1.In figure, lines AB and CD intersect at O. If \(\angle{AOC}\) + \(\angle{BOE}\) = \(70^\circ\) and
\(\angle{BOD}\) = \(40^\circ\), find \(\angle{BOE}\) and reflex \(\angle{COE}\).

Here, \(\angle{AOC}\) and \(\angle{BOD}\) are vertically opposite angles.

Therefore, \(\angle{AOC}\) = \(\angle{BOD}\)

Hence, \(\angle{AOC}\) = \(40^\circ\), [Since, \(\angle{BOD}\) = \(40^\circ\)] ....(i)

It is given that, \(\angle{AOC}\) + \(\angle{BOE}\) = \(70^\circ\)

Hence, from Eq. (i),

\(40^\circ\) + \(\angle{BOE}\) = \(70^\circ\)

i.e., \(\angle{BOE}\)= \(70^\circ\) - \(40^\circ\)

Therefore, \(\angle{BOE}\) = \(30^\circ\)

Now, by Linear pair axiom,

\(\angle{AOC}\) + \(\angle{COE}\) + \(\angle{BOE}\) = \(180^\circ\)

By substituting the values,

we get, \(40^\circ\) + \(\angle{COE}\) + \(30^\circ\) = \(180^\circ\)

i.e., \(\angle{COE}\) = \(180^\circ\) - \(40^\circ\) - \(30^\circ\)

Therefore, \(\angle{COE}\) = \(110^\circ\)

Now, so as to find the reflex angle,

\(\angle{COE}\) + reflex \(\angle{COE}\) = \(360^\circ\)

i.e., \(110^\circ\) + reflex \(\angle{COE}\) = \(360^\circ\) .....(proved)

i.e., reflex \(\angle{COE}\) = \(360^\circ\) - \(110^\circ\)

Therefore, reflex \(\angle{COE}\) = \(250^\circ\)

2. In figure, lines XY and MN intersect at O. If \(\angle{POY}\) = \(90^\circ\) and a :b = 2:3. Find c.

It is given that, \(\angle{POY}\) = \(90^\circ\)

Now, by Linear pair axiom,

\(\angle{POY}\) + \(\angle{POX}\) = \(180^\circ\)

i.e., \(\angle{POX}\) = \(180^\circ\) - \(\angle{POY}\)

Therefore, \(\angle{POX}\) = \(90^\circ\)

So, we get, a + b = \(90^\circ\) ....(i)

It is also given that, a :b = 2:3

Let a = 2k and b = 3k,

From eq. (i), 2k + 3k = \(90^\circ\)

i.e., 5k = \(90^\circ\)

Therefore, k = \(18^\circ\)

So, a = 2 × \(18^\circ\) = \(36^\circ\)

and b = 3 × \(18^\circ\) = \(54^\circ\)

Again by Linear pair axiom,

\(\angle{MOX}\) + \(\angle{XON}\) = \(180^\circ\)

i.e., b + c = \(180^\circ\)

i.e., c = \(180^\circ\) - \(54^\circ\)

Therefore, c = \(126^\circ\)

3. In figure,\(\angle{PQR}\) = \(\angle{PRQ}\), then prove that\(\angle{PQS}\) = \(\angle{PRT}\) .

By Linear pair axiom,

\(\angle{PQS}\) + \(\angle{PQR}\) = \(180^\circ\)...(i)

Similarly, again by Linear pair axiom,

\(\angle{PRQ}\) + \(\angle{PRT}\) = \(180^\circ\) ...(ii)

Thus by Eq. (i) and (ii),

we get, \(\angle{PQS}\) + \(\angle{PQR}\) = \(\angle{PRQ}\) + \(\angle{PRT}\)

As it is given, \(\angle{PQR}\) = \(\angle{PRQ}\)

Therefore, \(\angle{PQS}\) + \(\angle{PRQ}\) = \(\angle{PRQ}\) + \(\angle{PRT}\)

By cancelling equal terms,

we get, \(\angle{PQS}\) = \(\angle{PRT}\)

Hence, proved.

4.In figure, if x + y = w + z, then prove that AOB is a line.

Since, x,y,w and z are angles at a single point,

x + y + w + z = \(360^\circ\)

But it is given that, x + y = w + z

Hence, x + y + x + y = \(360^\circ\)

i.e., 2(x + y) = \(360^\circ\)

Therefore, (x + y) = \(180^\circ\)

Hence, by converse of linear pair axiom, it is proved that AOB is a straight line.

5.In figure below, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR.

Prove that \(\angle{ROS}\) = 1/2(\(\angle{QOS}\) - \(\angle{POS}\)).

Prove that \(\angle{ROS}\) = 1/2(\(\angle{QOS}\) - \(\angle{POS}\)).

Since it is given that, OR is perpendicular to PQ, we have,\(\angle{PQR}\) = \(\angle{ROQ}\) = \(90^\circ\)

Also, we can say that, \(\angle{POS}\) + \(\angle{ROS}\) = \(90^\circ\)

So, \(\angle{ROS}\) = \(90^\circ\) - \(\angle{POS}\) ...(i)

Now, by adding, \(\angle{ROS}\) on both the side, we get,

2\(\angle{ROS}\) = \(90^\circ\) - \(\angle{POS}\) + \(\angle{ROS}\)

i.e., 2\(\angle{ROS}\) = (\(90^\circ\) + \(\angle{ROS}\)) - \(\angle{POS}\)

2\(\angle{ROS}\) = \(\angle{QOS}\) - \(\angle{POS}\)

Therefore, \(\angle{ROS}\) = 1/2(\(\angle{QOS}\) - \(\angle{POS}\)).

Hence, proved.

6. It is given that \(\angle{XYZ}\) = \(64^\circ\) and XY is produced to point P. Draw a figure from the given information.

If ray YQ bisects \(\angle{ZYP}\) , find \(\angle{XYQ}\) and reflex\(\angle{QYP}\).

If ray YQ bisects \(\angle{ZYP}\) , find \(\angle{XYQ}\) and reflex\(\angle{QYP}\).

Given that, ray YQ bisects \(\angle{ZYP}\)

Thus, \(\angle{ZYQ}\) = \(\angle{QYP}\) = 1/2(\(\angle{ZYP}\)) ...(i)

By Linear pair axiom,

\(\angle{XYZ}\) + \(\angle{ZYQ}\) + \(\angle{QYP}\) = \(180^\circ\)

But, it is given that, \(\angle{XYZ}\) = \(64^\circ\) ...(ii)

Therefore, from (i) and (ii), we get,

i.e., \(64^\circ\) + \(\angle{ZYQ}\) + \(\angle{ZYQ}\) = \(180^\circ\)

i.e., 2\(\angle{ZYQ}\) = \(180^\circ\) - \(64^\circ\)

i.e., 2\(\angle{ZYQ}\) = \(116^\circ\)

Therefore, \(\angle{ZYQ}\) = \(58^\circ\)

Now, since, \(\angle{XYQ}\) = \(\angle{XYZ}\) + \(\angle{ZYQ}\)

i.e., \(\angle{XYQ}\) = \(64^\circ\) + \(58^\circ\) = \(122^\circ\)

Now, \(\angle{QYP}\) + reflex\(\angle{QYP}\) = \(360^\circ\)

i.e., \(58^\circ\) + reflex\(\angle{QYP}\) = \(360^\circ\)

Therefore, reflex\(\angle{QYP}\) = \(302^\circ\)

1. In figure, find the values of x and y and then show that AB ||CD.

Since, x + \(50^\circ\) = \(180^\circ\) .....(Linear pair)

i.e., x = \(130^\circ\)

Therefore, y = \(130^\circ\) .....(vertically opposite pair)

Here, these are corresponding angles for lines AB and CD.

Hence, it is proved that AB || CD.

2. In figure, if AB || CD, CD || EF and y : z = 3:7 , find x.

It is given that, y : z = 3:7

Let, y = 3k and z = 7k,

x = \(\angle{CHG}\) (Corresponding angles) ....(i)

Z = \(\angle{CHG}\) (Alternate angles) ....(ii)

From (i) and (ii), we get,

Therefore, x = z ....(iii)

Now, x + y = \(180^\circ\) ....(Internal angles on the same side of the transversal)

From eq.(iii), z + y = \(180^\circ\)

Therefore, 3k + 7k = \(180^\circ\)

i.e., 10k = \(180^\circ\)

i.e., k = \(18^\circ\)

Therefore, y = 3 × \(18^\circ\) and z = 7 × \(18^\circ\)

Therefore, y = \(54^\circ\) and z = \(126^\circ\)

Hence, x = \(126^\circ\)

3. In figure, if AB || CD, EF is perpendicular to CD and \(\angle{GED}\) = \(126^\circ\) , find \(\angle{AGE}\),\(\angle{GEF}\) and \(\angle{FGE}\)

Since, \(\angle{AGE}\) = \(\angle{GED}\) ....(Alternate interior angles)

But it is given that, \(\angle{GED}\) = \(126^\circ\)

Therefore, \(\angle{AGE}\) = \(126^\circ\) ....(i)

Also, \(\angle{GEF}\) + \(\angle{FED}\) = \(126^\circ\)

Since, EF is perpendicular to CD, \(\angle{GEF}\) + \(90^\circ\) = \(126^\circ\)

Therefore, \(\angle{GEF}\) = \(36^\circ\)

Also, by linear pair axiom, we get,

\(\angle{AGE}\) + \(\angle{FGE}\) = \(180^\circ\)

i.e., \(126^\circ\) + \(\angle{FGE}\) = \(180^\circ\) ....(from (i))

\(\angle{FGE}\) = \(54^\circ\)

4. In figure, if PQ || ST, \(\angle{PQR}\) = \(110^\circ\) and \(\angle{RST}\) = \(130^\circ\) , find \(\angle{QRS}\) .

Construction : Draw a line parallel to ST through R.

As it is given that, PQ || ST,\(\angle{PQR}\) = \(110^\circ\) and \(\angle{RST}\) = \(130^\circ\)

We can also say that, AB || PQ || ST

Since, \(\angle{PQR}\) + \(\angle{QRA}\) = \(180^\circ\) ....(interior angles on the same side of
transversal)

so, \(110^\circ\) + \(\angle{QRA}\) = \(180^\circ\)

i.e., \(\angle{QRA}\) = \(70^\circ\)

Since, \(\angle{ARS}\) = \(130^\circ\) ....(Alternate Interior angle)

As, \(\angle{RST}\) = \(130^\circ\)

Now, so as to find \(\angle{QRS}\), We have,

\(\angle{ARS}\) = \(\angle{ARQ}\) + \(\angle{QRS}\)

i.e., \(130^\circ\) = \(70^\circ\) + \(\angle{QRS}\)

Therefore, \(\angle{QRS}\) = \(60^\circ\)

5. In figure, if AB || CD, \(\angle{APQ}\) = \(50^\circ\) and \(\angle{PRD}\) = \(127^\circ\), find x and y.

We have, AB || CD

\(\angle{PQR}\) = \(\angle{APQ}\) ....(Alternate interior angles)

i.e.,x = \(50^\circ\) ....(i)(given (\(\angle{APQ}\) = \(50^\circ\)))

Now, as we know that, Exterior angle is equal to sum of interior opposite angles of a triangle.

Therefore, \(\angle{PQR}\) + \(\angle{QPR}\) = \(127^\circ\)

from (i), we get,

\(50^\circ\) + \(\angle{QPR}\) = \(127^\circ\)

Therefore, y = \(77^\circ\)

6. In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray
AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror
RS at C and again reflects back along CD. Prove that AB || CD.

Draw perpendiculars BE and CF on PQ and RS, respectively.

Therefore, we can say that BE || CF

As angle of incidence = angle of reflection,

we get, \(\angle{a}\) = \(\angle{b}\) ....(i)

Similarly, we get, \(\angle{x}\) = \(\angle{y}\) ....(ii)

By Alternate angles theorem, \(\angle{b}\) = \(\angle{y}\)

Now, doubling the angles, we get, 2\(\angle{b}\) = 2\(\angle{y}\)

i.e., \(\angle{b}\) + \(\angle{b}\) = \(\angle{y}\) + \(\angle{y}\)

from (i) and (ii), \(\angle{a}\) + \(\angle{b}\) = \(\angle{x}\) + \(\angle{y}\)

Hence, \(\angle{ABC}\) = \(\angle{DCB}\)

Thus by converse of alternate angles theorem,

we get, AB || CD

Hence, it is proved.

1. In figure, sides QP and RQ of \(\angle{PQR}\) are produced to points S and T, respectively. If
\(\angle{SPR}\) = \(135^\circ\) and \(\angle{PQT}\) = \(110^\circ\) , find \(\angle{PRQ}\).

By linear pair axiom, \(\angle{RPS}\) + \(\angle{RPQ}\) = \(180^\circ\)

i.e., \(135^\circ\) + \(\angle{RPQ}\) = \(180^\circ\)

i.e., \(\angle{RPQ}\) = \(180^\circ\) - \(135^\circ\)

Therefore, \(\angle{RPQ}\) = \(45^\circ\)

We also know that, Sum of interior opposite angles is equal to exterior angles.

Thus, \(\angle{RPQ}\) + \(\angle{PRQ}\) = \(\angle{PQT}\)

i.e.,\(45^\circ\) + \(\angle{PRQ}\) = \(110^\circ\)

i.e., \(\angle{PRQ}\) = \(110^\circ\) - \(45^\circ\)

Therefore, \(\angle{PRQ}\) = \(65^\circ\).

2. In figure, \(\angle{X}\) = \(60^\circ\) , \(\angle{XYZ}\) = \(54^\circ\) , if YO and ZO are the bisectors of \(\angle{XYZ}\) and \(\angle{XZY}\) respectively of \(\triangle{XYZ}\) , find \(\angle{OZY}\) and \(\angle{YOZ}\).

In \(\triangle{XYZ}\), \(\angle{X}\) + \(\angle{Y}\) + \(\angle{Z}\) = \(180^\circ\)

because, Sum of all angles of triangle is equal to 180

Therefore, \(62^\circ\) + \(\angle{Y}\) + \(\angle{Z}\) = \(180^\circ\)

i.e., \(\angle{Y}\) + \(\angle{Z}\) = \(118^\circ\)

Now, so as to find bisected angles, multiply both sides by 1/2,

We get, 1/2[\(\angle{Y}\) + \(\angle{Z}\)] = 1/2 × \(118^\circ\) = \(59^\circ\)

Thus we get, \(\angle{OYZ}\) + \(\angle{OZY}\) = \(59^\circ\)....(as YO and ZO are the bisectors of \(\angle{XYZ}\) and \(\angle{XZY}\))

Also, it is given that, \(\angle{XYZ}\) = \(54^\circ\) and we have \(\angle{OYZ}\) = 1/2 × \(\angle{XYZ}\)

\(\angle{OZY}\) + 1/2 × \(54^\circ\) = \(59^\circ\)

\(\angle{OZY}\) = \(59^\circ\) - \(27^\circ\) = \(32^\circ\)

Also, in \(\triangle{YOZ}\), \(\angle{OYZ}\) + \(\angle{YOZ}\) + \(\angle{OZY}\) = \(180^\circ\)

because, Sum of all angles of triangle is equal to 180

Therefore, \(27^\circ\) + \(\angle{YOZ}\) + \(32^\circ\) = \(180^\circ\)

i.e., \(\angle{YOZ}\) + \(59^\circ\) = \(180^\circ\)

i.e., \(\angle{YOZ}\) = \(180^\circ\) - \(59^\circ\)

Therefore, \(\angle{YOZ}\) = \(121^\circ\)

3. In figure, if AB || DE , \(\angle{BAC}\) = \(35^\circ\) and \(\angle{CDE}\) = \(53^\circ\) , find \(\angle{DCE}\).

We have, AB || DE,

Therefore, by alternate angles theorem, we get,

\(\angle{AED}\) = \(\angle{BAE}\)

Also, given that, \(\angle{BAC}\) = \(35^\circ\) and \(\angle{BAC}\) = \(\angle{BAE}\)

Therefore, \(\angle{AED}\) = \(35^\circ\)

Now, in \(\triangle{DCE}\),

\(\angle{DCE}\) + \(\angle{CED}\) + \(\angle{CDE}\) = \(180^\circ\)

because, Sum of all angles of triangle is equal to 180.

Thus, by putting given values, we get,

\(\angle{DCE}\) + \(35^\circ\) + \(53^\circ\) = \(180^\circ\)

i.e., \(\angle{DCE}\) = \(180^\circ\) - \(88^\circ\)

Therefore, \(\angle{DCE}\) = \(92^\circ\)

4. In figure, if lines PQ and RS interest at point T, such that \(\angle{PRT}\) = \(40^\circ\) ,\(\angle{RPT}\) = \(95^\circ\) and \(\angle{TSQ}\) = \(75^\circ\) , find \(\angle{SQT}\).

Since, we know, exterior angle is equal to sum of interior opposite angles

Thus, we get, \(\angle{PTS}\) = \(\angle{RPT}\) + \(\angle{PRT}\)

i.e., \(\angle{PTS}\) = \(95^\circ\) + \(40^\circ\)

Therefore, \(\angle{PTS}\) = \(135^\circ\) .....(given \(\angle{PRT}\) = \(40^\circ\) and \(\angle{RPT}\) = \(95^\circ\))

Similarly, \(\angle{TSQ}\) + \(\angle{SQT}\) = \(\angle{PTS}\)

i.e., \(75^\circ\) + \(\angle{SQT}\) = \(135^\circ\)

i.e., \(\angle{SQT}\) = \(135^\circ\) - \(75^\circ\)

Therefore, \(\angle{SQT}\) = \(60^\circ\)

5.In figure, if PQ is perpendicular to PS , PQ || SR , \(\angle{SQR}\) = \(28^\circ\) and \(\angle{QRT}\) = \(65^\circ\) , then find the values of x and y.

Here, as PQ || SR, by alternate angles axiom,

we get, \(\angle{PQR}\) = \(\angle{QRT}\)

It is given that, \(\angle{PQS}\) = x, \(\angle{SQR}\) = \(28^\circ\) and \(\angle{QRT}\) = \(65^\circ\)

i.e., \(\angle{PQS}\) + \(\angle{SQR}\) = \(\angle{QRT}\)

i.e., x + \(28^\circ\) = \(65^\circ\)

Therefore, x = \(37^\circ\) ...(i)

Now, considering right angled triangle PQS, \(\angle{SPQ}\) = \(90^\circ\)

\(\angle{SPQ}\) + x + y = \(180^\circ\) ....(Since Sum of all angles of a triangle is equal to 180)

i.e., \(90^\circ\) + \(37^\circ\) + y = \(180^\circ\)

i.e., y = \(180^\circ\) - \(127^\circ\)

Therefore, y = \(53^\circ\)

6. In figure, the side QR of \(\angle{PQR}\) is produced to a point S. If the bisectors of \(\angle{PQR}\)
and \(\angle{PRS}\) meet at point T, then prove that \(\angle{QTR}\) = (1/2) \(\angle{QPR}\)

In \(\triangle{PQS}\), we have,

\(\angle{QPR}\) + \(\angle{PQR}\) = \(\angle{PRS}\) ...(i)

As we know, Sum of interior opposite angles is equal to exterior angle.

Similarly in \(\triangle{TQR}\), we have,

\(\angle{QTR}\) + \(\angle{TQR}\) = \(\angle{TRS}\) ...(ii)

As, QT and RT are the bisectors of \(\angle{PRS}\) and \(\angle{PQR}\), respectively, we get,

\(\angle{TRS}\) = (1/2) \(\angle{PRS}\) ...(iii)

\(\angle{TQR}\) = (1/2) \(\angle{PQR}\) ...(iv)

By multiplying both sides of eq.(i), we get,

(1/2)\(\angle{QPR}\) + (1/2)\(\angle{PQR}\) = (1/2)\(\angle{PRS}\)

from eq. (iii) and (iv), (1/2)\(\angle{QPR}\) + \(\angle{TQR}\) = \(\angle{TRS}\) ...(v)

Thus, from eq.(ii) and (v), we get,

\(\angle{QTR}\) + \(\angle{TQR}\) = (1/2)\(\angle{QPR}\) + \(\angle{TQR}\)

cancelling equal terms, we get,

\(\angle{QTR}\) = (1/2)\(\angle{QPR}\)

Hence, proved.