CBSE Important Questions Class 9 Maths Chapter 6

Important Questions Class 9 Mathematics Chapter 6 – Lines and Angles

Class 9 Mathematics Chapter 6, Lines And Angles, exposes you to fundamental geometry with a particular emphasis on the characteristics of the angles created when two lines cross one another and when a line intersects two or more parallel lines at different points. 

You can prepare for the upcoming board exams and improve your grade in class by using the Chapter 6 Class 9 Mathematics important questions. Extramarks has concentrated on getting you ready for the Class 9 exam using the CBSE curriculum. Your mathematical knowledge will be sharpened by solving these important questions Class 9 Mathematics Chapter 6, which will also help you grasp the topic better.

These important questions Class 9 Mathematics Chapter 6 provide a sample of the kinds of questions that are frequently asked in board exams. Learning about these can also give you more assurance as you take the examinations. Students can practice all types of questions from the chapters with the aid of these important questions Class 9 Mathematics chapter 6. 

Extramarks experts have created the important questions Class 9 Mathematics Chapter 6 in a well-structured format to offer a variety of potential approaches to solving problems and guarantee a thorough comprehension of the concepts. For their exams, it is advised that the students thoroughly practice all of these solutions. Additionally, it will assist students in laying the groundwork for more challenging courses.

Students are given other online learning resources, like revision notes, sample papers, and previous years question papers in addition to the NCERT Solutions, which are accessible in Extramarks. These materials were created with consideration for the NCERT and CBSE curricula. Additionally, it is suggested that students practice the important CBSE questions in Class 9 Mathematics Chapter 6 to get a sense of the final exam’s question format.

Important Questions Class 9 Mathematics Chapter 6 – With Solutions

The students may easily prepare all the concepts included in the CBSE Syllabus in a much better and more effective method with the help of Extramarks important questions Class 9 Mathematics Chapter 6. These resources include a thorough explanation, key formulas, are also offered to students to assist them in getting a quick review of all the topics. 

A few Important Questions Class 9 Mathematics Chapter 6 are provided here, along with their answers:

Question 1: If one angle of the triangle is equal to the sum of the other two angles, then the triangle is

(A) An equilateral triangle

(B) An obtuse triangle

(C) An isosceles triangle

(D) A right triangle

Solution 1: (D) A right triangle

Explanation:

We suppose the angles of △ABC be ∠A, ∠B and ∠C

Given, ∠A= ∠B+∠C …(equation 1)

But, in any △ABC,

Using the angle sum property, we have,

∠A+∠B+∠C=180o …(equation 2)

From equations (eq1) and (eq2), we get

∠A+∠A=180o

⇒2∠A=180o

⇒∠A=180o/2 = 90o

⇒∠A = 90o

Thus, we get that the triangle is a right-angled triangle

Question 2. The exterior angle of the triangle is 105°, and its two interior opposite angles are equal. Each of these equal angles is

(A)  75° 

(B) 52 ½o

(C) 72 ½o

(D)37 ½o

Solution 2: (B) 52 ½o

Explanation:

As per the question,

The exterior angle of the triangle will be = 105°

We suppose the two interior opposite angles of the triangle = x

We know that,

The exterior angle of a triangle will be = the sum of interior opposite angles

Thus, we have,

105° = x + x

2x = 105°

x = 52.5°

x = 52½

Question 3: The angles of the triangle are in the ratio 5 : 3: 7. The triangle is

(A) An acute angled triangle

(B) An isosceles triangle

(C) A right triangle

(D) An obtuse-angled of triangle

Solution 3: (A) An acute angled triangle

Explanation:

As per the question,

The angles of the triangle are in the ratio of 5 : 3: 7

Let the ratio 5:3:7 be 5x, 3x and 7x

Using the angle sum property of the triangle,

5x + 3x +7x =180

15x=180

x=12

Putting the value of x, i.e., x = 12, in 5x, 3x and 7x we have,

5x = 5×12 = 60o

3x = 3×12 = 36o

7x = 7×12 = 84o

As all the angles are less than 90o, the triangle will be an acute-angled triangle.

Question 4: In the given figure, if PQ || RS, then find the measure of angle m.

Solution 4:

Here, PQ || RS, PS is a transversal.

⇒ ∠PSR = ∠SPQ = 56°

Also, ∠TRS + m + ∠TSR = 180°

14° + m + 56° = 180°

⇒ m = 180° – 14 – 56 = 110°

Question 5:  In Figure, the lines AB and CD intersect at the point O. If ∠AOC +∠BOE = 70° and ∠BOD = 40°, find ∠BOE and the reflex ∠COE.

Solution 5:

From the diagram, we have

(∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) forming a straight line.

Then, ∠AOC+∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180°

Now, by substituting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we have,

∠COE = 110° and ∠BOE = 30°

So, reflex ∠COE = 360o – 110o = 250°

Question 6: In the given figure, POQ is the line. The ray OR is the perpendicular to the line PQ. OS is another ray lying between the rays OP and OR. Prove that ∠ROS = 12 (∠QOS – ∠POS).

Solution 6:

Given that OR is perpendicular to PQ

⇒ ∠POR = ∠ROQ = 90°

∴ ∠POS + ∠ROS = 90°

⇒ ∠ROS = 90° – ∠POS

Adding ∠ROS to both sides, we have

∠ROS + ∠ROS = (90° + ∠ROS) – ∠POS

⇒ 2∠ROS = ∠QOS – ∠POS

⇒ ∠ROS = 12 (∠QOS – ∠POS).

Question 7:  In Figure, the lines XY and MN intersect at the point O. If ∠POY = 90° and a: b = 2 : 3, find c.

Solution 7:

We know, the sum of the linear pair is always equal to 180°

Thus,

∠POY +a +b = 180°

Putting the value of ∠POY = 90° (given in the question), we have,

a+b = 90°

Now, given, a: b = 2 : 3, so,

We suppose a be 2x, and b be 3x

∴ 2x+3x = 90°

Solving this equation, we get

5x = 90°

So, x = 18°

∴ a = 2×18° = 36°

In the similar manner, b can be calculated, and the value will be

b = 3×18° = 54°

From the given diagram, b+c also forms a straight angle, so,

b+c = 180°

c+54° = 180°

Therefore, c = 126°

Question 8: In the Figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution 8:

As ST is a straight line so,

∠PQS+∠PQR = 180° (since it is a linear pair) and

∠PRT+∠PRQ = 180° (since it is a linear pair)

Now, ∠PQS + ∠PQR = ∠PRT+∠PRQ = 180°

We know, ∠PQR =∠PRQ (as given in the question)

∠PQS = ∠PRT. (Hence proved).

Question 9: In the Figure, if x+y = w+z, then prove that AOB is a line.

Solution 9:

To prove AOB is a straight line, we will first have to prove that x+y is a linear pair

i.e. x+y = 180°

We know, the angles around a point are 360° so,

x+y+w+z = 360°

In the question, it is given that,

x+y = w+z

So, (x+y)+(x+y) = 360°

2(x+y) = 360°

∴ (x+y) = 180° (Hence proved).

Question 10: In Figure, POQ is a line. The ray OR is perpendicular to the line PQ. OS is another ray lying between the rays OP and OR. Prove that ∠ROS = ½ (∠QOS – ∠POS).

Solution 10:

Given that (OR ⊥ PQ) and ∠POQ = 180°

Thus, ∠POS+∠ROS+∠ROQ = 180°

Now, ∠POS+∠ROS = 180°- 90° (As ∠POR = ∠ROQ = 90°)

∴ ∠POS + ∠ROS = 90°

Again, ∠QOS = ∠ROQ+∠ROS

Given, ∠ROQ = 90°,

∴ ∠QOS = 90° +∠ROS

Or, ∠QOS – ∠ROS = 90°

As ∠POS + ∠ROS = 90° and ∠QOS – ∠ROS = 90°, we have

∠POS + ∠ROS = ∠QOS – ∠ROS

2 ∠ROS + ∠POS = ∠QOS

Or, ∠ROS = ½ (∠QOS – ∠POS) (Hence proved).

Question 11: In the figure, find the values of x and y and show that AB || CD.

Solution 11:

We know, a linear pair is equal to 180°.

Thus, x+50° = 180°

∴ x = 130°

We also know, vertically opposite angles are equal.

Thus, y = 130°

In the two parallel lines, the alternate interior angles are equal. Here,

x = y = 130°

This proves that the alternate interior angles are equal, and thus, AB || CD.

Question 12: In the given figure, PQ || RS and EF || QS. If ∠PQS = 60°, then find the value of ∠RFE.

Solution 12:

Given PQ || RS

Thus, ∠PQS + ∠QSR = 180°

⇒ 60° + ∠QSR = 180°

⇒ ∠QSR = 120°

Now, EF || QS ⇒ ∠RFE = ∠QSR [corresponding ∠s]

⇒ ∠RFE = 120°

Question 13: In the figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Solution 13:

We know, AB || CD and CD||EF

Since the angles on the same side of the transversal line sum up to 180°,

x + y = 180° —–equation (i)

Also,

∠O = z (Since corresponding angles)

and, y +∠O = 180° (Since linear pair)

So, y+z = 180°

Now, let y = 3w and thus, z = 7w (As y : z = 3 : 7)

Therefore, 3w+7w = 180°

Or, 10 w = 180°

Thus, w = 18°

Now, y = 3×18° = 54°

and, z = 7×18° = 126°

Now, the angle x can be calculated from equation (i)

x+y = 180°

Or, x+54° = 180°

∴ x = 126°

Question 14:  In the figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Solution 14:

As AB || CD, GE is a transversal.

Given that ∠GED = 126°

So, ∠GED = ∠AGE = 126° (Since they are alternate interior angles)

And,

∠GED = ∠GEF +∠FED

As EF⊥ CD, ∠FED = 90°

∴ ∠GED = ∠GEF + 90°

Or, ∠GEF = 126° – 90° = 36°

Again, ∠FGE +∠GED = 180° (Since transversal)

Substituting the value of ∠GED = 126° we get,

∠FGE = 54°

So,

∠AGE = 126°

∠GEF = 36° and

∠FGE = 54°

Question 15: In figure, if AB || CD. If ∠ABR = 45° and ∠ROD = 105°, then find ∠ODC.

Solution 15:

Through the point O, we draw a line ‘l’ parallel to AB.

⇒ line I will also be parallel to CD, then

∠1 = 45°[alternate int. angles]

∠1 + ∠2 + 105° = 180° [straight angle]

∠2 = 180° – 105° – 45°

⇒ ∠2 = 30°

Now, ∠ODC = ∠2 [alternate int. angles]

= ∠ODC = 30°

Question 16: In the figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

[Hint: Draw a line parallel to ST through the point R.]

Solution 16:

First, we construct a line XY parallel to PQ.

We know, the angles on the same side of the transversal are equal to 180°.

Thus, ∠PQR+∠QRX = 180°

Or, ∠QRX = 180°-110°

∴ ∠QRX = 70°

In the similar manner,

∠RST +∠SRY = 180°

Or, ∠SRY = 180°- 130°

Therefore, ∠SRY = 50°

Now, from the linear pairs on the line XY-

∠QRX+∠QRS+∠SRY = 180°

Putting the values, we have,

∠QRS = 180° – 70° – 50°

Hence, ∠QRS = 60°

Question 17: In the figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Solution 17:

From the above diagram,

∠APQ = ∠PQR (Since Alternate interior angles)

Now, substituting the value of ∠APQ = 50° and ∠PQR = x, we ,

x = 50°

Also,

∠APR = ∠PRD (i.e., alternate interior angles)

Or, ∠APR = 127° (Given ∠PRD = 127°)

We know,

∠APR = ∠APQ+∠QPR

Now, substituting the values of ∠QPR = y and ∠APR = 127° we get,

127° = 50°+ y

Or, y = 77°

Thus, the measure of x and y are  as follows:

x = 50° and y = 77°

Question 18: In the given figure, p ll q, find the value of x.

Solution 18:

We extend the line p to meet RT at S.

Such that MS || QT

Now, in ARMS, we have

∠RMS = 180° – ∠PMR (Since linear pair]

= 180° – 120°

= 60°

∠RMS + ∠MSR + ∠SRM = 180° [i.e., by angle sum property of a ∆]

⇒ 60° + ∠MSR + 30o = 180°

⇒ MSR = 90°

Now, PS || QT – ∠MSR = ∠RTQ

⇒ ∠RTQ = x = MSR = 90° (Since corresponding ∠s]

Question 19:  In the figure, PQ and RS are the two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Solution 19:

Firstly, we draw the two lines, BE and CF, such that BE ⊥ PQ and CF ⊥ RS.

Now, since PQ || RS,

So, BE || CF

We know,

The angle of incidence = Angle of reflection (By the law of reflection)

So,

∠1 = ∠2 and

∠3 = ∠4

We also know, the alternate interior angles are equal. Here, BE ⊥ CF and the transversal line BC cuts them at points B and C.

So, ∠2 = ∠3 (Since they are alternate interior angles)

Here, ∠1 +∠2 = ∠3 +∠4

Or, ∠ABC = ∠DCB

So, AB || CD (since alternate interior angles are equal)

Question 20: In figure, the sides QP and RQ of ΔPQR are produced to the points S and T, respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution 20:

Given that TQR is a straight line, and thus, the linear pairs (i.e. ∠TQP and ∠PQR) will add up to 180°

So, ∠TQP +∠PQR = 180°

Now, substituting the value of ∠TQP = 110° we have,

∠PQR = 70°

We consider the ΔPQR,

The side QP is extended to the point S, and so ∠SPR forms the exterior angle.

Therefore, ∠SPR (∠SPR = 135°) is equal to the sum of the interior opposite angles. (By triangle property)

Or, ∠PQR +∠PRQ = 135°

Now, substituting the value of ∠PQR = 70° we get,

∠PRQ = 135°-70°

Hence, ∠PRQ = 65°

Question 21:  In the figure, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY, respectively of Δ XYZ, find ∠OZY and ∠YOZ.

Solution 21:

We know, the sum of the interior angles of the triangle is 180

So, ∠X +∠XYZ +∠XZY = 180°

Putting the given values in the question, we have,

62°+54° +∠XZY = 180°

Or, ∠XZY = 64°

Now, we know that ZO is the bisector, so,

∠OZY = ½ ∠XZY

Therefore, ∠OZY = 32°

In the similar manner, YO is a bisector, and so,

∠OYZ = ½ ∠XYZ

Or, ∠OYZ = 27° (As ∠XYZ = 54°)

Now, the sum of the interior angles of the given triangle,

∠OZY +∠OYZ +∠O = 180°

Putting their respective values, we get,

∠O = 180°-32°-27°

Hence, ∠O = 121°

Question 22:  In the figure, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution 22:

We know, AE is the transversal since AB || DE

Here ∠BAC and ∠AED are the alternate interior angles.

Hence, ∠BAC = ∠AED

Given, ∠BAC = 35°

∠AED = 35°

Now considering the triangle CDE. We know that the sum of the interior angles of the triangle is 180°.

∴ ∠DCE+∠CED+∠CDE = 180°

Putting the values, we get

∠DCE+35°+53° = 180°

Hence, ∠DCE = 92°

Question 23: To protect the poor people from cold weather, Ram Lal. has given his land to make a shelter home for them. In the given figure, ‘the sides QP and RQ of ∆PQR are produced to points S and T, respectively. If ∠PQT = 110° and ∠SPR = 135°, find the value of ∠PRQ.

Solution 23:

Here,

∠SPR + ∠QPR = 180° [i.e., a linear pair]

135° + ∠QPR = 180° [∵ ∠SPR = 135°]

⇒ ∠QPR = 180° – 135° = 45°

In ∆PQR, by the exterior angle property, we have

∠QPR + ∠PRQ = ∠PQT

45° + ∠PRQ = 110°

∠PRQ = 110° – 45° = 65°

Question 24: In the figure, if the lines PQ and RS intersect at point T, in such a way that ∠PRT = 40°, ∠RPT = 95°, and ∠TSQ = 75°, find ∠SQT.

Solution 24:

We consider the triangle PRT.

∠PRT +∠RPT + ∠PTR = 180°

Thus, ∠PTR = 45°

Now, ∠PTR will be equal to ∠STQ as they are the vertically opposite angles.

So, ∠PTR = ∠STQ = 45°

Again, in triangle STQ,

∠TSQ +∠PTR + ∠SQT = 180°

Solving this equation, we get,

74° + 45° + ∠SQT = 180°

∠SQT = 60°

Question 25:  In the figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution 25:

x +∠SQR = ∠QRT (Because they are alternate angles and QR is the transversal)

Thus, x+28° = 65°

∴ x = 37°

It is also known that the alternate interior angles are the same, and so,

∠QSR = x = 37°

Also,

∠QRS +∠QRT = 180° (Since linear pair)

Or, ∠QRS+65° = 180°

So, ∠QRS = 115°

Using the angle sum property in Δ SPQ,

∠SPQ + x + y = 180°

Putting their respective values, we get,

90°+37° + y = 180°

y = 1800 – 1270 = 530

Hence, y = 53°

Question 26:  In the figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at the point T, then prove that ∠QTR = ½ ∠QPR.

Solution 26:

We consider the ΔPQR. ∠PRS is the exterior angle, and ∠QPR and ∠PQR are the interior angles.

So, ∠PRS = ∠QPR+∠PQR (According to the triangle property)

Or, ∠PRS -∠PQR = ∠QPR ———–equation(i)

Now, considering the ΔQRT,

∠TRS = ∠TQR+∠QTR

Or, ∠QTR = ∠TRS-∠TQR

We know, QT and RT bisect ∠PQR and ∠PRS, respectively.

So, ∠PRS = 2 ∠TRS and ∠PQR = 2∠TQR

Here, ∠QTR = ½ ∠PRS – ½∠PQR

Or, ∠QTR = ½ (∠PRS -∠PQR)

From equation (i), we know, ∠PRS -∠PQR = ∠QPR

Therefore, ∠QTR = ½ ∠QPR (hence proved).

Question 27: For what value of x + y in the figure  will ABC be a line? Justify the answer.

Solution 27:

The value of x + y should be 180o for ABC to be a line.

Justification:

From the figure, we can state that,

BD is a ray intersecting AB and BC at point B, which implies

∠ABD = y

and, ∠DBC = x

We know,

If a ray stands on the line, then sum of the two adjacent angles formed will be 180°.

⇒ If the sum of the two adjacent angles is 180°, then a ray stands on the line.

So, for ABC to be a line,

Then, the sum of ∠ABD and ∠DBC should be equal to 180°.

⇒ ∠ABD + ∠DBC = 180°

⇒ x + y = 180°

Thus, the value of x + y should be equal to 180° for ABC to be a line.

Question 28: Can a triangle have all angles less than 60°? Give a reason for your answer.

Solution 28:

No. A triangle cannot have all the angles less than 60°

Justification:

As per the angle sum property,

We know the sum of all the interior angles of a triangle should be = 180°.

We suppose all the angles are 60o,

Then we get, 60o + 60o + 60o = 180o.

Now, considering angles less than 60o,

We suppose 59o to be the highest natural number, less than 60o.

Then we get,

59o +59o + 59o = 177o ≠ 180o

Thus, we can say that if all the angles are less than 60o, the measure of the angles won’t be satisfying the angle sum property.

Therefore, a triangle cannot have all the angles less than 60o.

Question 29: Can a triangle have two obtuse angles? Give a reason for your answer.

Solution 29:

No. A triangle cannot have two obtuse angles.

Justification:

According to the angle sum property,

We know, the sum of all the interior angles of the triangle should be = 180o.

An obtuse angle is one with a value greater than 90° but less than 180°.

We consider the two angles to be equal to the lowest natural number greater than 90o, i.e., 91o.

As per the question,

If the triangle has two obtuse angles, then there are two angles that would be at least 91° each.

By adding the two angles, we get

Sum of the two angles = 91° + 91°

⇒ Sum of the two angles = 182°

The sum of the two angles already exceeds the sum of the three angles of the triangle, even before taking into consideration the third angle.

Thus, a triangle cannot have two obtuse angles.

Question 30: How many triangles can be drawn having angles as 45°, 64° ,and 72°? Give a reason for your answer.

Solution 30:

No such triangle can be drawn having its angles 45°, 64° and 72°.

Justification:

According to the angle sum property,

We know the sum of all the interior angles of a triangle should be = 180o.

But, as per the question,

We have the angles as 45°, 64° and 72°.

Sum of these angles is = 45° + 64° + 72°

= 181o, which is greater than 180o.

So, the angles do not satisfy the angle sum property of a triangle.

Therefore, no triangle can be drawn having angles 45°, 64° and 72°.

Question 31: How many triangles can be drawn having their angles as 53°, 64° and 63°? Give a reason for your answer.

Solution 31:

Infinitely many triangles can be drawn having angles as 53°, 64° and 63°.

Justification:

As per the angle sum property,

We know the sum of all the interior angles of the triangle should be = 180o.

As per the question,

We have the angles as 53°, 64°, and 63°.

Sum of these angles = 53° + 64° + 63°

= 180o

Thus, the angles satisfies the angle sum property of the triangle.

Therefore, infinitely many triangles may be drawn, having their angles as 53°, 64° and 63°.

 

Question 32: In the figure, OD is the bisector of ∠AOC, and OE is the bisector of ∠BOC and OD ⊥ OE. Show that points A, O and B are collinear.

Solution 32:

According to the question,

In the figure,

OD ⊥ OE,

OD and OE are the bisectors of ∠AOC and ∠BOC.

To prove: The points A, O and B are collinear.

i.e., AOB is a straight line.

Proof:

As OD and OE bisect angles ∠AOC and ∠BOC, respectively.

∠AOC = 2∠DOC …(equation 1)

And ∠COB = 2∠COE …(equation 2)

Adding (equation 1) and (equation 2), we have,

∠AOC = ∠COB = 2∠DOC + 2∠COE

∠AOC +∠COB = 2(∠DOC +∠COE)

∠AOC + ∠COB = 2∠DOE

Since OD⊥OE

We get,

∠AOC +∠COB = 2×90o

∠AOC +∠COB =180o

∠AOB =180o

So, ∠AOC + ∠COB form a linear pair.

Thus, AOB is a straight line.

Therefore, the points A, O and B are collinear.

Question 33: In the figure, OP bisects ∠BOC and OQ bisects ∠AOC. Prove that ∠POQ = 90°

Solution 33:

∵ OP bisects ∠BOC

∴ ∠BOP = ∠POC = x (say)

Also, OQ bisects. ∠AOC

∠AOQ = ∠COQ = y (say) .

∵ Ray OC stands on ∠AOB

∴ ∠AOC + ∠BOC = 180° [linear pair]

⇒ ∠AOQ + ∠QOC + ∠COP + ∠POB = 180°

⇒ y + y + x + x = 180°.

⇒ 2x + 2y = 180°

⇒ x + y = 90°

Now, ∠POQ = ∠POC + ∠COQ

= x + y = 90°

Question 34: In the figure, ∠1 = 60° and ∠6 = 120°. Show that the lines m and n are parallel.

Solution 34:

As per the question,

We have from figure ∠1 = 60° and ∠6 = 120°

As, ∠1 = 60° and ∠6 = 120°

Here, ∠1 = ∠3 [i.e.,vertically opposite angles]

∠3 = ∠1 = 60°

Now, ∠3 + ∠6 = 60° + 120°

⇒ ∠3 + ∠6 = 180°

We know,

If the sum of the two interior angles on the same side of l is 180°, then the lines are parallel.

Therefore, m || n

Question 35:  AP and BQ are the bisectors of the two alternate interior angles which are formed by the intersection of the transversal t with the parallel lines l and m (figure). Show that AP || BQ.

Solution 35:

l || m and t is the transversal

∠MAB = ∠SBA [alternate angles]

⇒ ½ ∠MAB = ½ ∠SBA

⇒ ∠PAB = ∠QBA

⇒ ∠2 = ∠3

But, ∠2 and ∠3 are alternate angles.

Hence, AP||BQ.

Question 36: If in the Figure, the bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m.

Solution 36:

AP is the bisector of ∠MAB

BQ is the bisector of ∠SBA.

Given: AP||BQ.

As AP||BQ,

We have,

So ∠2 = ∠3 [Alternate angles]

2∠2 = 2∠3

⇒ ∠2 + ∠2 = ∠3 +∠3

From the figure, we have ∠1= ∠2and ∠3 = ∠4

⇒ ∠1+ ∠2 = ∠3 +∠4

⇒ ∠MAB = ∠SBA

But we also know that these are the alternate angles.

Therefore, the lines l and m are parallel, i.e., l ||m.

Question 37:  In the figure, BA || ED and BC || EF. Show that ∠ABC = ∠DEF [Hint: Produce DE to intersect BC at P (say)].

Solution 37:

Construction:

We extend DE to intersect BC at point P.

Given that EF||BC and DP are transversal,

∠DEF = ∠DPC …(equation 1) [Since corresponding angles]

Also given, AB||DP and BC is a transversal,

∠DPC = ∠ABC …(equation 2) [Since Corresponding angles]

From (equation 1) and (equation 2), we get

∠ABC = ∠DEF

Hence, Proved.

Question 38: In the given figure, AB || CD, ∠FAE = 90°, ∠AFE = 40°, find ∠ECD.

Solution 38:

In AFAE,

external ∠FEB = ∠A + F

= 90° + 40° = 130°

As AB || CD

Therefore, ∠ECD = FEB = 130°

Hence, ∠ECD = 130°.

Question 39:  If the two lines intersect each other, prove that the vertically opposite angles are equal.

Solution 39:

We have,

AB and CD intersect each other at the point O.

We suppose the two pairs of vertically opposite angles be,

1st pair – ∠AOC and ∠BOD

2nd pair – ∠AOD and ∠BOC

To prove:

The vertically opposite angles are equal,

i.e., ∠AOC = ∠BOD, and ∠AOD = ∠BOC

Again,

The ray AO stands on the line CD.

We know,

If a ray lies on the line, then the sum of the adjacent angles is equal to 180°.

⇒ ∠AOC + ∠AOD = 180° (i.e., By linear pair axiom) … equation (i)

In the similar manner, the ray DO lies on the line AOB.

⇒ ∠AOD + ∠BOD = 180° (i.e.,By linear pair axiom) … equation (ii)

From equations (i) and (ii),

We get,

∠AOC + ∠AOD = ∠AOD + ∠BOD

⇒ ∠AOC = ∠BOD – – – – equation (iii)

In the similar manner, the ray BO lies on the line COD.

⇒ ∠DOB + ∠COB = 180° (By using linear pair axiom) – – – – equation (iv)

Also, the ray CO is lying on the line AOB.

⇒ ∠COB + ∠AOC = 180° (By using linear pair axiom) – – – – equation (v)

From equations (iv) and (v),

We have,

∠DOB + ∠COB = ∠COB + ∠AOC

⇒ ∠DOB = ∠AOC – – – – equation (vi)

Therefore, from equation (iii) and equation (vi),

We have,

∠AOC = ∠BOD, and ∠DOB = ∠AOC

Thus, we get vertically opposite angles are equal.

Hence Proved.

Question 40: Bisectors of the interior ∠B and the exterior ∠ACD of a Δ ABC intersect at point T.

Prove that ∠ BTC = ½ ∠ BAC.

Solution 40:

Given: In△ ABC, we produce BC to D and the bisectors of ∠ABC and ∠ACD meet at point T.

To prove:

∠BTC = ½ ∠BAC

Proof:

In △ABC, ∠ACD is an exterior angle.

We know that,

The exterior angle of the triangle is equal to the sum of two opposite angles,

Here,

∠ACD = ∠ABC + ∠CAB

Now, dividing the L.H.S and R.H.S by 2,

⇒ ½ ∠ACD = ½ ∠CAB + ½ ∠ABC

⇒ ∠TCD = ½ ∠CAB + ½ ∠ABC …equation (1)

[∵CT is the bisector of ∠ACD⇒ ½ ∠ACD = ∠TCD]

We know,

The exterior angle of the triangle is equal to the sum of two opposite angles,

Then in △ BTC,

∠TCD = ∠BTC +∠CBT

⇒ ∠TCD = ∠BTC + ½ ∠ABC …(2)

[∵BT is bisector of △ ABC ⇒∠CBT = ½ ∠ABC ]

From equations (1) and (2),

We get,

½ ∠CAB + ½ ∠ABC = ∠BTC + ½ ∠ABC

⇒ ½ ∠CAB = ∠BTC or ½ ∠BAC = ∠BTC

Hence, proved.

Question 41: A transversal intersects the two parallel lines. Prove that the bisectors of any pair of the corresponding angles so formed are parallel.

Solution 41:

We suppose,

AB ║ CD

EF be the transversal that passes through the two parallel lines at the point P and Q, respectively.

PR and QS are the bisectors of ∠EPB and ∠PQD.

We know, corresponding angles of the parallel lines are equal,

So, ∠EPB = ∠PQD

½ ∠EPB = ½ ∠PQD

∠EPR = ∠PQS

But we also know that they are the corresponding angles of PR and QS

Since the corresponding angles are equal,

We have,

PR ║ QS

Hence Proved.

Question 42: The lines AB and CD intersect at the point O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and the reflex ∠COE.

Solution 42:

We have,

∠AOC, ∠BOE, ∠COE and ∠COE, ∠BOD,  ∠BOE form a straight line each.

Thus, ∠AOC + ∠BOE +∠COE = ∠COE +∠BOD + ∠BOE = 180°

Here, by putting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get:

70° +∠COE = 180°

∠COE = 110°

Similarly,

110° +  40° + ∠BOE = 180°

∠BOE = 30°

Question 43: The lines XY and MN intersect at O. If ∠POY = 90° and a: b = 2 : 3, find c.

Solution 43:

We know, the sum of a linear pair is always equal to 180°

So,

∠POY + a + b = 180°

Substitutg the value of ∠POY = 90° (as given), we get,

a + b = 90°

Now, we know from the question a: b = 2 : 3, so,

We suppose  a be 2x, and b be 3x.

∴ 2x + 3x = 90°

By solving the equation, we get

5x = 90°

So, x = 18°

∴ a = 2 × 18° = 36°

In the similar manner, b can be calculated, and the value will be

b = 3 × 18° = 54°

Here, b + c also forms a straight angle, so,

b + c = 180°

=> c + 54° = 180°

∴ c = 126°

Question 44: Given, POQ is a line. The ray OR is perpendicular to the line PQ. OS is another ray lying between the rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).

Solution 44:

Given in the question that (OR ⊥ PQ) and ∠POQ = 180°

So, ∠POS + ∠ROS + ∠ROQ = 180°    (Since linear pair of angles)

Now, ∠POS + ∠ROS = 180° – 90°      (As ∠POR = ∠ROQ = 90°)

∴ ∠POS + ∠ROS = 90°

Now, ∠QOS = ∠ROQ + ∠ROS

Again, given ∠ROQ = 90°,

Therefore, ∠QOS = 90° + ∠ROS

Or, ∠QOS – ∠ROS = 90°

Since ∠POS + ∠ROS = 90° and ∠QOS – ∠ROS = 90°, we get

∠POS + ∠ROS = ∠QOS – ∠ROS

=>2 ∠ROS + ∠POS = ∠QOS

Or, ∠ROS = ½ (∠QOS – ∠POS) (Hence proved).

Question 45: If AB || CD, EF ⊥ CD and ∠GED = 126°, find the value of  ∠AGE, ∠GEF and ∠FGE.

Solution 45:

Since AB || CD GE is a transversal.

It is given that ∠GED = 126°

So, ∠GED = ∠AGE = 126° (alternate interior angles)

Also,

∠GED = ∠GEF + ∠FED

As

EF ⊥ CD, ∠FED = 90°

∴ ∠GED = ∠GEF + 90°

Or, ∠GEF = 126° – 90° = 36°

Again, ∠FGE + ∠GED = 180° (Since it is the transversal)

Substituting the value of ∠GED = 126° we get,

∠FGE = 54°

So,

∠AGE = 126°

∠GEF = 36° and

∠FGE = 54°

Question 46: If ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY, respectively of Δ XYZ, find ∠OZY and ∠YOZ.

Solution 46:

We know, the sum of the interior angles of the triangle is 180°.

Then, ∠X +∠XYZ + ∠XZY = 180°

Substituting the values given in the equation we get,

62° + 54° + ∠XZY = 180°

Or, ∠XZY = 64°

Now, ZO is the bisector, so,

∠OZY = ½ ∠XZY

∴ ∠OZY = 32°

In the similar manner, YO is the bisector, and thus,

∠OYZ = ½ ∠XYZ

Or, ∠OYZ = 27° (As ∠XYZ = 54°)

Now, as the sum of interior angles of the triangle,

So, ∠OZY +∠OYZ + ∠O = 180°

Putting their respective values, we get,

∠O = 180° – 32° – 27°

Or, ∠O = 121°

Question 47: If AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then find the value of ∠QRS.

Solution 47:

We have,

AB || CD || EF

PQ || RS

∠RQD = 25°

∠CQP = 60°

PQ || RS.

We know,

If the transversal intersects the two parallel lines, then each pair of the alternate exterior angles are equal.

Now, as PQ || RS

⇒ ∠PQC = ∠BRS

We have ∠PQC = 60°

⇒ ∠BRS = 60° … equation (i)

We know,

If the transversal intersects the two parallel lines, then each pair of the alternate interior angles is equal.

Now, since AB || CD

⇒ ∠DQR = ∠QRA

We have ∠DQR = 25°

⇒ ∠QRA = 25° … equation (ii)

By using the linear pair axiom,

We get,

∠ARS + ∠BRS = 180°

⇒ ∠ARS = 180° – ∠BRS

⇒ ∠ARS = 180° – 60° (From equation (i), ∠BRS = 60°)

⇒ ∠ARS = 120° … equation (iii)

Now, ∠QRS = ∠QRA + ∠ARS

From equations (ii) and (iii), we have,

∠QRA = 25° and ∠ARS = 120°

Hence, the above equation can be written as:

∠QRS = 25° + 120°

⇒ ∠QRS = 145°

Question 48: If an angle is half of its complementary angle, then find its degree measure.

Solution 48:

We suppose that the required angle be x

∴ Its complement = 90° – x

Now, according to the given statement, we obtain

x = 12(90° – x)

⇒ 2x = 90° – x

⇒ 3x = 90°

⇒ x = 30°

Hence, the required angle is 30°.

Question 49: The two complementary angles are in the ratio of 1: 5. Find the measures of the angles.

Solution 49:

We suppose that the two complementary angles be x and 5x.

∴ x + 5x = 90°

⇒ 6x = 90°

⇒ x = 15°

Hence, the two complementary angles are 15° and 5 × 15°, i.e., 15° and 75°.

Question 50: If an angle is 14o more than its complement, then find its measure.

Solution 50:

Let the required angle be x

∴ Its complement = 90° – x

Now, according to the given statement, we obtain

x = 90° – x + 14°

⇒ 2x = 104°

⇒ x = 52°

Hence, the required angle is 52o.

Question 51: If AB || EF and EF || CD, then find the value of x.

Solution 51:

Since EF || CD ∴ y + 150° = 180°

⇒ y = 180° – 150° = 30°

Now, ∠BCD = ∠ABC

x + y = 70°

x + 30 = 70

⇒ x = 70° – 30° = 40°

Hence, the value of x is 40°.

Question 52: In the given figure, the lines AB and CD intersect at O. Find the value of x.

Solution 52:

Here, the lines AB and CD intersect at O.

∴ ∠AOD and ∠BOD form a linear pair

⇒ ∠AOD + ∠BOD = 180°

⇒ 7x + 5x = 180°

⇒ 12x = 180°

⇒ x = 15°

Question 53: In the given figure, if x°, y° and z° are the exterior angles of ∆ABC, then find the value of x° + y° + z°.

Solution 53:

We know that an exterior angle of a triangle is equal to the sum of two opposite interior angles.

⇒ x° = ∠1 + ∠3

⇒ y° = ∠2 + ∠1

⇒ z° = ∠3 + ∠2

Adding all these, we have

x° + y° + z° = 2(∠1 + ∠2 + ∠3)

= 2 × 180°

= 360°

Question 54: In figure., AD and CE are the angle bisectors of ∠A and ∠C, respectively. If ∠ABC = 90°, then find ∠AOC.

Solution 54:

∵ AD and CE are the bisectors of ∠A and ∠C

∠OAC = 12∠A and

∠OCA 12∠C

∠OAC + ∠OCA = 12 (∠A + ∠C)

= 12 (1800 – ∠B) [ Since, ∠A + ∠B +∠C = 1800]

=12 (1800 – 900) [Since,  ∠ABC = 900]

= 12 x 900  = 450

In ∆AOC,

∠AOC + ∠OAC + ∠OCA = 180°

⇒ ∠AOC + 45o = 180°

⇒ ∠AOC = 180° – 45° = 135°.

Question 55: In the given figure, prove that m || n.

Solution 55:

In ∆BCD,

ext. ∠BDM = ∠C + ∠B

= 38° + 25° = 63°

Now, ∠LAD = ∠MDB = 63°

But these are corresponding angles. Hence,

m || n

Question 56: In the given figure, two straight lines, PQ and RS, intersect each other at O. If ∠POT = 75°, find the values of a, b, and c.

Solution 56:

Here, 4b + 75° + b = 180° [since a straight angle]

5b = 180° – 75° = 105°

b –  105∘5 = 21°

Therefore, a = 4b = 4 × 21° = 84° (i.e., vertically opp. ∠s]

Again, 2c + a = 180° [Since, a linear pair]

⇒ 2c + 84° = 180°

⇒ 2c = 96°

⇒ c = 96°2 = 48°

Therefore, the values of a, b and c are a = 84°, b = 21° and c = 48°.

Question 57: In the figure, ∠X = 72°, ∠XZY = 46°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY, respectively  ∆XYZ, find ∠OYZ and ∠YOZ.

Solution 57:

In ∆XYZ, we have

∠X + XY + ∠Z = 180°

⇒ ∠Y + ∠Z = 180° – ∠X

⇒ ∠Y + ∠Z = 180° – 72°

⇒ Y + ∠Z = 108°

⇒ 12 ∠Y + 12∠Z = 12 × 108°

∠OYZ + ∠OZY = 54°

[∵ YO and ZO are the bisectors of ∠XYZ and ∠XZY]

⇒ ∠OYZ + 12 × 46° = 54°

∠OYZ + 23° = 54°

⇒ ∠OYZ = 540 – 23° = 31°

Again, in ∆YOZ, we have

∠YOZ = 180° – (∠OYZ + ∠OZY)

= 180° – (31° + 23°) 180° – 54° = 126°

Question 58: Prove that if the two lines intersect each other, then the bisectors of the vertically opposite angles are in the same line.

Solution 58:

Let AB and CD be the two intersecting lines intersecting each other at the point O.

OP and OQ are the bisectors of ∠AOD and ∠BOC.

∴ ∠1 = ∠2 and ∠3 = ∠4 …equation (i)

Now, ∠AOC = ∠BOD [vertically opp. ∠s] ……equation (ii)

⇒ ∠1 + ∠AOC + ∠3 = ∠2 + ∠BOD + ∠4 [adding equation (i) and (ii)]

Also, ∠1 + ∠AOC + ∠3 + ∠2 + ∠BOD + ∠4 = 360° (Since ∠s at a point are 360°]

⇒ ∠1 + ∠AOC + ∠3 + ∠1 + ∠AOC + ∠3 = 360° [by using equation (i), (ii)]

⇒ ∠1 + ∠AOC + ∠3 = 180°

or ∠2 + ∠BOD + ∠4 = 180°

Therefore, OP and OQ are in the same line.

Question 59: In the given figure, AB || CD and EF || DG, find ∠GDH, ∠AED and ∠DEF.

Solution 59:

As AB || CD and HE is the transversal.

∴ ∠AED = ∠CDH = 40° [i.e., corresponding ∠s]

Now, ∠AED + ∠DEF + ∠FEB = 180° [since a straight ∠]

40° + CDEF + 45° = 180°

∠DEF = 180° – 45 – 40 = 95°

Again, given, EF || DG and HE is the transversal.

∴ ∠GDH = ∠DEF = 95° [i.e., corresponding ∠s]

Therefore, ∠GDH = 95°, ∠AED = 40° and ∠DEF = 95°

Question 60: In the figure, DE || QR and AP and BP are the bisectors of ∠EAB and ∠RBA, respectively. Find ∠APB.

Solution 60:

Here, AP and BP are the bisectors of ∠EAB and ∠RBA, respectively.

⇒ ∠1 = ∠2 and ∠3 = ∠4

Since DE || QR and the transversal n intersects DE and QR at A and B, respectively.

⇒ ∠EAB + ∠RBA = 180°

[Since co-interior angles are supplementary]

⇒ (∠1 + ∠2) + (∠3 + ∠4) = 180°

⇒ (∠1 + ∠1) + (∠3 + ∠3) = 180° (using equation (i)

⇒ 2(∠1 + ∠3) = 180°

⇒ ∠1 + ∠3 = 90°

Now, in ∆ABP, by angle sum property, we have

∠ABP + ∠BAP + ∠APB = 180°

⇒ ∠3 + ∠1 + ∠APB = 180°

⇒ 90° + ∠APB = 180°

⇒ ∠APB = 90°

Question 61: If the two parallel lines are intersected by a transversal, then prove that the bisectors of any of the two corresponding angles are parallel.

Solution 61:

Given: AB || CD and the transversal PQ meet these lines at E and F, respectively. EG and FH are

the bisectors of pair of the corresponding angles ∠PEB and ∠EFD.

To Prove: EG || FH 

Proof

∵ EG and FH are the bisectors of ∠PEB, respectively.

∠PEG = 12 ∠PEB ………equation (i)

And, ∠EFH = 12 ∠EFD  …..equation (ii)

Since, AB || CD and PQ is a transversal

Therefore, ∠PEB = ∠EFD

12 ∠PEB = 12 ∠EFD

∠PEG = ∠EFH

Which are the corresponding angles of EG and FH∴ EG || FH.

Question 62: In the given figure, m and n are the two plane mirrors perpendicular to each other. Show that the incident rays CA is parallel to the reflected ray BD.

Solution 62:

Let the normals at A and B meet at point P.

Since the mirrors are perpendicular to each other, therefore, BP || OA and AP || OB.

Thus, BP ⊥ PA, i.e., ∠BPA = 90°

Therefore, ∠3 + ∠2 = 90° [by angle sum property] …equation (i)

Also, ∠1 = ∠2 and ∠4 = ∠3 [Angle of incidence = Angle of reflection]

Thus, ∠1 + ∠4 = 90° [from equations (i)) …(ii]

Adding equation (i) and (ii), we have

∠1 + ∠2 + ∠3 + ∠4 = 180°

i.e., ∠CAB + ∠DBA = 180°

Hence, CA || BD

Question 63: If the two parallel lines are intersected by a transversal, prove that the bisectors of two pairs of the interior angles form a rectangle.

Solution 63:

Given: AB || CD and the transversal EF cut them at P and Q, respectively and the bisectors of

the pair of interior angles form a quadrilateral PRQS.

To Prove: PRQS is a rectangle.

Proof: Since PS, QR, QS and PR are the bisectors of angles

∠BPQ, ∠CQP, ∠DQP and ∠APQ, respectively.

∴∠1 =12 ∠BPQ, ∠2 = 12∠CQP,

∠3 = 12∠DQP and ∠4 = 12∠APQ

Now, AB || CD and EF is the transversal

∴ ∠BPQ = ∠CQP

⇒ ∠1 = ∠2 (∵∠1 = 12∠BPQ and ∠2 =  12∠QP)

But these are the pairs of alternate interior angles of PS and QR

∴ PS || QR

In the similar manner, we can prove ∠3 = ∠4 = QS || PR

∴ PRQS is the parallelogram.

Further ∠1 + ∠3 =  12∠BPQ +  12∠DQP =  12 (∠BPQ + ∠DQP)

=  12 × 180° = 90° (Since ∠BPQ + ∠DQP = 180°)

∴ In ∆PSQ, we have ∠PSQ = 180° – (∠1 + ∠3) = 180° – 90° = 90°

Therefore, PRRS is the parallelogram whose one angle ∠PSQ = 90°.

Hence, PRQS is a rectangle.

Question 64: If in ∆ABC, the bisectors of ∠B and ∠C intersect each other at point O. Prove that ∠BOC = 90° +  12∠ A.

Solution 64:

We suppose ∠B = 2x and ∠C = 2y

∵OB and OC bisect ∠B and ∠C, respectively.

∠OBC =  12∠B =  12 × 2x = x 

and ∠OCB =  12 ∠C =  12 × 2y = y

Now, in ∆BOC, we have

∠BOC + ∠OBC + ∠OCB = 180°

⇒ ∠BOC + x + y = 180°

⇒ ∠BOC = 180° – (x + y)

Again, in ∆ABC, we have

∠A + 2B + C = 180°

⇒ ∠A + 2x + 2y = 180°

⇒ 2(x + y) =   12(180° – ∠A)

⇒ x + y = 90° –  12∠A …..equation (ii)

From equation (i) and (ii), we have

∠BOC = 180° – (90° –  12∠A) = 90° +  12 ∠A

Question 65: In the figure, if I || m and ∠1 = (2x + y)°, ∠4 = (x + 2y)° and ∠6 = (3y + 20)°. Find ∠7 and ∠8.

Solution 65:

Here, ∠1 and ∠4 form a linear pair

∠1 + ∠4 = 180°

(2x + y)° + (x + 2y)° = 180°

3(x + y)° = 180°

x + y = 60

As I || m and n is a transversal

∠4 = ∠6

(x + 2y)° = (3y + 20)°

x – y = 20

Adding equation (i) and (ii), we have

2x = 80 = x = 40

From (i), we have

40 + y = 60 ⇒ y = 20

Now, ∠1 = (2 x 40 + 20)° = 100°

∠4 = (40 + 2 x 20)° = 80°

∠8 = ∠4 = 80° [corresponding ∠s]

∠1 = ∠3 = 100° [vertically opp. ∠s]

∠7 = ∠3 = 100° [corresponding ∠s]

Hence, ∠7 = 100° and ∠8 = 80°

Question 66: In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28o and ∠QRT = 65°. Find the values of x, y and z.

Solution 66:

Here, PQ || SR.

⇒ ∠PQR = ∠QRT

⇒ x + 28° = 65°

⇒ x = 65° – 28° = 37°

Now, in ∆SPQ, ∠P = 90°

∴ ∠P + x + y = 180° [i.e.,angle sum property]

∴ 90° + 37° + y = 180°

⇒ y = 180° – 90° – 37° = 53°

Now, ∠SRQ + ∠QRT = 180° [linear pair]

z + 65° = 180°

z = 180° – 65° = 115°

Question 67: In the figure, AP and DP are bisectors of two adjacent angles, A and D, of a quadrilateral ABCD. Prove that 2∠APD = ∠B + ∠C.

Solution 67:

In quadrilateral ABCD, we have

∠A + ∠B + ∠C + ∠D = 360°

12∠A + 12∠B + 12∠C + 12∠D = 12 X3600

12∠A + 12∠D = 1800 – + 12 (∠B + ∠C)

As, AP and DP are the bisectors of ∠A and ∠D.

Therefore, ∠PAD = 12∠A

and, ∠PDA = 12∠D

Now, ∠PAD + ∠PDA = 1800  – 12 (∠B + ∠C)     (i)

In APD, we have,

∠APD + ∠PAD +  ∠PDA   = 1800

∠APD + 1800 – 12 (∠B + ∠C) = 1800  [ Using (i)]

∠APD = 12 (∠B + ∠C)

2 ∠APD = (∠B + ∠C)

Question 68: In the figure, PS is the bisector of ∠QPR; PT ⊥ RQ and Q > R. Show that ∠TPS = 12(∠Q – ∠R).

Solution 68:

As PS is the bisector of ∠QPR

∴∠QPS = ∠RPS = x (suppose)

In ∆PRT, we have

∠PRT + ∠PTR + ∠RPT = 180°

⇒ ∠PRT + 90° + ∠RPT = 180°

⇒ ∠PRT + ∠RPS + ∠TPS = 90°

⇒ ∠PRT + x + ∠TPS = 90°

⇒ ∠PRT or ∠R = 90° – ∠TPS – x

Again in ΔΡQT, we have

∠PQT + ∠PTQ + ∠QPT = 180°

⇒ ∠PQT + 90° + ∠QPT = 180°

⇒ ∠PQT + ∠QPS – TPS = 90°

⇒ ∠PQT + x – ∠TPS = 90° [Since ∠QPS = x]

⇒ ∠PQT or ∠Q = 90° + ∠TPS – x …equation (ii)

Subtracting equation (i) from (ii), we get

⇒ ∠Q – ∠R = (90° + ∠TPS – x) – 190° – ∠TPS – x)

⇒ ∠Q – ∠R = 90° + ∠TPS – X – 90° + ∠TPS + x

⇒ 2∠TPS = 2Q- ∠R

⇒ ∠TPS = 12(Q – ∠R)

Question 69: In a triangle, if ∠A = 2∠B = 6∠C, find the measures of ∠A, ∠B and ∠C, also find the value of  ∠A+2∠B3∠C.

Solution 69:

We consider ∠A = 2∠B = 6∠C = x

∴∠A = x

2∠B = x = ∠B = 12

6 ∠ C = x ∠C = x6

We know that ∠A + ∠B + ∠C = 1800

[ using angle sum property of a ]

x + x2 + x6 = 1800

6x + 3x + x = 1800  x 6

10x = 10800 x = 1080

x = ∠A  = 1080

Also, ∠B = x2 = 1082 = 540

∠C = x6 = 1086 = 180

Thus, ∠A = 1080 , ∠B = 540, ∠C = 180

Now, ∠A + 2∠B3∠C = 108 + 1083 x 18 = 21654 = 40

Question 70: Students in a school are preparing banners for a rally. The parallel lines I and m are cut by the transversal t. If ∠4 = ∠5 and ∠6 = ∠7, what is the measure of angle 8?

Solution 70:

Here, given, ∠4 = ∠5 and ∠6 = ∠7

Now, I || m and t is the transversal

∴ ∠4 + ∠5 + ∠6 + ∠7 = 180° [Since co-interior angles are supplementary]

∠5 + ∠5 + ∠6 + ∠6 = 180° [using equation (i)]

2(∠5 + ∠6) = 180°

∠5 + ∠6 = 90°

We know, the sum of all the interior angles of a triangle is 180°

∴ ∠8 + ∠5 + ∠6 = 180°

⇒ ∠8 + 90° = 180° [using (ii)]

⇒ ∠8 = 180° – 90° = 90°

Benefits of Solving Important Questions Class 9 Mathematics Chapter 6

The chapter on Lines and Angles is divided into two sections; the first section teaches students about lines, while the second section teaches them about various angles. It’s crucial to master the principles of lines and angles in order to comprehend geometry in classes 9th and 10th.

Below are a few benefits for students to practise from our set of Important Questions Class 9 Mathematics Chapter 6:

• It is a one-stop solution, meaning students will get questions from many different sources. Consequently, they will save time during preparation by gathering the questions and responses in advance.
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