# Class 9 NCERT Solutions – Chapter 6 Lines And Angles – Exercise 6.1

### Question 1. In the given figure, lines AB and CD intersect at O. If âˆ AOC + âˆ BOE = 70Â° and âˆ BOD = 40Â°, find âˆ BOE and reflex âˆ COE?

Solution:

Given, AB and CD are straight lines.

âˆ AOC + âˆ BOE = 70Â°  —-eq(i)

âˆ BOD = 40Â° —-eq(ii)

Since, AB is a straight line, the sum of all angles made on it is 180Â°

=> âˆ AOC  + âˆ COE + âˆ BOE = 180Â° —eq(iii)

We can rearrange this equation as,

=> âˆ AOC + âˆ BOE + âˆ COE = 180Â°

=> 70Â° + âˆ COE = 180Â°  —from eq(i)

=> âˆ COE = 180Â° – 70Â° = 110Â°

=> âˆ COE = 110Â° —eq(iv)

Reflex âˆ COE = 360Â° – âˆ COE = 360Â° – 110Â° = 250Â°

Now, it is also given that CD is also a straight line, so the sum of all angles made on it is 180Â°

=> âˆ COE + âˆ BOE + âˆ BOD = 180Â° —eq(v)

We can rearrange this equation as,

=> âˆ COE + âˆ BOD + âˆ BOE = 180Â°

=> 110Â° + 40Â° + âˆ BOE = 180Â°  —from eq(ii) and eq(iv)

=> 150Â° + âˆ BOE = 180Â°

=> âˆ BOE = 180Â° – 150Â° = 30Â°

=> âˆ BOE = 30Â°

### Question 2. In the given figure, lines XY and MN intersect at O. Ifâˆ POY = 90Â° and a : b = 2 : 3, find c?

Solution:

Given, XY and  MN are straight lines.

âˆ POY = 90Â° –eq(i)

a : b = 2 : 3 –eq(ii)

âˆ POM = a

âˆ XOM = b

âˆ XON = c

Taking XY as a straight line, so the sum of all angles made on it is 180Â°

=> âˆ XOM + âˆ POM + âˆ POY = 180Â° —eq(iii)

=> b + a + 90Â° = 180Â°

=> 3x + 2x + 90Â° = 180Â° from eq(i) and eq(ii)

=> 5x + 90Â° = 180Â°

=> 5x = 180Â° – 90Â° = 90Â°

=> 5x = 90Â°

=> x = 18Â°

a : b = 2x : 3x = 2×18 : 3×18

a = 36Â°

b = 54Â°

Taking MN as a straight line so,the sum of all the angles made on it is 180Â°

=> âˆ XOM  + âˆ XON = 180Â°

=> 54Â° + âˆ XON = 180Â°  from above finding value

=> âˆ XON = 126Â° or c = 126Â°

### Question 3. In the given figure, âˆ PQR = âˆ PRQ, then prove that âˆ PQS = âˆ PRT?

Solution:

Given, âˆ PQR = âˆ PRQ

Taking ST is a straight line, so the sum of all angles made on it is 180Â°

=> âˆ PQS + âˆ PQR = 180Â° —-eq(i)

also,  âˆ PRQ +  âˆ PRT = 180Â° —eq(ii)

By equating both the equations because RHS of both the equation is equal So, LHS will also be equal.

=> âˆ PQS +  âˆ PQR  = âˆ PRQ  + âˆ PRT

=> âˆ PQS + âˆ PQR = âˆ PQR + âˆ PRT  –[ Given in question âˆ PQR = âˆ PRQ ]

=> âˆ PQS = âˆ PRT

### Question 4. In the given figure, if x + y = w + z, then prove that AOB is a line?

Solution:

Given, x + y  = w + z  –eq(i)

We know that , sum of all angles made along  a point is 360Â°

So, Taking O as a point âˆ AOC + âˆ BOC + âˆ BOD + âˆ AOD = 360Â°

=> y + x + w + z = 360Â° from the given figure

=> (x + y) + (x + y) = 360Â°  from eq(i)

=> 2x + 2y = 360Â°

=> 2(x + y) = 360Â°

=> x + y=180Â°

From this statement it is proved that AOB is a straight line because the sum of angles made on the line is 180Â°. So, AOB is a straight line.

### Question 5. 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 thatâˆ ROS = (1/2) (âˆ QOS â€“ âˆ POS)?

Solution:

Given POQ is a straight line

So, the sum of all angles made on it is 180Â°

=> âˆ POS + âˆ ROS + âˆ ROQ = 180Â°

=> âˆ POS + âˆ ROS + 90Â° = 180Â° [given âˆ ROQ = 90Â°]

=> âˆ POS + âˆ ROS = 90Â°

=> âˆ ROS = 90Â°  – âˆ POS –eq(i)

Now, âˆ ROS + âˆ ROQ = âˆ QOS   [from figure]

=> âˆ ROS + 90Â° = âˆ QOS

=> âˆ ROS = âˆ QOS – 90Â° –eq(ii)

Now Adding both the equations eq(i) + eq(ii)

=> âˆ ROS + âˆ ROS = 90Â° – âˆ POS + âˆ QOS – 90Â°

=> 2âˆ ROS =(âˆ QOS  – âˆ POS)

=> âˆ ROS = (1/2) (âˆ QOS – âˆ POS)

Hence Verified!!!

### Question 6. It is given that âˆ XYZ = 64Â° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects âˆ ZYP, find âˆ XYQ and reflex âˆ QYP?

Solution:

From the drawn figure, it is clearly shown that XYP is a straight line.

So, âˆ XYZ + âˆ ZYQ + âˆ QYP = 180Â°

=> 64Â°+ âˆ ZYQ + âˆ QYP = 180Â°  [ given âˆ XYZ = 64Â°]

=> 64Â° + 2âˆ QYP = 180Â°  [ YQ bisect âˆ ZYP so, âˆ QYP = âˆ ZYQ]

=> 2âˆ QYP = 180Â° – 64Â° = 116Â°

=> âˆ QYP = 58Â°

So, Reflex âˆ QYP = 360Â° – 58Â° = 302Â°

Since âˆ XYQ = âˆ XYZ + âˆ ZYQ

=> âˆ XYQ = 64Â° + âˆ QYP  [ given âˆ XYZ = 64Â° and âˆ ZYQ = âˆ QYP]

=> âˆ XYQ =64Â° + 58Â° = 122Â°

Thus, âˆ XYQ = 122Â° and Reflex âˆ QYP = 302Â°

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