = 35^\circ$, $\angle 6 = 55^\circ$, $\angle 7 = 55^\circ$, $\angle 8 = 35^\circ$, $\angle 9 = 110^\circ$, $\angle 10 = 70^\circ$, $\angle 11 = 110^\circ$, and $\angle 12 = 70^\circ$.