CHAPTER-6, LINES AND ANGLES

Explanation: 

145°

Here we have PQ||RS.

The right answer is c 145°

Here we have PQ || RS. Produce PQ to M.

∠CQP=∠MQD[Vertically opp. ∠S]

Therefore,60°=∠1+25°

This implies,∠1=35°

So now, QM || RS and QR will cut CD and AB

∠ARQ=∠RQD=25°[Alt. ∠S]

Therefore,∠1+(∠ARQ+∠RMP)=180°

This implies ∠1+(∠ARQ+∠ARS)=180°

This implies 35°+(25°+∠ARS)=180°

This implies ∠ARS=180°−60°=120°

Therefore ∠QRS=∠ARQ+∠ARS

 =25°+120°=145°.

Explanation:  

(d) A right triangle

In the triangle ABC, 

∠A= ∠B+∠C

So now the sum of the angles=180

∠A+∠B+∠C=180

∠A+∠A=180

Therefore, ∠A=90

Therefore, the triangle ABC is a right-angled triangle.

Explanation: 

The correct answer is 52.5°

From the properties of the triangles: Exterior angle is equal to the sum of the interior opposite angles.

If each interior opposite angles=x

So, 105=x+x,

Therefore, 105=2x

Therefore, x= 52.5°.

Explanation:

 (a) An acute angled triangle

Given, the angles of a triangle are in the ratio 5:3:7

 ∴ 5x+3x+7x=180° 

15x=180°

x=12°

5x=5 ×12=60°

3x=3 ×12=36° 

7x=7 ×12=42°

Therefore, all the angles will be less than 90°

Hence proved, the triangle is an acutely angled triangle.

Explanation:

(d) 155°

Let's consider a triangle ABC so that∠ BAC= 130 °

And bisectors of∠B and ∠C meet at O

In order  to find ∠BOC

Now in triangle ABC,

∠BAC+ ∠ABC+ ∠ ACB=180°                                        

130+∠ABC+ ∠ ACB=180°

∠ABC+∠ACB=50°

1 /2(∠ABC+∠ACB)=25°

∠OBC +∠OCB =25°(OB &OC bisect ∠ ABC & ∠ACB)

So, in triangle OBC, 

∠OBC +∠OCB + ∠BOC=180°

25+∠BOC=180°.   

Therefore, ∠BOC=155°.

Explanation:

 (a) 20°

Here, POQ is a line, therefore POQ=180°

40+4x+3x=180°

This implies, 7x=140°

Therefore, x=20°.

Explanation:

(c ) 60°

In the above figure, if we extend the point OP, so that a new triangle PQT will be formed, and T is the point where OP will cut QR after extending it.

In the above figure:

∠OPQ+∠QPT=180°

This implies, ∠QPT =180°-110°=70°

So, if OP||RS, ∠SRQ, and ∠UTQ will be corresponding in the angles hence,∠ UTQ= ∠SRQ =130°

So we have, ∠UTQ+∠PTQ=180°

This implies, ∠PTQ=180°-130°=50°

In triangle∠ PTQ,

∠PTQ +∠TQP+ ∠ QPT=180°

This implies, 50°+∠ TQP+70°=180°

So, ∠TQP=180°-120°=60°

Therefore ∠TQP=∠PQR=60°.

Explanation: As mentioned above, the angles of the triangle are in the ratio 2:4:3

If we take angles as, 2x,4x and,3x

And the angle sum property,2x+4x+3x=180

9x=180

So, x=20

So that  the angles are 40°,80°,60°

Therefore the smallest angle will be 40°.

Explanation:

 AOC is not a line, because ∠ AOB+ ∠COB=100°+82°=182°, so this is not equal to 180. Similarly, BOF is also not a line.

Explanation:

Two lines will not be parallel as the sum of two equal angles will not be equal to 180°. When each equal angle is equal to 90°, then the lines will become parallel. Hence, the two lines will not be parallel.

Explanation:

Here we have to find the value of x+y for which ABC will be a line. If the sum of 2 adjacent angles is 180°, then a ray stands on the line.

For ABC to become a line,

∠ABD + ∠DBC = 180°

Therefore : x+y =180°

So the value of x+y = 180°.

Explanation:

By the angle sum property we know that the sum of all angles of a triangle is equal to 180°

This implies 60°+60°+60°=180°.

If all the angles are less than 60° then the sum together will not be 180°. For Example, 30°, 40° and 50° are three angles of a triangle therefore 30+40+50=120° <180° therefore this is not a possible scenario.

Explanation:

 A triangle will not have 2 obtuse angles because the sum of the angles of a triangle is 180°, if we assume any 2 obtuse angles and even if we add an acute angle to it we will get the sum of more than 180°.

Explanation:

 Here the given angles are 45°,64° and 72° the sum of the angle = 45°+64°+72°=181° Hence a triangle is not possible.

Explanation:

Here the above-given angles are 53°,64° and 63° So the sum of the angles is 53°+64°+63°=180°. Therefore infinitely many triangles can be formed by using these three angles.

Explanation:

In the above figure, the 2 lines l and m are parallel 

Here n is a transversal line 

So ∠x and 44° will be co-interior angles 

The sum of the co-interior angle is 180°

Therefore ∠x +44°=180°

So, ∠x =136°.

Explanation:

 No, each of these angles will be a right angle only when they form a linear pair i.e. when the noncommon arms of the given 2 adjacent angles are 2 opposite rays and common arms are perpendicular to them.

Explanation: 

The two intersecting lines l and m make one right angle, So l and m are perpendicular to each other, So if we use the linear pair axiom, the other 3 angles will also be right angles.

Explanation:

In figure a the sum of 2 interior angles 132°+48°=180° now we see that the sum of 2 interior angles on the same side of “n” is 180°, then they are the parallel lines.

In figure b the sum of 2 interior angles i.e., 73°+106° =179° which is not equal to 180°

So, the sum of 2 interior angles on the same side of r is not equal to 180°, So they are not parallel lines.

Explanation:

No, Because the lines l and m are perpendicular to line n 

∠1=∠2=90°

This shows that these are corresponding angles 

Therefore l || m 

Explanation:

In the above figure,  ∠BOF =5y 

 So ∠EOA=∠BOF=5y

 And also ∠COE+∠EOA+∠AOD=180°

 Therefore, 2y+5y+2y=180°

 i.e., 9y=180°

 Hence y=180°/9=20°

 Hence the value of y is 20°.

Explanation:

 Here it is given that, x=y and a=b 

 Now we have to prove that l || m 

 x=y shows that the corresponding angles are equal.

 So, the lines l and m are parallel.

 And also a=b shows that the corresponding angles are equal.

 So, the lines m and n are parallel.

  Now we have observed that l || m and m || n 

  Therefore l || m || n

Explanation:

It is given in figure 6.9, OD ⊥ OE(i.e.,∠DOE=90°) and OD and OE are the bisector of ∠AOC and ∠BOC 

The points A, O and B are collinear that is AOB is a straight line 

Since OD and OE bisect the angle ∠AOC and ∠BOC respectively 

Therefore ∠AOC =2∠DOC……………(a)

∠COB = 2∠COE……………………… (b)

By adding (a) and (b) we get that 

∠AOC +∠COB = 2∠DOC+2∠COE

This implies, ∠AOC+∠COB=2(∠DOC+∠COE)

This implies ∠AOC+∠COB=2∠DOE

So ∠AOC+∠COB= 2×90°

[∵ OD ⊥ OE]

So,∠AOC+∠COB = 180°

Therefore ∠AOB = 180°

This proves that ∠AOC+∠COB are for forming linear pairs or AOB is a straight line 

Therefore points A, O and B are collinear.

Explanation:

Here we have ∠5+∠6=180° (linear pair angles) 

This implies, ∠5+120° =180°

This implies ∠5=180°-120°=60°

 Now ∠1=∠5(each = 60°)

 But these are corresponding angles 

 Therefore m and n are parallel.

Explanation:

∵ l||m  and t is the transversal 

∠MAB= ∠SBA[Alt. ∠s]

1/2 ∠MAB=1/2∠SBA

This implies ∠PAB=∠QBA

But ∠PAB and ∠QBA are alternate angles 

Therefore AP || BQ.

Explanation:

The bisectors AB and BQ of the alternate interior angles are parallel. Now we have to show that l and m are parallel

Now AP and BQ are parallel

Here we know that, if a transversal intersects two parallel lines, then the alternate interior angles are equal i.e., ∠CAB and ∠ABF

As AP || BQ and t are transversal 

Therefore ∠PAB=∠ABQ 

Now let us multiply 2 on both sides 

i.e.,2∠PAB=2∠ABQ

As the alternate interior angles are equal, the lines are parallel

So that the l and m are parallel.

Explanation:

Given the lines BA and ED are parallel and also the lines BC and Ef are parallel

 Now we have to show that ∠ABC =∠DEF now extent DE to meet BC at P 

Here EF || BC and DP are the transversal

“If 2 parallel lines are cut by a transversal, then the corresponding angle are equal”

∠DEF=∠CPD………………….(a)

Now AB || DP and BC are the  transversal 

So that the corresponding angles are equal

So ∠CPD=∠ABC……………..(b)

From (a) and  (b) ∠DEF = ∠ABC

Therefore it is proved that ∠DEF = ∠ABC.

Explanation:

 Because BA||ED

∠1 = ∠2 ………..(a)

(Corresponding angles)

Also BC||EF

This implies ∠2+∠3=180° …(b)

The sum of the interior angles on the same sides of a transversal is 180°

From (a) and (b) we get that ∠1+∠3=180°

Therefore ∠ABC+∠DEF=180°

Explanation:

 It is given DE || QR

 Therefore ∠EAB + ∠RBA=180°

 The sum of the interior angles on the same side of the transversal is 180°

 This implies ½∠EAB+½ ∠RBA= 90° ………..(a)

 Now AP and BP are the Bisectors of ∠EAB and ∠RBA respectively 

So ½ ∠EAB = ∠PAB

 And ½ ∠RBA=∠PAB

Because in triangle APB,∠PAB + ∠PBA + ∠APB=180°

This implies {½ ∠EAB+½ ∠RBA}+∠APB=180°

This implies 90°+∠APB=180°

Therefore ∠APB=90°.

Explanation:

 The angle some property of a triangle is 180° Here it is given that the angles are in the ratio of 2:3:4

So now let us consider that the angles are 2x,3x,4x 

If 2x+3x+4x=9x

The angle sum of a triangle is 180°

So 9x=180° or x=180°/9= 20°

Therefore 2x=2x20°=40°

And also 3x=3x20°=60°

And also 4x=4x20°=80°

Therefore the angles of a triangle are 40°,60° and 80°.

Explanation: 

In triangle ABC let's consider ∠ABC = x 

Therefore ∠BCA = 90° - x…………(a)

Now consider triangle BAL if ∠ABL = x(=∠ABC)

Therefore ∠BAL = 90° - x ………..(b)

So from (a) and (b)

It is proved that ∠ACB = ∠BAL

Explanation:

From the above figure let's consider the 2 parallel lines m and n 

Here we know that p is perpendicular to m and q is perpendicular to n so we get that ∠1=∠2=90°

We also know that m || n and p is a transversal from the above figure we know that ∠1 and ∠3 are corresponding angles so ∠1=∠3

We also know that ∠2+∠3=90°

∠2 and ∠3 are corresponding angles when the transversal n cuts p and q  so we get that p || q 

Therefore the 2 lines which are perpendicular to 2 parallel lines are parallel to each other.

Explanation:

Let normals A and B meet at P 

As the mirrors are perpendicular to each other BP is parallel to OA and AP is parallel to OB

Therefore BP perpendicular to PA i.e.,∠BPA=90°

So ∠3+∠2=90°(Angle sum property).......(a)

Also ∠1=∠2 and ∠4=∠3 (Angle of incidence = Angle of reflection)...(b)

Therefore ∠1+∠4=90°(From a and b adding a and b we have ∠1+∠2+∠3+∠4=180° i.e.,

∠CAB+∠DBA=180°

Hence CA||BD.

Explanation:

Let us consider a triangle ABC

Now draw a line l through A, parallel to BC

Now we have to prove that  ∠A+ ∠B+ ∠C= 180°

Now we have to measure A= ∠1 and B= ∠2 and C=v3

Given l is parallel to BC

We know that the alternate angles are equal to parallel lines.

So,  ∠2= ∠4………..(a)

∠ 3=∠5………….(b)

By adding (a )and (b)  ∠2+∠3=∠4+∠5

And by adding ∠1 to both the sides 

∠1+∠2+∠3=∠4+∠5+∠1

From the above figure, the sum of the angles at a point on a line is =180° 

So ∠1+∠2+∠3 = 180°

Therefore ∠A+ ∠B+ ∠C= 180°.

Explanation: 

In the above figure Bisectors of angle B and C of a triangle ABC intersect each other at point O 

Now we have to prove that ∠BOC =90°+½ ∠A 

BO is the bisector of angle B such that ∠OBC = (∠1)

CO is the Bisector of the ∠C such that ∠OCB = (∠2)

By considering the triangle BOC 

By angles sum property ∠OBC+∠BOC+∠OCB=180

∠1+∠BOC+∠2=180° ………………………………(a)

By considering triangle ABC 

By angles and property ∠A+∠B+∠C=180°

∠A+2(∠1)+2(∠2)=180°

Now by dividing 2 into both the sides 

∠A/2 + ∠1+∠2 = 180°/2

∠A/2+∠1+∠2=90°

∠1+∠2=90°-∠A/2………………………………….(b)

By Substituting b in a 

∠BOC+90°-∠A/2=180°

∠BOC-∠A/2=180°-90°

∠BOC-∠A/2=90°

Therefore ∠BOC=90°+∠A/2.

Explanation:

If 2 lines AB and CD intersect each other then they have 2 pairs of opposite angles 

I.e., ∠AOC,∠DOB,∠AOB,∠COB

Now we have to prove that ∠AOC = ∠DOB and ∠AOD = ∠COB

Proof : ∠AOC +∠AOD =180° …….(a)

∠AOD + ∠BOD=180° …………(b)

By taking equations a and b 

∠AOC +∠AOD =∠AOD+∠BOD

i.e.,(∠AOC=∠BOD) and similarly (∠AOD=∠COB) Hence proved 

Explanation:

In Triangle ABC, produce BC to D and the Bisectors of ∠ABC and ∠ACD meet at the point T 

Now we have to prove ∠BTC = 1/2∠BAC

In a triangle ABC ∠C is the exterior angle 

Therefore ∠ACD = ∠ABC+∠CAB

This implies 1/2∠ACD = 1/2∠CAB+1/2∠ABC

This implies ∠TCD=1/2∠CAB+1/2∠ABC  ….. (a)

(Because CT is the bisector of the ∠ACD which implies 1/2∠ACD=∠TCD)

In triangle BTC, ∠TCD= ∠BTC + ∠CBT

This implies ∠TCD =∠BTC+1/2∠ABC…….(b)

(Because BT bisector of ∠ABC which implies ∠CBT = 1/2∠ABC) 

From a and b 1/2∠CAB+1/2∠ABC = ∠BTC+1/2∠ABC

This implies ∠BTC=1/2∠CAB or ∠BTC=1/2∠BAC

Explanation:

The transversal AD intersects at 2 lines PQ and RS at points B and C 

BE is the Bisector of ∠ABQ and CF is the Bisector of ∠BCS

As BE is the Bisector of ∠ABQ then ∠ABE = 1/2∠ABQ and ∠BCF=1/2∠BCS

Since BE and CF are parallel and AD is transversal therefore by corresponding angle axiom ∠ABE=∠BCF

i.e.,1/2∠ABQ=1/2∠BCS

∠ABQ=∠BCS

So by the converse of corresponding angle axiom, PQ is parallel to RS

Explanation:

From the point P, a perpendicular PM is drawn to the given line AB therefore ∠PNB =90°

If possible, we can draw another perpendicular PN to the line AB So ∠PNB=90°

Therefore ∠PNB =∠PNB

This is possible when PM and PN coincide with each other

Therefore through a given point, we can only draw one perpendicular to a given line 

Explanation:

The lines l and m are 2 intersecting lines again n and p the other 2 lines are perpendicular to the intersecting lines that meet at point D  now we have to prove that 2 lines n and p intersect at point D

Proof: 

 Suppose lines n and p are not intersecting then it means they are parallel to each other i.e., n||p ……..(a)

Since n and p are perpendicular to m and l respectively 

But equation a n||p implies that l||m 

So it is a contradiction So our assumption is wrong. Therefore the lines n and p intersect at a point.

Explanation:

The sum of the angle of a triangle is 180° 

Let's assume that only one angle has to be acute i.e. 90°

Then other 2 angles will have to be right or obtuse (90° or >90°)

Let's assume both are 90° then the third angle (X)

Which is an acute angle that will be 180° = 90°+90° not equal to x

x=180°-90°-90°

x=0°

We know that the angle of a triangle cannot be a zero 

Hence our assumption was wrong, So the triangle has at least 2 acute angle

Explanation:

PA is the bisector of the ∠QPR therefore ∠QPA = ∠APR ……(a)

In triangle PQM ∠Q+∠PMQ+∠QPM=180°

(Angles and properties of triangle)

This implies ∠Q+90° + ∠QPM=180°(∠PMQ=90°)

This implies ∠Q=90° -∠QPM…… (b)

In Triangle PMR ∠PMR+∠R+∠RPM=180°

(Angles and properties of triangle)

This implies 90° +∠R +∠RPM =180°(∠PMR=90°)

This implies∠R=180°-90°-∠RPM………(c)

From equation c and b we get ∠Q - ∠R=(90°-∠QPM)-(90°-∠RPM)

This implies∠Q-∠R=∠RPM-∠QPM

This implies ∠Q-∠R(∠RPA+∠APM)-(∠QPA-∠APM).... (d)

This implies ∠Q-∠R=∠APM+∠APM

[Using equation a ]

This implies ∠Q-∠R= 2∠APM

Therefore ∠APM =½(∠Q-∠R).