# Class 9 RD Sharma Solutions – Chapter 8 Introduction to Lines and Angles- Exercise 8.2 | Set 2

**Question 11. In Figure, ACB is a line such that âˆ DCA = 5x and âˆ DCB = 4x. Find the value of x.**

**Solution:**

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

Thus, âˆ ACD and âˆ BCD form a linear pair.

Therefore, their sum must be equal to 180Â°.

Or we can say that

âˆ ACD + âˆ BCD = 180Â°

Also, âˆ ACD = 4x and âˆ BCD = 5x.

This further simplifies to:

4x + 5x = 180

9x = 180

x =

x = 20Â°

**Question 12. In the given figure, âˆ POR = 3x and âˆ QOR = 2x + 10, find the value of x for which POQ will be a line.**

**Solution:**

Here we have POQ as a line

So, âˆ POR and âˆ QOR form a linear pair.

Therefore, their sum must be equal to 180Â°

Or

âˆ POR + âˆ QOR = 180Â°

It is given that âˆ POR = (3x)Â° and âˆ QOR = (2x + 10)Â°. On substituting these values above we get,

3x + (2x + 10) = 180Â°

3x + 2x + 10 = 180Â°

5x + 10 = 180Â°

5x = 180 – 10

5x = 170

x =

x = 34Â°

**Question 13. In the given figure, a is greater than b by **one-third** of a right-angle. Find the values of a and b.**

**Solution:**

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

i.e.

a – b =

a – b = 30Â° ……..(i)

Also a and b form a linear pair. Therefore, their sum must be equal to 180Â°.

We can say that:

a + b = 180Â° ……….(ii)

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

2a = 180 + 30

2a = 210

a =

a = 105Â°

On putting a = 105 in (i)

105 – b = 30

-b = 30 – 105

-b = -75

b = 75Â°

Hence, a = 105Â° and b = 75Â°

**Question 14. What value of y would make AOB a line in the given figure, if âˆ AOC = 4y and âˆ BOC = (6y + 30)**

**Solution:**

Let us assume, AOB as a straight line.

This makes âˆ AOC and âˆ BOC to form a linear pair. Therefore, their sum must be equal to 180Â°.

We can say that:

âˆ AOC + âˆ BOC = 180Â°

Also, âˆ AOC = 4y and âˆ BOC = 6y + 30. This further simplifies to:

4y + (6y + 30) = 180

10y + 30 = 180

10y = 180 – 30

10y = 150

y =

y = 15Â°

Thus, the value of y = 15Â° make AOB as a line.

**Question 15. If the given figure, âˆ AOF and âˆ FOG form a linear pair.**

**âˆ EOB = âˆ FOC = 90Â° and âˆ DOC = âˆ FOG = âˆ AOB = 30Â°**

**(i) Find the measure of âˆ FOE, âˆ COB and âˆ 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.**

**Solution:**

The given figure is as follows:

(i)It is given that âˆ AOB, âˆ FOE, âˆ EOB and âˆ FOG form a linear pair.Therefore, their sum must be equal to 180Â°

i.e.

âˆ AOB + âˆ FOE + âˆ EOB + âˆ FOG = 180Â°

It is given that:

âˆ FOG = 30Â°

âˆ AOB = 30Â°

âˆ EOB = 90Â° in equation above, we get:

âˆ AOB + âˆ FOE + âˆ EOB + âˆ FOG = 180Â°

30Â° + âˆ FOE + 90Â° + 30Â° = 180Â°

âˆ FOE + 150Â° = 180Â°

âˆ FOE = 180Â° – 150Â°

âˆ FOE = 30Â°

It is given that

âˆ FOC = 90Â°

From the above figure:

âˆ FOE + âˆ DOE + âˆ COD = 90Â°

30Â° + âˆ DOE + 30Â° = 90Â°

âˆ DOE + 60Â° = 90Â°

âˆ DOE = 90Â° – 60Â°

âˆ DOE = 30Â°

Similarly, we have:

âˆ EOB = 90Â°

From the above figure:

âˆ DOE + âˆ DOC + âˆ COB = 90Â°

30Â° + 30Â° + âˆ COB = 90Â°

âˆ COB + 60Â° = 90Â°

âˆ COB = 90Â° – 60Â°

âˆ COB = 30Â°

(ii)We have:âˆ FOG = 30Â°

âˆ FOE = 30Â°

âˆ EOD = 30Â°

âˆ COD = 30Â°

âˆ COB = 30Â°

âˆ AOB = 30Â°

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

Two right angles are already given as âˆ FOC and âˆ EOB

(iii)We have to find the three pair of adjacent complementary angles.We know that âˆ EOB is a right angle.

Therefore,

âˆ EOC and âˆ COB are complementary angles.

Similarly, âˆ AOD is a right angle.

Therefore,

âˆ AOC and âˆ COD are complementary angles.

(iv)We have to find the three pair of adjacent supplementary angles.Since, âˆ AOG is a straight line.

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

âˆ AOB and âˆ BOG

âˆ AOC and âˆ COG

âˆ AOD and âˆ DOG

(v)We have to find pair of adjacent angles, which are as follows:âˆ AOB and âˆ BOC

âˆ COD and âˆ DOE

âˆ EOF and âˆ FOG

**Question 16. In the given figure, OP, OQ , OR**,** and OS are four rays. Prove that:**

**âˆ POQ + âˆ QOR + âˆ SOR + âˆ POS = 360Â°.**

**Solution:**

Let us draw TOP as a straight line.

Since, TOP is a line, therefore, âˆ POQ, âˆ QOR and âˆ ROT form a linear pair.

Also, âˆ POS and âˆ SOT form a linear pair.

Thus, we have:

âˆ POQ + âˆ QOR + âˆ ROT = 180Â° ……(i)

and

âˆ POS + âˆ SOT = 180Â° …….(ii)

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

(âˆ POQ + âˆ QOR + âˆ ROT) + (âˆ POS + âˆ SOT) = 180Â° + 180Â°

âˆ POQ + âˆ QOR + (âˆ ROT + âˆ SOT) + âˆ POS = 360Â°

âˆ POQ + âˆ QOR + âˆ SOR + âˆ POS = 360Â°

Hence proved.

**Question 17. In the given figure, ray OS stand on a line POQ, Ray OR and ray OT are angle bisectors of âˆ POS andâˆ SOQ respectively. If âˆ POS = x, find âˆ ROT.**

**Solution:**

In the figure given below, we have

Ray OR as the bisector of âˆ POS

Therefore,

âˆ POR = âˆ ROS

or,

âˆ POS = 2âˆ ROS ………..(i)

Similarly, ray OT as the bisector of âˆ SOQ

Therefore,

âˆ TOQ = âˆ TOS

or,

âˆ QOS = 2âˆ TOS ……….(ii)

Also, Ray OS stand on a line POQ. Therefore, âˆ POS and âˆ QOS form a linear pair.

Thus,

âˆ POS + âˆ QOS = 180Â°

From (i) and (ii)

2âˆ ROS + 2âˆ TOS = 180Â°

2(âˆ ROS + âˆ TOS) = 180Â°

âˆ ROS + âˆ TOS =

âˆ ROT = 90Â°

**Question 18. In the given figure, lines PQ and RS intersect each other at point O. If âˆ POR: âˆ ROQ = 5:7, find all the angles.**

**Solution:**

Let âˆ POR and âˆ ROQ be 5x and 7x respectively.

Since, Ray OR stand on line POQ. Thus, âˆ POR and âˆ ROQ form a linear pair.

Therefore, their sum must be equal to 180Â°.

Or,

âˆ POR + âˆ ROQ = 180Â°

5x + 7x = 180Â°

12x = 180Â°

x =

x = 15Â° …….(i)

Thus,

âˆ POR = 5x

= 5(15)

= 75

âˆ POR = 75Â°

Thus,

âˆ ROQ = 7x

= 7(15)

=105

âˆ ROQ = 105Â°

It is evident from the figure, that âˆ QOS and âˆ POR are vertically opposite angles.

And we know that vertically opposite angles are equal.

Therefore,

âˆ QOS = âˆ POR

âˆ QOS = 75Â°

Similarly, âˆ POS and âˆ ROQ are vertically opposite angles.

And we know that vertically opposite angles are equal.

Therefore,

âˆ POS = âˆ ROQ

âˆ POS = 105Â°

**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**

**âˆ ROS = 1212 (âˆ QOS âˆ’ POS).**

**Solution:**

The given figure shows:

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

âˆ ROQ = 90Â°

âˆ POR = 90Â°

From the figure above, we get

âˆ ROS + âˆ POS = 90Â° ………(i)

âˆ POS and âˆ QOS form a linear pair.

Therefore,

âˆ QOS + âˆ POS = 180Â° ……(ii)

From (i) and (ii) equation we get:

âˆ QOS + âˆ POS = 2 Ã— 90 âˆ QOS + âˆ POS = 2 Ã— 90

âˆ QOS + âˆ POS = 2(âˆ ROS + âˆ POS)

2âˆ ROS = âˆ QOS – âˆ POS

âˆ ROS = (âˆ QOS – âˆ POS)

Hence, proved.