RD Sharma 2020 solution class 9 chapter 10 Lines and Angles Exercise 10.2

Exercise 10.2

Page-10.14

Question 1:

In the given figure, OA and OB are opposite rays:


(i) If x = 25°, what is the value of y?

(ii) If y = 35°, what is the value of x?

Answer 1:

In figure:

Since OA and OB are opposite rays. Therefore, AB is a line. Since, OC stands on line AB.

Thus,and form a linear pair, therefore, their sum must be equal to.

Or, we can say that

From the given figure:

and

On substituting these two values, we get

                             ...(i)

(i) On puttingin (i), we get:

Hence, the value of y is.

(ii) On putting in in equation (A), we get:

Hence, the value of x is.

Question 2:

In the given figure, write all pairs of adjacent angles and all the linear pairs.

Answer 2:

The figure is given as follows:

The following are the pair of adjacent angles:

and

and

The following are the linear pair:

and

and

Question 3:

In the given figure, find x. Further find ∠BOC, ∠COD and∠AOD

Answer 3:

In the given figure:

AB is a straight line. Thus,, and form a linear pair.

Therefore their sum must be equal to.

We can say that

(i)

It is given that, and.

On substituting these values in (i), we get:

It is given that:

Therefore,

Also,

Therefore,

Therefore,

Question 4:

In the given figure, rays OA, OB, OC, OD and OE have the common end point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.

Answer 4:

Let us draw a straight line.

,and form a linear pair. Thus, their sum should be equal to.

Or, we can say that:

(I)

Similarly,,and form a linear pair. Thus, their sum should be equal to.

Or, we can say that:

(II)

On adding (I) and (II), we get:

Hence proved.

Page-10.15

Question 5:

In the given figure, ∠AOC and ∠BOC form a linear pair. If a − 2b = 30°, find a and b.

Answer 5:

In the figure given below, it is given thatand forms a linear pair.

Thus, the sum of and should be equal to.

Or, we can say that:

From the figure above, and

Therefore,

It is given that:

On comparing (i) and (ii), we get:

Putting in (i), we get :

Hence, the values for a and b areand respectively.

Question 6:

How many pairs of adjacent angle are formed when two lines intersect in a point?

Answer 6:

Suppose we have two lines, say AB and CD intersect at a point, O as shown in the figure below:

 

Then there are 4 pairs of adjacent angles formed, namely:

1. and

2. and

3. and

4. and

Question 7:

How many pairs of adjacent angles, in all, can you name in the given figure.

Answer 7:

In the given figure,

We have 10 adjacent angle pairs, namely:

1. and

2. and

3. and

4. and

5. and

6. and

7. and

8. and

9. and

10. and

Question 8:

In the given figure, determine the value of x.

Answer 8:

In the given figure:

is a straight line. Thus,and form a linear pair.

Therefore their sum must be equal to.

We can say that

It is given that, substituting this value in equation above, we get:

Question 9:

In the given figure, AOC is a line, find x.

Answer 9:

It is given that AOC is a line. Therefore, and form a linear pair. Thus, the sum of and must be equal to .

Or, we can say that

Also, and. On putting these values in the equation above we have:

Hence, the required value of is.

Question 10:

In the given figure, POS is a line, find x.

Answer 10:

The figure is given as follows:

It is given that POS is a line.

Therefore,,and form a linear pair. Thus, their sum must be equal to.

It is given that, and. Therefore, we get:

$$\begin{gathered} {60}^0+4x+{40}^0={180}^0 \\ 4x+{100}^0={180}^0 \\ 4x={180}^0-{100}^0 \\ 4x={80}^0 \\ x=\frac{{80}^0}{4} \\ x=\boxed{{20}^0} \end{gathered}$$
Hence, the required value of x is.

Question 11:

In the given figure, ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the values of ∠DCA and ∠DCB.

Answer 11:

It is given that ACB is a line in the figure given below.

Thus,and form a linear pair.

Therefore, their sum must be equal to.

Or, we can say that

Also, and. This further simplifies to :

$$\begin{gathered} \therefore \angle DCA=5x=5\times 20°=100° \\ \angle DCB=4x=4\times 20°=80° \end{gathered}$$

Hence, the values of ∠DCA and ∠DCB are 100^∘ and 80^∘ respectively.

Question 12:

In the given figure, ∠POR = 3x and ∠QOR = 2x + 10, find the value of x for which POQ will be a line.

Answer 12:

Here we have POQ as a line

So, andform a linear pair.

Therefore, their sum must be equal to.

Or, we can say that

It is given that and .On substituting these values above, we get :

Hence, the value of x is .

Question 13:

What value of y would make AOB a line in the given figure, if ∠AOC = 4y and ∠BOC = (6y + 30)

Answer 13:

Let us assume,as a straight line.

This makes and to form a linear pair. Therefore, their sum must be equal to.

We can say that:

Also, and. This further simplifies to:

Hence, the value of makesas a line.

Page-10.16

Question 14:

In the given figure, OP, OQ, OR and OS are four rays. Prove that:
∠POQ + ∠QOR + ∠SOR + ∠POS = 360°

Answer 14:

Let us draw as a straight line.

 

Since,is a line, therefore,, and form a linear pair.

Also, and form a linear pair.

Thus, we have:

(i)

And

(ii)

On adding (i) and (ii), we get :

Hence proved.

Question 15:

In the given figure, ray OS stand on a line POQ, Ray OR and ray OT are angle bisectors of ∠POS and∠SOQ respectively. If ∠POS = x, find ∠ROT.

Answer 15:

In the figure given below, we have

Rayas the bisector of

Therefore,

Or,

(I)

 

Similarly, Rayas the bisector of

Therefore,

Or,

(II)

Also, Raystand on a line. Therefore,and form a linear pair.

Thus,

From (I) and (II):

Hence, the value of is 90°.

Question 16:

In the given figure, lines PQ and RS intersect each other at point O. If ∠POR: ∠ROQ = 5 : 7, find all the angles.

Answer 16:

Let andbe and respectively.

Since, Ray stand on line.Thus, and form a linear pair.

Therefore, their sum must be equal to.

Or,

Thus,

Thus,

It is evident from the figure, thatand are vertically opposite angles.

And we know that vertically opposite angles are equal.

Therefore,

Similarly,and are vertically opposite angles.

And we know that vertically opposite angles are equal.

Therefore,

Question 17:

In the given figure, a is greater than b by one third of a right-angle. Find the values of a and b.

Answer 17:

It is given that in the figure given below; a is greater than b by one-third of a right angle.

Or we can say that, the difference between a and b is.

That is;

Also a and b form a linear pair. Therefore, their sum must be equal to.

We can say that:

On adding (i) and (ii), we get:

On putting, in (i):

Hence, the values are and.

Question 18:

If the given figure, ∠AOF and ∠FOG form a linear pair.
∠EOB = ∠FOC = 90° and ∠DOC = ∠FOG = ∠AOB = 30°


(i) Find the measure of ∠FOE, ∠COB and ∠DOE.
(ii) Name all the right angles.
(iii) Name three pairs of adjacent complementary angles.
(iv) Name three pairs of adjacent supplementary angles.
(v) Name three pairs of adjacent angles.

Answer 18:

The given figure is as follows:

 

(i)

It is given that,,and form a linear pair .

Therefore, their sum must be equal to .

That is ,

It is given that :

,

and

in equation above, we get:

It is given that:

From the above figure:

Similarly, we have:

From the above figure:

(ii)

We have:

From the figure above and the measurements of the calculated angles we get two right angles as and.

Two right angles are already given asand.

(iii)

We have to find the three pair of adjacent complementary angles.

We know that is a right angle.

Therefore,

and are complementary angles.

Similarly, is a right angle.

Therefore,

and are complementary angles.

Similarly, is a right angle.

Therefore,

and are complementary angles.

(iv)

We have to find the three pair of adjacent supplementary angles.

Since,is a straight line.

Therefore, following are the three linear pair, which are supplementary:

and ;

and and

and

(v)

We have to find three pair of adjacent angles, which are as follows:

and

and

and

Page-10.17

Question 19:

In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = $\frac{1}{2}$ (∠QOS − POS).

Answer 19:

The given figure is as follows:

We have POQ as a line. Ray OR is perpendicular to line PQ. Therefore,

From the figure above, we get:

 (i)

and  form a linear pair. Therefore,

 (ii)

From (i) and (ii) equation we get:
$\angle QOS+\angle POS=2\times 90$

Hence proved.