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## Download RS Aggarwal Solutions Class 9 Maths Chapter 7 Lines and Angles PDF

RS Aggarwal Solutions Class 9 Maths Chapter 7 Lines and Angles

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### RS Aggarwal Solutions Class 9 Maths Chapter 7 Lines and Angles Ex 7.1

**1:** Define the following terms:

**(i) Angle**

When two rays originated from same point, then an angle is formed.

**(ii) Interior of an angle**

The area between two rays which is originated from same point, is called Interior of an angle.

**(iii) Obtuse angle**

An angle which is measured more than 90^{0} but less than 180^{0} is called Obtuse angle.

**(iv) Reflex angle**

An angle which is measured more than 180^{0} but less than 360^{0} is called Reflex angle.

**(v) Complementary angle**

Two angle are said to be complementary if, the sum of both angles are 90^{0} .

**(vi) Supplementary angle**

Two angle are said to be supplementary, if the sum of both angles are 180^{0}.

**2: **Find the complement of each of the following angles:

*(i)* 55^{0} *(ii)* 16^{0} *(iii)* 90^{0} *(iv)* 2/3 of a right angle

Complement of 55^{0} = 90^{0} – 55^{0} = 35^{0}

Complement of 16^{0} = 90^{0} – 16^{0}= 74^{0}

Complement of 90^{0} = 90^{0} – 90^{0} = 0^{0}

Complement of 2/3 of a right angle

= 90^{0} – ( 90^{0} × 2/3) = 90^{0} – 60^{0} =30^{0}

**3: **Find the supplementary of each of the following angles:

*(i) *42^{0} *(ii)* 90^{0} *(iii)* 124^{0} *(iv)* 3/5 of a right angle

Supplementary of 42^{0} = 180^{0} – 42^{0 }=138^{0}

Supplementary of 90^{0} = 180^{0} – 90^{0 }= 90^{0}

Supplementary of 124^{0} = 180^{0} – 124^{0 }= 56^{0}

Supplementary of 3/5 of a right angle

= 180^{0} – ( 90^{0} × 3/5) = 180^{0} – 54^{0}= 126^{0}

**4: **Find the measure of an angle which is

*(i)* equal to its complement,

Let the angle is x^{0} , then

⇒ x^{0} = 90^{0} – x^{0}

⇒ 2x^{0} = 90^{0}

⇒ x^{0} = 45^{0}

Hence the angle is 45^{0} which is equal to its complement.

*(ii)* equal to its supplement

Let the angle is x^{0} then

⇒ x^{0} = 180^{0} – x^{0}

⇒ 2x^{0} = 180^{0}

⇒ x^{0} = 90^{0}

Hence the angle is 90^{0} which is equal to its supplement.

**5: **Find the measure of an angle which is 36^{0} more than its complement.

Let the angle is x^{0} , then its complement angle will be 90^{0} – x^{0}

According to question,

⇒ x^{0} – 36^{0} = 90^{0} – x^{0}

⇒ 2x^{0} = 90^{0} + 36^{0} = 126^{0}

⇒ x^{0} = 63^{0}

Hence the angle is 63^{0} .

**6:** Find the measure of an angle which is 30^{0} less than its supplement.

Let the angle is x^{0}, then its supplement angle will be 180^{0} – x^{0}

According to question,

⇒ x^{0} +30^{0} = 180^{0} – x^{0}

⇒ 2x^{0} = 180^{0} – 30^{0} = 150^{0}

⇒ x^{0} = 75^{0}

Hence the angle is 75^{0}.

**7: **Find the angle which is four times its complement.

Let the angle is x^{0}, then its complement angle will be 90^{0} – x^{0}

According to question,

⇒ x^{0} = 4(90^{0} – x^{0} )

⇒ x^{0} = 360^{0} – 4x^{0}

⇒ 5x^{0} = 360^{0}

⇒ x^{0}= 72^{0}

Hence the angle is 72^{0}

**8:** Find the angle which is five times its supplement.

Let the angle is x^{0}, then its supplementary angle will be 180^{0} – x^{0}

According to question,

⇒ x^{0} = 5(180^{0} – x^{0} )

⇒ x^{0} = 900^{0} – 5x^{0}

⇒ 6x^{0} = 900^{0}

⇒ x^{0} = 150^{0}

Hence the angle is 150^{0} .

**9: **Find the angle whose supplement is four times its complement.

Let the angle is x0, then its supplement angle will be 180^{0} – x^{0}

and complement will be 90^{0} – x^{0}

According to question,

⇒ (180^{0} – x^{0} ) = 4(90^{0} – x^{0} )

⇒ 180^{0} – x^{0} = 360^{0} – 4x^{0}

⇒ 4x^{0} – x^{0} = 360^{0} – 180^{0} = 180^{0}

⇒ 3x^{0}=180^{0}

⇒ x^{0} = 60^{0}

**10:** Find the angle whose complement is one third of its supplement.

Let the angle is x^{0} , then its supplement angle will be 180^{0} – x^{0}

and complement will be 90^{0} – x^{0}

According to question,

⇒ (90^{0} – x^{0}) = 1/3 (180^{0} – x^{0})

⇒270^{0} – 3x^{0}= 180^{0} – x^{0}

⇒ – 3x^{0} + x^{0} = 180^{0} – 270^{0} = –90^{0}

⇒ – 2x^{0} = – 90^{0}

⇒ x^{0} = 45^{0}

Hence the angle is 45^{0} .

**11: **Two complementary angles are in the ratio 4 : 5. Find the angles.

Hence the angles are 40^{0} and 50^{0} .

**12: **Find the value of x for which the angles (2x – 5)^{0}

and (x – 10)^{0} are complementary angles.

Since the given angle are complementary to each other, therefore sum of both angle will be 90^{0}

So, (2x – 5)^{0} + (x – 10)^{0} = 90^{0}

⇒ (3x – 15)^{0} = 90^{0}

⇒3x = 90^{0} +15^{0} = 105^{0}

⇒ x^{0} = 35^{0}

Hence the angle is 35^{0}.

### RS Aggarwal Solutions Class 9 Maths Chapter 7 Lines and Angles Ex 7.2

**1: **In the adjoining figure, AOB is a straight line. Find the value of x.

Solution:

From figure,

⇒∠AOC+∠COB= (180)^{0}

⇒ (62)^{0} + x^{0} = (180)^{0}

⇒ x^{0} = (180)^{0} – (62)^{0} = (118)^{0}

**2: **In the adjoining figure, AOB is a straight line. Find the value of x.

Hence, find ∠AOC and ∠BOD.

**3: **In the adjoining figure, AOB is a straight line. Find the value of x.

Hence, find ∠AOC,∠COD and ∠BOD.

**4: **In the adjoining figure, x : y : z = 5 : 4 : 6. If XOY is a straight line,

find the values of x, y and z.

**5:** In the adjoining figure, what value of x will make AOB, a straight line?

**6:** Two lines AB and CD intersect at O. If ∠AOC= (50)^{0} , find ∠AOD,∠BOD and ∠BOC.

**7: **In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O,

forming angles as shown. Find the values of x, y, z and t.

**8:** In In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O.

Find the value of x. Hence, find ∠AOD,∠COE and ∠AOE.

**9: **two adjacent angles on a straight line are in the ratio 5 : 4.

Find the measure of each one of these angles.

**10: **If two straight lines intersect each other in such a way that one of the angles formed measures (90)^{0}, show that each of the remaining angles (90)^{0}.

**11:** Two lines AB and CD intersect at a point O such that

∠BOC+∠AOD= (280)^{0} , as shown in the figure. Find all the four angles.

**12:** Two lines AB and CD intersect each other at a point O such that ∠AOC∶∠AOD=5:7.

Find all the angles.

**13: **In the given figure, three lines AB, CD and EF intersect at a point O such that

∠AOE= (35)^{0} and ∠BOD= (40)^{0} .

Find the measure of ∠AOC,∠BOF,∠COF and ∠DOE.

**14: **In the given figure, two lines AB and CD intersect at a point O such that

∠BOC= (125)^{0} . Find the value of x, y and z.

**15: **If two straight lines intersect each other then prove that the ray opposite

to the bisector of one of the angles so formed bisects the vertically opposite angle.

**16:** Prove that the bisectors of two adjacent supplementary angles include a right angle.

### RS Aggarwal Solutions Class 9 Maths Chapter 7 Lines and Angles Ex 7.3

**1:** In the given figure, l ∥ m and a transversal t cuts them. If ∠1 = 120^{0}, find the measure of each of the remaining marked angle.

**2:** In the given figure, l ∥ m and a transversal t cuts them. If ∠7 = 80^{0}, find the measure of each the remaining marked angles.

**3:** in the given figure, l ∥ m and a transversal t cuts them. If ∠1 : ∠2 = 2:3, find the measure of each of the marked angle.

**4: **For what value of x will the lines l and m be parallel to each other?

**5:** for what value of x will the lines l and m be parallel to each other?

**6: **In the given figure, AB ∥ CD and BC∥ ED. Find the value of x.

**7:** In the given figure, AB ∥ CD∥ EF. Find the value of x.

**8:** In the given figure, AB ∥ CD. Find the value of x, y and z.

**9: **In each of the figures given below, AB∥ CD. Find the value of x in each case.

**10:** In the given figure, AB ∥ CD. Find the value of x.

**11: **8: In the given figure, AB ∥ PQ. Find the value of x, and y.

**12: **In the given figure, AB ∥ CD. Find the value of x.

**13: **In the given figure, AB ∥ CD. Find the value of x.

**14:** In the given figure, AB ∥ CD. Find the value of x, y and z.

**15: ** In the given figure, AB ∥ CD. Prove that ∠BAE – ∠ECD = ∠AEC

**16: **In the given figure, AB ∥ CD. Prove that p+q – r = 180

**17: **In the given figure AB∥ CD and EF∥ GH. Find the value of x, y and z.

**18:** In the given figure, AB∥ CD and a transversal t cuts them at E and F respectively. If EG and FG are the bisectors of ∠BEF and ∠EFD respectively, prove that ∠EGF =90^{0.}

**19: **In the given figure, AB ∥ CD and a transversal t cuts them at E and F respectively. If FP and EQ are the bisectors of ∠AEF and ∠EFD respectively, prove that EP ∥ FQ.

**20:** In the given figure, BA∥ ED and BC ∥ EF. Show that ∠ABC =∠DEF.

**21: **In the given figure, BA ∥ ED and BC ∥ EF. Show that ∠ABC+∠DEF = 180^{0}.

**22:** In the given figure, m and n are two plane mirrors perpendiculars to each other. Show that the incident ray CA is parallel to the reflected ray BD.

**23:** Inn the figure given below, state which lines are parallel and why?

**24: **Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

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