Lines and Angles: Important Question

Very Short Answer Questions [1 Marks Question]

Ques. Measurement of reflex angle is

1. 90°
2. between 0º and 90º
3. between 90º and 180º
4. between 180º and 360º

Ans: We already know that the reflex angle lie between 180º to 360º . Therefore (d) between 180º and 360º

Read more: Reflex Angle

Ques. The sum of angle of a triangle is

1. 0º
2. 90º
3. 180º
4. none of these

Ans: (c) Since we already are aware of the property of the triangle i.e, sum of the angle of a triangle is 180º so therefore (c) 180º

Also Check: Properties of a Triangle

Ques. If two lines intersect each other then

1. Vertically opposite angles are equal
2. Corresponding angle are equal
3. Alternate interior angle are equal
4. None of these

Ans: (a) Vertically opposite angles are equal

Ques. The measure of Complementary angle of 63º is

1. 30º
2. 36º
3. 27º
4. none of these

Ans: (c) 27º

Ques. If two angles of a triangle is 30º and 45º what is measure of third angle

1. 95º
2. 90º
3. 60º
4. 105º

Ans: (d) 105º

Ques. The measurement of Complete angle is

1. 0º
2. 90º
3. 180º
4. 360º

Ans: (d) 360º

Ques. The measurement of sum of linear pair is

1. 180º
2. 90º
3. 270º
4. 360º

Ans: (a) 180º

Ques. The difference of two complementary angles is 40º. The angles are

1. 65º, 35º
2. 70º, 30º
3. 25º, 65º
4. 70º, 110º

Ans: (c) 25º, 65º

Ques. If an angle is half of its complementary angle, then find its other angle.

Ans: Let the required angle assumed to be x

∴ Its complement = 90° – x

Now, according to given statement, we obtain

x = 1/2 (90° – x)

⇒ 2x = 90° – x

⇒ 3x = 90°

⇒ x = 30°

Hence, the required angle is 30º

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

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

⇒ ∠PSR = ∠SPQ = 56°

Also, ∠TRS + m + ∠TSR = 180°

14° + m + 56° = 180°

⇒ m = 180° – 14 – 56 = 110°

Short Answer Questions [2 Marks Question]

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

Ans:

AB and CD are assumed to be two intersecting lines intersecting themselves at O.

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

∴ ∠1 = ∠2 and ∠3 = ∠4 …(i)

Now, ∠AOC = ∠BOD [vertically opposite angle ∠s] ……(ii)

⇒ ∠1 + ∠AOC + ∠3 = ∠2 + ∠BOD + ∠4 [adding (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° [using (i), (ii)]

⇒ ∠1 + ∠AOC + ∠3 = 180°

or ∠2 + ∠BOD + ∠4 = 180°

Hence, OP and OQ are in the same line.

Ques. In figure, OP bisects ∠BOC and OQ bisects ∠AOC. Prove that ∠POQ = 90°

Ans: We know that OP bisects ∠BOC

∴ ∠BOP = ∠POC = x (say)

Also, OQ bisects. ∠AOC

∠AOQ = ∠COQ = y (say) .

and the 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°

Ques. In given figure, AB || CD and EF || DG, find ∠GDH, ∠AED and ∠DEF.

Ans: Since AB || CD and HE is a transversal.

∴ ∠AED = ∠CDH = 40° [corresponding ∠s]

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

40° + CDEF + 45° = 180°

∠DEF = 180° – 45 – 40 = 95°

Again, given that EF || DG and HE is a transversal. .

∴ ∠GDH = ∠DEF = 95° [corresponding ∠s]

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

Ques. In figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA respectively. Find ∠APB.

Ans:

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

⇒ ∠1 = ∠2 and ∠3 = ∠4

Since DE || QR and 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 (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°

Ques. 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 ∠ROS = 1/2 (∠QOS – ∠POS).

Ans:

Since given 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 = 1/2 (∠QOS – ∠POS).

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

Ans:

Let normals at A and B meet at P.

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

So, BP ⊥ PA i.e., ∠BPA = 90°

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

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

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

Adding (i) and (ii), we have

∠1 + ∠2 + ∠3 + ∠4 = 180°

i.e., ∠CAB + ∠DBA = 180°

Hence, CA || BD

Ques. if x + y = w + z, then prove AOB is a line.

Ans: Since we have to prove that AOB is a line.

According to the given question,we know that x+y=w+z.

and we already know that the sum of all the angles around a fixed point is 360º

Therefore, we can easily determine that

∠AOC+∠BOC+∠AOD+∠BOD=360º, or

y+x+z+w=360º

But, it's already given that x+y=w+z (According to the question).

Therefore, 2(y+x)=360º

y+x=180º

we can determine that y and x form a linear pair.

we are also aware that if a ray stands on a straight line, then the sum of all the angles of linear pair formed by the ray with respect to the line is 180º

So, y+x=180º

Hence, we can determine that AOB is a line.

Long Answer Questions [3 Marks Question]

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

Ans:

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

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.

Therefore ∠1 = ½ ∠BPQ, ∠2 = ½ ∠CQP,

∠3 = ½ ∠DQP and ∠4 = ½ ∠APQ

Now, AB || CD and EF is a transversal

Therefore ∠BPQ = ∠CQP

⇒ ∠1 = ∠2 (Since ∠1 = ½ ∠BPQ and ∠2 = ½ ∠CQP)

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

Therefore PS || QR

Similarly, we can prove ∠3 = ∠4 = QS || PR

Therefore PRQS is a parallelogram.

Further ∠1 + ∠3 = ½ ∠BPQ + ½ ∠DQP = ½ (∠BPQ + ∠DQP)

= ½ × 180° = 90° (∠BPQ + ∠DQP = 180°)

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

Thus, PRRS is a parallelogram whose one angle ∠PSQ = 90°.

Hence, PRQS is a rectangle.

Ques. If in triangle ABC, the bisectors of ∠B and ∠C intersect each other at O. Prove that ∠BOC = 90° + ½ ∠A.

Ans: Let ∠B = 2x and ∠C = 2y

Therefore OB and OC bisect ∠B and ∠C respectively.

∠OBC =½ ∠B = ½ × 2x = x

and ∠OCB = ½ C = ½ × 2y = y

Now, in Triangle BOC, we have

∠BOC + ∠OBC + ∠OCB = 180°

⇒ ∠BOC + x + y = 180°

⇒ ∠BOC = 180° – (x + y)

Now, in Triangle ABC, we have

∠A + 2B + C = 180°

⇒ ∠A + 2x + 2y = 180°

⇒ 2(x + y) = ½ (180° – ∠A)

⇒ x + y = 90° –½ ∠A …..(ii)

From (i) and (ii), we have

∠BOC = 180° – (90° – ½ ∠A) = 90° + ½ ∠A

Ques. In figure, if l || m and ∠1 = (2x + y)°, ∠4 = (x + 2y)° and ∠6 = (3y + 20)°. Find ∠7 and ∠8.

Ans:

Here, ∠1 and ∠4 are forming a linear pair

∠1 + ∠4 = 180°

(2x + y)° + (x + 2y)° = 180°

3(x + y)° = 180°

x + y = 60

Since I || m and n is a transversal

∠4 = ∠6

(x + 2y)° = (3y + 20)°

x – y = 20

Adding (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 opposite ∠s]

∠7 = ∠3 = 100° [corresponding ∠s]

Hence, ∠7 = 100° and ∠8 = 80°

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

Ans:

Here, PQ || SR .

⇒ ∠PQR = ∠QRT

⇒ x + 28° = 65°

⇒ x = 65° – 28° = 37°

Now we see that in the triangle SPQ, ∠P = 90°

∴ ∠P + x + y = 180° [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°

Ques. In figure, PS is bisector of ∠QPR ; PT ⊥ RQ and Q > R. Show that ∠TPS = ½ (∠Q – ∠R).

Ans:

We know PS is the bisector of ∠QPR

Therefore let ∠QPS = ∠RPS = x 

In Triangle 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

In Triangle ΡQT, 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 …(ii)

Subtracting (i) from (ii), we have

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

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

⇒ 2∠TPS = 2Q – ∠R

⇒ ∠TPS = ½ (Q – ∠R)

Very Long Answer Questions [5 Marks Question]

Ques. The side AB and AC of Triangle ABC are produced to point E and D respectively. If bisector BO And CO of ∠CBE and ∠BCD respectively meet at point O, then prove that ∠BOC = 90º − ½ ∠BAC

Ans: Ray BO bisects ∠CBE

So, ∠CBO = ½ ∠CBE

= ½ (180º − y) (Because, ∠CBE + y = 180º)

= 90º − y/2……(1)

Similarly, ray CO bisects ∠BCD

∠BCO = ½ ∠BCD

= ½ (180º − z)

= 90º − z/2……(2)

In Triangle BOC,

∠BOC + ∠BCO + ∠CBO=180º

∠BOC = ½ (y+z)

But x + y + z = 180º

y + z = 180º − x

∠BOC = ½ (180º − x) = 90º − x/2

∠BOC = 90º − ½ ∠BAC

Hence proved.

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

Ans:

In the question, it is given that (OR ⊥ PQ) and ∠POQ = 180º

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

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

∴ ∠POS + ∠ROS = 90°

Now, ∠QOS = ∠ROQ + ∠ROS

Since It is given that ∠ROQ = 90°,

∴ ∠QOS = 90° + ∠ROS

Or, ∠QOS – ∠ROS = 90°

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

∠POS + ∠ROS = ∠QOS – ∠ROS

=> 2 ∠ROS + ∠POS = ∠QOS

Or, ∠ROS = ½ (∠QOS – ∠POS) ( proved).

Ques: It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Ans:

Given that, XP is a straight line

So, ∠XYZ +∠ZYP = 180°

so by substituting the value of ∠XYZ = 64° we will get,

64° +∠ZYP = 180°

∴ ∠ZYP = 116°

From the above diagram, we also get to know that ∠ZYP = ∠ZYQ + ∠QYP

since, YQ bisects ∠ZYP,

∠ZYQ = ∠QYP

Or, ∠ZYP = 2∠ZYQ

∴ ∠ZYQ = ∠QYP = 58°

Again, ∠XYQ = ∠XYZ + ∠ZYQ

So by substituting the value of ∠XYZ = 64° and ∠ZYQ = 58° we get.

∠XYQ = 64° + 58°

Or, ∠XYQ = 122°

Now, reflex angle of ∠QYP = 180° + ∠XYQ

Therefore we calculate that the value of ∠XYQ = 122°. So,

∠QYP = 180° + 122°

∴ ∠QYP = 302°

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