NCERT solution for class 9 maths lines and angles ( Chapter 6)

Get it on Google Play

Solution for Exercise 6.1

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 \(\angle{COE}\).

Open in new tab

Answer :

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.

Open in new tab

Answer :

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\(\angle{PQS}\) = \(\angle{PRT}\) .

Open in new tab

Answer :

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.

Open in new tab

Answer :

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(\(\angle{QOS}\) - \(\angle{POS}\)).

Open in new tab

Answer :

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\(\angle{QYP}\).

Open in new tab

Answer :

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\(\angle{QYP}\) = \(302^\circ\)

Solution for Exercise 6.2

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

Open in new tab

Answer :

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.

Open in new tab

Answer :

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 \(\angle{FGE}\)

Open in new tab

Answer :

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 \(\angle{QRS}\) .

Open in new tab

Answer :

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.

Open in new tab

Answer :

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.

Open in new tab

Answer :

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.

Solution for Exercise 6.3

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 \(\angle{PRQ}\).

Open in new tab

Answer :

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 \(\angle{YOZ}\).

Open in new tab

Answer :

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 \(\angle{DCE}\).

Open in new tab

Answer :

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 \(\angle{SQT}\).

Open in new tab

Answer :

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.

Open in new tab

Answer :

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) \(\angle{QPR}\)

Open in new tab

Answer :

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.