Class 9 RD Sharma Solutions – Chapter 8 Introduction to Lines and Angles- Exercise 8.2 | Set 1

Question 1: In the below Fig. OA and OB are opposite rays:

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

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

Solution:

(i) Given:

x = 25

In the figure;

âˆ AOC and âˆ BOC are forming a linear pair

So, âˆ AOC + âˆ BOC = 180Â°

In the figure;

âˆ AOC = 2y + 5 and âˆ BOC = 3x

âˆ AOC + âˆ BOC = 180Â°

(2y + 5) + 3x = 180

(2y + 5) + 3 (25) = 180

2y + 5 + 75 = 180

2y + 80 = 180

2y = 100

y =  = 50

Hence,

y = 50Â°

(ii) Given:

y = 35Â°

In the figure;

âˆ AOC + âˆ BOC = 180Â° {Linear pair angles}

(2y + 5) + 3x = 180

(2(35) + 5) + 3x = 180

75 + 3x = 180

3x = 105

x = 35

Therefore, x = 35Â°

Question 2. In the below figure, write all pairs of adjacent angles and all the linear pairs.

Solution

In the figure;

(âˆ AOC, âˆ COB);

(âˆ AOD, âˆ BOD);

(âˆ AOD, âˆ COD);

(âˆ BOC, âˆ COD)

âˆ AOD + âˆ BOD = 180Â° {Linear pair}

and

âˆ AOC+ âˆ BOC = 180Â° {Linear pair}

Question 3. In the given figure, find x. Further find âˆ BOC, âˆ COD, and âˆ AOD.

Solution:

In the figure;

âˆ AOD and âˆ BOD are forming linear pair,

Thus,

âˆ AOD+ âˆ BOD = 180Â°

And,

âˆ AOD + âˆ BOC + âˆ COD = 180Â°

Given:

âˆ AOD = (x+10)Â°,

âˆ COD = xÂ°

and

âˆ BOC = (x + 20)Â°

(x + 10) + x + (x + 20) = 180Â°

3x + 30 = 180Â°

3x = 180 â€“ 30

x =

x = 50Â°

Here,

âˆ AOD = (x+10) = 50 + 10 = 60

âˆ COD = x = 50Â°

âˆ BOC = (x+20) = 50 + 20 = 70Â°

Therefore,

âˆ AOD = 60Â°,

âˆ COD = 50Â°

and

âˆ BOC=70Â°

Question 4. In figure, rays OA, OB, OC, OD, and OE have the common endpoint 0. Show that âˆ AOB+âˆ BOC+âˆ COD+âˆ DOE+âˆ EOA=360Â°.

Solution:

Given:

Rays OA, OB, OC, OD and OE have the common endpoint O.

Construct: Draw an opposite ray OX to ray OA, which make a straight line AX.

In the figure:

âˆ AOB + âˆ BOX = 180Â° {Linear pair}

Or,

âˆ AOB + âˆ BOC + âˆ COX = 180Â° ……….(i)

Also,

âˆ AOE + âˆ EOX  = 180Â°  {Linear pair}

Or,

âˆ AOE + âˆ DOE + âˆ DOX = 180Â° ………..(ii)

After adding equations, (i) and (ii), we will get;

âˆ AOB + âˆ BOC + âˆ COX + âˆ AOE + âˆ DOE + âˆ DOX = 180Â° + 180Â°

âˆ AOB + âˆ BOC + âˆ COD + âˆ DOE + âˆ EOA = 360Â°

Hence, Proved.

Question 5. In figure, âˆ AOC and âˆ BOC form a linear pair. If a â€“ 2b = 30Â°, find a and b?

Solution:

Given:

âˆ AOC and âˆ BOC are forming a linear pair.

a + b = 180Â° â€¦….(i)

a â€“ 2b = 30Â° â€¦….(ii) {given}

After subtracting equation (ii) from (i), we will get

a + b â€“ a + 2b = 180 â€“ 30

3b = 150

b =

b = 50Â°

Thus,

a â€“ 2b = 30Â°

a â€“ 2(50) = 30Â°

a = 30 + 100

a = 130Â°

Hence,

a = 130Â°

b = 50Â°

Question 6. How many pairs of adjacent angles are formed when two lines intersect at a point?

Solution

Here, the four pairs of adjacent angles are formed when two lines intersect each other at a single point.

So here Let two lines AB and CD intersect at point O as shown below in the figure

Thus, the 4 pair of adjacent angles are :

(âˆ AOD, âˆ DOB),

(âˆ DOB, âˆ BOC),

(âˆ COA, âˆ AOD)

and

(âˆ BOC, âˆ COA)

Question 7. How many pairs of adjacent angles, in all, can you name in the figure given?

Solution

The number of Pairs of adjacent angles, from the following figure are;

âˆ EOC and âˆ DOC

âˆ EOD and âˆ DOB

âˆ DOC and âˆ COB

âˆ EOD and âˆ DOA

âˆ DOC and âˆ COA

âˆ BOC and âˆ BOA

âˆ BOA and âˆ BOD

âˆ BOA and âˆ BOE

âˆ EOC and âˆ COA

âˆ EOC and âˆ COB

Thus, we have 10 pairs of adjacent angles.

Question 8. In figure, determine the value of x.

Solution:

As we know that, the sum of all the angles around a point O is equal to 360Â°.

Thus,

3x + 3x + 150 + x = 360Â°

7x = 360Â° â€“ 150Â°

7x = 210Â°

x =

x = 30Â°

Hence, the value of x is 30Â°.

Question 9. In figure, AOC is a line, find x.

Solution:

In the following figure,

âˆ AOB + âˆ BOC = 180Â° {Linear pairs}

70 + 2x = 180

2x = 180 â€“ 70

2x = 110

x =

x = 55Â°

Hence, the value of x is 55Â°.

Question 10. In figure, POS is a line, find x.

Solution:

In the following figure;

âˆ POQ + âˆ QOS = 180Â° {Linear pair}

âˆ POQ + âˆ QOR +âˆ SOR = 180Â°

60Â° + 4x + 40Â° = 180Â°

4x = 180Â° -100Â°

4x = 80Â°

x = 20Â°

Thus, the value of x is 20Â°.

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