42 + Interview Questions in LINES-AND-ANGLES IN GENERAL AWARENESS

1.	In the given figure, ?1 and ?2 are supplementary angles. If ?1 is decreased, what changes should take place in ?2 so that both the angles still remainsupplementary.
Answer» Sum of angles=180 `angle1+angle2=180^0` if angle 1 decreases then angle 2 should rise to still remain supplementary.

Discussion

2.	In the given figure, `AB\|\|CD` and `EF\|\|GH`. Find the values of x, y, z and t.
Answer» Correct Answer - `x=60, y=60, z=70, t=70` `x=60` [vert. opp `angles`] ltbr. `angleQPR=(180^(@)-110^(@))=70^(@)=angleBQS=z^(@)` [corresponding `angles`] `x^(@)+angleQRS=110^(@)` [alt. int. `angles`] `implies 60^(@)+angleQRS=110^(@)` `implies angleQRS=50^(@)`. `x^(@)+angleQRS+t^(2)=180^(@)` [consecutive int. `angles`] `implies 110^(@)+t^(@)=180^(@) implies t=70` `y^(@)+angleQRS+t^(@)=180^(@) implies y^(@)+50^(@)+70^(@)=180^(@) implies y=60`.

Discussion

3.	The supplement of an angle is one-third of itself.Determine the angle and its supplement.
Answer» Let the qangle is `theta`. Then, its supplement will be `180-theta`. It is given that, `180-theta = 1/3 theta` `=>540-3theta = theta` `=>4theta = 540` `=>theta = 135` So, angle is `135^@` and its supplement is `45^@.`

Discussion

4.	Find the measure of an angle which is `32^(@)` more than its complement.
Answer» Let the measure of the reauired angle be `x^(@)` Then, measure of its complent =`(90-x)^(@)` `:. x=(90-x)+32^(@)` `rArr 2x=122^(@)` `rArr x=61^(@)` Hence, the measure of the angle is `61^(@)`

Discussion

5.	The supplement of an angle is one third of the given angle. Find the measures of the given angle and its supplement.
Answer» Let the measure of the given angle be `x^(@)`. Then, the measure of its supplement `=(180-x)^(@)`. `:. 180-x=1/3 x implies implies 3(180-x)=x` `implies 540-3x=x` `implies 4x=540 implies x=135`. Hence, the measure of the given angle is `135^(@)` and the measure of its supplement is `(180-135)^(@)=45^(@)`.

Discussion

6.	The supplement of an angle is `10^(@)` more than three times its complement. Find the angle.
Answer» Let the measure of the reauired angle be `x^(@)` `:.` Its supplement `=180=x^(@)-x` and its complement =`90-x^(@)` According to question, `180-x=3(90-x)+10` `rArr180-x=270-3x+10` `rArr 3x0x=270+10-180` `rArr2x=100` `rArrx=50^(@)` Hence, the measure of the angle is `50^(@)`

Discussion

7.	Find the angle whose supplement is four times its complement.
Answer» Correct Answer - `60^(@)` `(180-x)=4(90-x)`. Find x.

Discussion

8.	In the adjoining figure, `x : y : z=5 : 4 : 6`. If XOY is a straight line, find the values of x, y and z.
Answer» Correct Answer - `x=60, y=48, z=72` Let `x=(5t)^(@), y=(4t)^(@) and z=(6t)^(@)`. `x+y+z=180^(@) implies 5t+4t+6t=180 implies 15t=180 implies t=12`. `:. x=(5xx12)^(@)=60^(@), y=(4xx12)^(@)=48^(@) and z=(6xx12)^(@)=72^(@)`.

Discussion

9.	Find the angle whose complement is one third of its supplement.
Answer» Correct Answer - `45^(@)` `(90-x)=1/3 (180-x) implies 270-3x=180-x`. Find x.

Discussion

10.	Find the values of x for which the angles `(2x-5)^(@)` and `(x-10)^(@)` are the complementary angles.
Answer» Correct Answer - `x=35` `(2x-5)+(x-10)=90`. Find x.

Discussion

11.	Two complementary angles are in the ratio `4 : 5`. Find the angles.
Answer» Correct Answer - `40^(@), 50^(@)` `4x+5x=90`. Find x.

Discussion

12.	Which of the following statements is false ?A. Through a given point, only one straight line can be drawn.B. Through two given points, it is possible to draw one and only one straight line.C. Two straight lines can intersect only at one point.D. A line segment can be produced to any desired length.
Answer» Correct Answer - A

Discussion

13.	For what value of x + y in figure will ABC be a line? Justify your answer.
Answer» For ABC to be a line, the sum of the adjacent angles must be `180^(@) i.e., x + y = 180^(@)`.

Discussion

14.	Two complementary angles are such that twice the measure of the one is equal to three times the measure of the other. The largest of the two measures.A. `72^(@)`B. `54^(@)`C. `63^(@)`D. `36^(@)`
Answer» Correct Answer - B

Discussion

15.	Can a triangle have two obtuse angles ? Give reason for your answer.
Answer» No, because it the triangle have two obtuse angles i.e., more the `90^(@)` angle, then the sum of all the three angles of a triangle will not be equal to `180^(@)`.

Discussion

16.	Can a triangle have all angles less than `60^(@)`? Given reason for your answer.
Answer» No, a triangle connot have all angles less then`60^(@)`, because if all angles will be less than `60^(@)`, then their sum will not be equal to `180^(@)` . Hence, it will not be a triangles.

Discussion

17.	In the adjoining figure, AOB is a straight line. Find the value of x.
Answer» Correct Answer - 118

Discussion

18.	An exterior angle of a triangle is `110^(@)` and its two interior opposite angles are equal. Each of these equal angles isA. `70^(@)`B. `55^(@)`C. `35^(@)`D. `27 1/2^(@)`
Answer» Correct Answer - B

Discussion

19.	The angle between the tangents drawn at the points `(5,12)` and `(12,-5)` to the circle `x^2+y^2=169` is:
Answer» `x^2+ y^2=169` Differentiating w.r.t x `2x+ 2ydy/dx=0` `dy/dx= -x/y` `dy/dx at (5,12)=-5/12=m1` `dy/dx at (12,-5)= 12/5=m2` As, `m1*m2=(-5/12)(12/5)=-1`So, Angle between tangents= `pi/2`

Discussion

20.	When the sum of the measure of two angles is `90^(@)`, the angles are called-A. Supplementary anglesB. Complementary anglesC. Adjacent anglesD. none of these
Answer» Correct Answer - A

Discussion

21.	The measure of the angle which is equal to its complement is-A. `30^(@)`B. `60^(@)`C. `45^(@)`D. `90^(@)`
Answer» Correct Answer - D

Discussion

22.	The measure of the complement of `80^(@)` is-A. `10^(@)`B. `100^(@)`C. `36^(@)`D. `20^(@)`
Answer» Correct Answer - A

Discussion

23.	In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find `angleAOC` and `angleBOD`.
Answer» Correct Answer - `x=28, angleAOC=77^(@), angleBOD=48^(@)`

Discussion

24.	Two lines AB and CD intersect at O. If `angleAOC=50^(@)`, find `angleAOD, angleBOD` and `angleBOC`.
Answer» Correct Answer - `angleAOD=130^(@), angleBOC=130^(@), angleBOD=50^(@)`

Discussion

25.	If the angles are in the ratio `5 : 3 : 7,` then the triangle isA. an acute angled triangleB. an obtuse angled triangleC. a right angled triangleD. an isosceles triangle
Answer» Correct Answer - a Given, the ratio of angles of a triangle is `5 : 3 : 7` Let angles of a triangle be `/_A, /_B and /_C.` Then, `" "/_A = 5X,/_B = 3X and /_C = 7x` In `Delta ABC, " " /_A + /_B + /_C = 180^(@) ["since, sum of all angles of a triangle is" 180^(@)]` `:. " " 5x + 3x +7x = 180^(@)` `rArr " " 15x = 180^(@)` `rArr " "x = 180^(@)/15 = 12^(@)` `:. " " /_A =5x = 5xx12^(@) = 60^(@)` `/_B = 3x = 3xx12^(@) = 36^(@)` and `" " /_C = 7x = 7xx12^(@) = 84^(@)` Since, all angles are less then `90^(@)`, hence the triangle is an acute angled triangle .

Discussion

26.	If in a `Delta ABC, /_A=45^@,/_B=75^@,` then
Answer» In `Deltaxyz ` `anglex+ angley+anglez=180^(@)" "` (Angle sum property) or `45^(@)+75^(@)+z=180^(@)` or ` z=60^(@)`

Discussion

27.	The length of the common chord of the two circles `x^2+y^2-4y=0` and `x^2+y^2-8x-4y+11=0` is
Answer» `x^2+(y-2)^2=4 -(1)` `(x-4)^2+(y-2)^2=9 -(2)` from diagram `CC_1=4` area of `/_ACC_1=sqrt((9/2)(5/2)(3/2)(1/2))` =`sqrt(135)/4` area can also be find by=`1/2ADCC_1` so, it can be said that `sqrt(135)/4=1/2AD*4` AD=`sqrt135/8` length of chord=2AD=`sqrt135/4`

Discussion

28.	In the given figure, `angleXYZ=56^(@)` and XY is produced to a point P. If ray YQ bisects `angleZYP`, find `angleXYQ` and reflex `angleQYP`.
Answer» Since XY is produced to point P, it follows that XP is a straight line and ray YZ stands on it. `:. angleXYZ+angleZYP=180^(@)` `implies 56^(@)+angleZYP=180^(@)` `implies angleZYP=(180^(@)-56^(@))=124^(@)` `implies angleZTQ=1/2 angleZTP=(1/2xx124^(@))=62^(@)`. `:. angleXYQ=angleXYZ+angleZYQ=(56^(@)+62^(@))=118^(@)`. Also, `angleQYP=1/2 angleZYP=(1/2xx124^(@))=62^(@)` `:.` reflex `angleQYP=360^(@)-angleQYP=(360-62)^(@)=298^(@)`.

Discussion

29.	Bisectors of interior `/_`B and exterior `/_`ACD of a `Delta`ABC intersect at the point T. prove that `/_BTC = 1/2 /_`BAC.
Answer» Given In `Delta` ABC, produce BC to DC and the bisectors of `/_`ABC and `/_`ACD meet at point T. To prove `/_`BTC = 1/2 `/_`BAC Proof In `Delta`ABC, `/_`C is an exterior angle. `:." "/_ACD = /_ABC + /_CAB `[exterior angle of a triangle is equal to the sum of two opposite angles] `rArr" "1/2/_ACD = 1/2/_CAB + 1/2/_ABC " "`[dividing both sides by 2] `rArr" "/_TCD= 1/2/_CAB + 1/2/_ABC " "`....(i) [`:. CT is a bisector of /_ACD rArr1/2 /_ACD = /_`TCD ] In `DeltaBTC, " " /_TCD = /_BTC + C/_`CBT [exterior angles of a equal to the sum of two opposite interior angles] `rArr" "/_TCD = /_BTC + 1/2/_ABC " "` ....(ii) `[:.BT bisects of /_ABC rArr /_CBT = 1/2 /_`ABC] From Eqs. (i) and (ii) , 1/2`/_CAB + 1/2/_ABC = /_BTC + 1/2/_`ABC `rArr" " /_BTC = 1/2 /_`CAB `or" " /_BTC = 1/2/_`ABC

Discussion

30.	In Figure, ray `O S`stand on a line `P O Qdot`Ray `O R`and ray `O T`are angle bisectors of `/_P O S a n d /_S O Q`respectively. If `/_P O S=x ,`find `/_R O T`
Answer» `/_POS+/_SOQ=180^0-(1)` `/_POS and /_SOQ` is linear pair of angles. `/_POR=/_ROS /_SOT+/_TOQ=/_SOQ` `/_SOT=/_TOQ` `x+/_SOQ=180^0` `/_SOQ=180^0-x-(2)` `2/_SOT=180^0-x` `/_SOT=(180^0-x)/2`-(3) `/_ROS=/_(POS)/2` `/_ROS=x/2` `/_ROT=/_SOT+/_ROS` `(180-x)/2+x/2` =`90^0`.

Discussion

31.	In the figure, DE`\|""\|` QR and BP are bisectors of `/_`EAB and `/_`RBA, respectively. Find `/_`APB.
Answer» Given, DE`\|\|` OR and AP and PB are the bisectors of `/_`EAB and `/_`RBA, respectively. We know that. Interior angles on the same side of transversal are supplementary. `:." " /_EAB + /_RBA = 180^(@)` `rArr" " 1/2/_EAB +1/2/_RBA = 180^(@)/2" " ` [dividing both sides by 2] `rArr" " 1/2/_EAB +1/2/_RBA = 90^(@)" "`...(i) Since, AP and BP are the bisectors of `/_EAB and /_`RBA, respectively. `:. " " /_BAP = 1/2 /_`EAB ....(ii) and `" " /_ABP = 1/2 /_RBA " "` .........(iii) On adding Eqs. (ii) and (iii), we get `/_BAP + /_ABP = 1/2/_EAB +1/2/_`RBA From Eqs. (i), `rArr" "/_BAP + /_ABP = 90^(@) " "....(iv)` In `DeltaAPB, " "/_BAP + /_ABP + /_APB = 180^(@) " "["sum of all angles of a triangle is" 180^(@)]` `rArr" "90^(@)+/_APB = 180^(@) " "["from Eq".(iv)]` `rArr" "/_APB = 180^(@)-90^(@)=90^(@)`

Discussion

32.	In the given figure, AB is a mirror, PQ is the incident ray and QR is the relfected ray. If `anglePQR=108^(@)` then `angleAQP=?`A. `72^(@)`B. `18^(@)`C. `36^(@)`D. `54^(@)`
Answer» Correct Answer - C Angle of incidence = angle of reflection `=x^(@)` (say). Draw QM as normal to AB at Q.

Discussion

33.	In figure POQ is a line. Raw Oris perpendicular to line PQ .OS is another ray lying between rays OP and OR. Prove that `angleROS(1)/(2) (angleQOS-anglePOS)i.e., angle1=(1)/(2)(angle2-angle3)`
Answer» Since, Oris perpendicular to PQ `:. angle1+angle3=angle4` Now, `angle1=angle2-angle4 " " ("each" 90^(@)) .....(1)` `rArrangle1=angle2-(angle1+angle3)" "` (from the figure) `rArr angle1= angle2-angle1-angle3 " "` [ from (1)] `rArr 2angle1=angle2-angle3` `rArrangle1=(1)/(2)(angle2-angle3)`

Discussion

34.	In the given figure, `AB\|\|CD, angleA=90^(@)` and `angleAEC=40^(@)`. Find `angleECD`.
Answer» Through C draw `CF\|\|AE`, cutting AB at G. Now, `CF\|\|AE` and EC is the transversal. `:. angleECG=angleAEC=40^(@)` [alternate int. `angles`]. Now, `CF\|\|AE` and AGB is a trasversal. `:. angleFGB=angleEAG=90^(@)` [corresponding `angles`] Again, `AB\|\|CD` and GC is the transversal. `:. angleGCD=angleFGB=90^(@)` [corresponding `angles`] `:. angleECD=angleECG+angleGCD=(40^(@)+90^(@))=130^(@)`. Hence, `angleECD=130^(@)`.

Discussion

35.	In the given figure, m and n are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.
Answer» Let the normals at A and B meet at P. Since angle of incidence = angle of reflection, we have `angle1=angle2 and angle4=angle3` As the mirrors are perpendicular to each other, we have `BP bot PA`, we have `angleAPB=90^(@)`. `:. angle2+angle3=90^(@)` `:. angle1+angle4=angle2+angle3=90^(@)`. So, `angle1+angle2+angle3+angle4=180^(@)`. `:. angleCAB+angleABD=180^(@)`. Hence, `AC\|\|BD`.

Discussion

36.	In the given figure, `AB\|\|CD` and `angleCDP=35^(@)`. PD is produced downwards to meet AB at E and `angleBEF=75^(@)`. If `DQ\|\|EF`, find `angleAED, angleDEF` and `anglePDQ`.
Answer» `AB\|\|CD` and PDF is the transversal. `:. angleAED=angleCDP=35^(@)` [corresponding `angles`]. Now, AB is a straight line and ray PDE stands on it. `:. angleAED+angleDEF+angleBEF=180^(@)` `implies 35^(@)+angleDEF+75^(@)=180^(@)` `implies angleDEF=(180^(@)-110^(@))=70^(@)`. Now, `DQ\|\|EF` and PDE is the transversal. `:. anglePDQ=angleDEF=70^(@)` [corresponding `angles`]. Thus, `angleAED=35^(@), angleDEF=70^(@)` and `anglePDQ=70^(@)`.

Discussion

37.	In the figure, find the value of x for which the lines l and m are parallel.
Answer» In the given figure,l`\|"\|` m and we know that, it a transveral intersects two parallel lines, then sum of interior angles on the same side of a transversal is supplementary . `:. " " x + 44^(@) = 180^(@)` `rArr " " x = 180^(@)- 44^(@)` `rArr " " x = 136^(@)`

Discussion

38.	In the given figure, `angleABC=80^(@)` and `angleDEF=45^(@)`. The arms DE and EF of `angleDEF` cut BC at P and Q respectively, Prove that `PD\|\|BA`.
Answer» Let `angleEPQ=x^(@)` We known that the sum of the angles of a triangle is `180^(@)`. `:. angleEPQ+anglePEQ+angleEQP=180^(@)` `implies angleEPQ+45^(@)+35^(@)=180^(@)` `implies angleEPQ=(180^(@)-80^(@))=100^(@)`. Now, EPD is a straight line. `:. angleEPQ+angleDPQ=180^(@) implies 100^(@)+angleDPQ=180^(@)` `implies angleDPQ=(180^(@)-100^(@))=80^(@)`. Thus, `angleABP=angleDPQ` [each equal to `80^(@)`] But, these are corresponding angles. Hence, `PD\|\|BA`.

Discussion

39.	In the given figure, `AB\|\|CD` and `BC\|\|ED`. Find the value of x.
Answer» Correct Answer - `x=105` `angleACD+angleCDE=180^(@) implies angleBCD+75^(@)=180^(@) implies angleBCD=105^(@)`. `angleABC=angleBCD` [alternate interior `angles`] `implies x^(@)=105^(@)`.

Discussion

40.	In the figure, `/_1 =60^(@)` and `/_6 =120^(@)` Show that the lines m and n are parallel.
Answer» Given IN the figure `/_1 =60^(@)` and `/_6 =120^(@)` To show `" " m\|""\|n` Proof Since,` " " /_1 =60^(@) and /_6 =120^(@)` Here,` " "/_1 = /_3 " " `[vertically opppsite angles] `:. " "/_3 = /_1 = 60^(@)` Now ` " "/_3 + /_6 = 60^(@) +120^(@)` `rArr " "/_3 + /_6 = 180^(@)` We konw that, if the sum of two interior angles on same side of l is `180^(@)`, then lines are parallel. Hence, m`\|""\|`n

Discussion

41.	In the adjoining figure, identify(i) the pairs of corresponding angles.(ii) the pairs of alternate interior angles.(iii) the pairs of interior angles on the sameside of the transversal.(iv) the vertically opposite angles.
Answer» 1)`angle4,angle8;angle1,angle5;angle3,angle7;angle2,angle6` 2)`angle3,angle5;angle2,angle8` 3)`angle3,angle8;angle2,angle5` 4)`angle1,angle3;angle2,angle4;angle5,angle7;angle6,angle8`

Discussion

42.	In the given figure, `AB\|\|CD` and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of `angleAEF` and `angleEFD` respectively, prove that `EP\|\|FQ`.
Answer» `AB\|\|CD` and t is the transversal. `:. angleAEF=angleEFD` [alt. int. `angles`] `implies 1/2 angleAEF=1/2 angleEFD` `implies angleFEP=angleEFQ` But these are alternate interior angles. Hence, `EP\|\|FQ`.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

In the given figure, ?1 and ?2 are supplementary angles. If ?1 is decreased, what changes should take place in ?2 so that both the angles still remainsupplementary.

In the given figure, `AB||CD` and `EF||GH`. Find the values of x, y, z and t.

The supplement of an angle is one-third of itself.Determine the angle and its supplement.

Find the measure of an angle which is `32^(@)` more than its complement.

The supplement of an angle is one third of the given angle. Find the measures of the given angle and its supplement.

The supplement of an angle is `10^(@)` more than three times its complement. Find the angle.

Find the angle whose supplement is four times its complement.

In the adjoining figure, `x : y : z=5 : 4 : 6`. If XOY is a straight line, find the values of x, y and z.

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

Find the values of x for which the angles `(2x-5)^(@)` and `(x-10)^(@)` are the complementary angles.

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

Which of the following statements is false ?A. Through a given point, only one straight line can be drawn.B. Through two given points, it is possible to draw one and only one straight line.C. Two straight lines can intersect only at one point.D. A line segment can be produced to any desired length.

For what value of x + y in figure will ABC be a line? Justify your answer.

Two complementary angles are such that twice the measure of the one is equal to three times the measure of the other. The largest of the two measures.A. `72^(@)`B. `54^(@)`C. `63^(@)`D. `36^(@)`

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

Can a triangle have all angles less than `60^(@)`? Given reason for your answer.

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

An exterior angle of a triangle is `110^(@)` and its two interior opposite angles are equal. Each of these equal angles isA. `70^(@)`B. `55^(@)`C. `35^(@)`D. `27 1/2^(@)`

The angle between the tangents drawn at the points `(5,12)` and `(12,-5)` to the circle `x^2+y^2=169` is:

When the sum of the measure of two angles is `90^(@)`, the angles are called-A. Supplementary anglesB. Complementary anglesC. Adjacent anglesD. none of these

The measure of the angle which is equal to its complement is-A. `30^(@)`B. `60^(@)`C. `45^(@)`D. `90^(@)`

The measure of the complement of `80^(@)` is-A. `10^(@)`B. `100^(@)`C. `36^(@)`D. `20^(@)`

In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find `angleAOC` and `angleBOD`.

Two lines AB and CD intersect at O. If `angleAOC=50^(@)`, find `angleAOD, angleBOD` and `angleBOC`.

If the angles are in the ratio `5 : 3 : 7,` then the triangle isA. an acute angled triangleB. an obtuse angled triangleC. a right angled triangleD. an isosceles triangle

If in a `Delta ABC, /_A=45^@,/_B=75^@,` then

The length of the common chord of the two circles `x^2+y^2-4y=0` and `x^2+y^2-8x-4y+11=0` is

In the given figure, `angleXYZ=56^(@)` and XY is produced to a point P. If ray YQ bisects `angleZYP`, find `angleXYQ` and reflex `angleQYP`.

Bisectors of interior `/_`B and exterior `/_`ACD of a `Delta`ABC intersect at the point T. prove that `/_BTC = 1/2 /_`BAC.

In Figure, ray `O S`stand on a line `P O Qdot`Ray `O R`and ray `O T`are angle bisectors of `/_P O S a n d /_S O Q`respectively. If `/_P O S=x ,`find `/_R O T`

In the figure, DE`|""|` QR and BP are bisectors of `/_`EAB and `/_`RBA, respectively. Find `/_`APB.

In the given figure, AB is a mirror, PQ is the incident ray and QR is the relfected ray. If `anglePQR=108^(@)` then `angleAQP=?`A. `72^(@)`B. `18^(@)`C. `36^(@)`D. `54^(@)`

In figure POQ is a line. Raw Oris perpendicular to line PQ .OS is another ray lying between rays OP and OR. Prove that `angleROS(1)/(2) (angleQOS-anglePOS)i.e., angle1=(1)/(2)(angle2-angle3)`

In the given figure, `AB||CD, angleA=90^(@)` and `angleAEC=40^(@)`. Find `angleECD`.

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

In the given figure, `AB||CD` and `angleCDP=35^(@)`. PD is produced downwards to meet AB at E and `angleBEF=75^(@)`. If `DQ||EF`, find `angleAED, angleDEF` and `anglePDQ`.

In the figure, find the value of x for which the lines l and m are parallel.

In the given figure, `angleABC=80^(@)` and `angleDEF=45^(@)`. The arms DE and EF of `angleDEF` cut BC at P and Q respectively, Prove that `PD||BA`.

In the given figure, `AB||CD` and `BC||ED`. Find the value of x.

In the figure, `/_1 =60^(@)` and `/_6 =120^(@)` Show that the lines m and n are parallel.

In the adjoining figure, identify(i) the pairs of corresponding angles.(ii) the pairs of alternate interior angles.(iii) the pairs of interior angles on the sameside of the transversal.(iv) the vertically opposite angles.

In the given figure, `AB||CD` and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of `angleAEF` and `angleEFD` respectively, prove that `EP||FQ`.