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KC Sinha Mathematics Solution Class 9 Chapter 8 Lines and Angles exercise 8.3

Page 8.43

Exercise 8.3

Type 1

Question 1 

Fill up the blanks:

(i) In a triangle sum of measures of three angles is __

(ii) In a triangle , maximum number of acute angles is __

(iii) In a triangle, minimum number of acute angles is __

(iv) In figure (i) , if ∠ACD=130° , ∠ABC=48° , then ∠BAC=__

(v) If a side of a triangle is extended , then the exterior angle so formed is equal to the __ of the opposite interior angles .


Question 2

Which of the following statements are true (T) and which are false (F):

(i) Sum of the three angles of a triangles is 180°

(ii) Sum of the four angles of a quadrilateral is three right angles.

(iii) In a triangle, there may be two right angles.

(iv) In a triangle, exterior angle is smaller than each interior opposite angle.

(v) In a triangle, there may be two obtuse angles.

(vi) In a triangle, there may be two acute angles.

(vii) In a triangle, exterior angle is equal to the sum of interior opposite angles.

 


Question 3

(i) In a triangle, one angle measures 70°,then write the sum of two remaining angles in degrees

(ii) In a ΔABC , ∠C=40° ,∠B=80° , find the value of ∠A 

(iii) In a ΔABC , ∠B=105° and ∠C=50° , find the value of ∠A 

 


Question 4

Give reasons for your answer in each of the following cases. Can in a triangle there be ?

(i) Two right angles

(ii) Two obtuse angles

(iii) Two acute angles

(iv) Each angle greater than 60°

(v) Each angle smaller than 60°

(vi) Each angle equal to 60°


Question 5

(i) At least how many acute angles are there in a triangle ?

(ii) What Will be the maximum numbers of acute angles in  a triangles ?

(iii) At most how many obtuse angles can there be in a triangle ?

 


Question 6

(i) If α be an angle of a triangle such that 90°<α<180° , then what type of triangle will it be ?

(ii) In a right angled triangle, how many acute angles are there ?


Question 7

(i) In a right angled triangle, one acute angle is 40°  , then find the measure of the other acute angle in degrees.

(ii) A quadrilateral has three angles as 110° , 40° and 50° . Find the value of the fourth angle.


Question 8

How many right angles is the sum of the interior angles of a rhombus ?

Sol :

 


Question 9

(i) In the given f‌igure side BC is produced to M, ∠ACM=100° and ∠ABC=45° , find ∠BAC.

Sol :

 

(ii) In the given f‌igure, if ∠MBC=140° , ∠BCA=40° ,  find ∠BAC.

 

(iii) If θ be the exterior angle of a triangle and sum of the two opposite interior angles, one is β , then what is the relation between θ and β ?

 

(iv) If in ΔABC (figure below) , ∠ACD=105° and ∠BAC=35° , find the value of ∠ABC.

<fig to be added>

 


Question 10

(i) In the following f‌igures, f‌ind the values of each angle.

(a)

 

(b)

 

(ii) In the given f‌igure, AM and DM are the bisectors of ∠A and ∠D respectively of quadrilateral ABCD . Find the value of ∠AMD in degrees.

 


Page 8.45

Question 11

(i) In a triangle, two angles are equal and third angle exceeds each of these angles by 30°. Determine all angles of the triangle

(ii) In a triangle, an exterior angle is 115° and one of the opposite interior angles 35° , then find the remaining angles 

(iii) An angle of a triangle measures 65° , then f‌ind the other two angles if their difference is 25°

(iv) In a right angled triangle, the bigger acute angle is two times the smaller acute angle, then f‌ind the measure of bigger acute angle.

(v) In a triangle, one of the three angles twice the smallest angle and other is three times the smallest angle. Find the angles.

(vi) Sum of two angles of a triangle is 80° and their difference is 20° . Find all angles of the triangle .

 


Question 12

Find the angle formed by the bisectors of two acute of a right-angled triangle.

Sol :

 


Question 13

(i) If the angles of a triangle are in the ratio 2:3:4 , then find the values of the biggest and smallest angles.

(ii) If the angles of a triangle are in the ratio 1:2:3 , then find the value of the greater angle of the triangles.

(iii) If angles of a triangle are in the ratio 2:3:5 , then f‌ind the measure in degree of the smallest angle.

(iv) if angles of a quadrilateral are in the ratio 1:2:3:4 , then find the angles of the quadrilateral


Type 2

Question 14

(i) In figure (i) below , side QR of a ΔPQR is produced to S . If ∠P:∠Q:∠R=3:2:1 and RT⊥PR , find ∠TRS


Sol :

 

 

(ii) In quadrilateral ABCD, AD||BC and bisectors of interior ∠A and ∠B meet at O , find the measure of ∠AOB in degree [fig. (ii)]


Sol :

 

(iii) In figure (iii) , AB||CD , then find the value of α+β+γ.

[Hint : Draw a line OE through point O parallel to AB or CD , then ∠BOE=180°-α [alternate angles]

∠DOE=180°-β [alternate angle]

Now , ∠BOE+∠DOE=γ=180°-α+180°-β

or α+β+γ=360°]

 

(iv) In figure (iv) , l||m , find the value of x

<fig to be added>

[Hint: l||m and they are intersected by a transversal

∴ 60°+y=180°⇒y=180°-60°=120°

∴ x=y+50°=120°+50°=170°]

(v) In figure (v) , ABC is an isosceles triangles whose side AB=AC and XY||BC. If ∠A=30° , then find the value if ∠BXY in degrees.

[Hint: AB=AC [∴ ΔABC is an isoceles triangle]

∴ ∠BXY+∠ABC=180° or ∠BXY + 75° = 180°

∴ ∠BXY=180°-75° = 105° ]

 

(vi) In figure (a)  and  (b) , l||m , then find the value of x 

(a)

 

(b)

 

(vii) In f‌igure (a) and (b) below, f‌ind x and y as required.

(a)

 

(b)

[Hint: In figure (a)

From ΔAEC, x=30°+42°=72°

Also, y=180°-42°=138°

In figure (b) exterior ∠CBD=30°+34°=64°=∠EBD

∴ Exterior angle x=∠EBD+∠EDB=64°+45°=109° ]

 

(viii) In the given f‌igure, sides BC, CA and BA of ΔABC are produced to D, Q and P respectively. If ∠ACD=100° and ∠QAP=35° , then f‌ind all angles of the triangle.

[Hint: ∠BAC=∠QAP=35°

Now , exterior ∠DCA=∠CAB+∠ABC

⇒100°=35°+∠ABC or ∠ABC=100°-35°=65°

Also , ∠ACB=180°-100°=80° ]

 

(ix) In ΔABC , ∠B=45° , ∠C=55° and bisector of ∠A meets BC in D.  Find ∠ADB and ∠ADC

[Hint: ∠A=180°-(45°+55°)=80°

∵ AD is the bisector of ∠A ,


∴ ∠BAD=\angle DAC=\dfrac{80^{\circ}}{2}=40^{\circ}

Now exterior ∠ADB=∠DAC+∠DCA=40°+55°=95°

and exterior ∠ADC=∠ABD+∠BAD=45°+40°=85° ]


Type 3

Question 15

In the given figure , prove that P||m

 


Question 16

Prove that the sum of the exterior angles formed on producing sides of a triangle in the same order is 360° or four right angles .

 


Question 17

Prove that each angle of a triangle will be 60°, if all its angles are of the same measure.

Sol :

 


Question 18

If an angle of a triangle is equal to the sum of remaining two angles, prove that triangle is right angled triangle.

Sol :

 

 


Question 19

In ΔABC,∠A = 40°, if bisectors of ∠B and ∠C meet at point O, then prove ∠BOC=110°

Sol :

[Hint: See the given f‌igure


Since BO and CO are bisectors of ∠ABC

and ∠ACB respectively ,

∴ ∠ABO=∠OBC=∠1 (say)

and ∠ACO=∠OCB=∠2 (say)

Now , in ΔOBC,

∠1+∠2+∠BOC=180°

In ΔABC , ∠ABC+∠ACB+∠BAC=180°..(i)

or 2∠1+2∠2+40°=180°..(ii) [∵ ∠BAC=40°]

∴ ∠1+∠2=\dfrac{180^{\circ}-40^{\circ}}{2}=70^{\circ}

∴ From (1) , 70°+∠BOC=180°

∠BOC=180°-70°=110°]

 


Question 20

In right angled ΔABC, ∠A=90° , and bisectors of ∠B and ∠C meet at O, prove that ∠BOC = 135°

[Hint: Proceed as in question 19

∠1+∠2+∠BOC=180°

Also , ∠A+∠B+∠C=180°

90°+2∠1+2∠2=180°

⇒∠1+∠2=\dfrac{180^{\circ}-90^{\circ}}{2}=45^{\circ}

From (i) , ∠BOC=180°-(∠1+∠2)

=180°-45°

=135°]

 


Question 21

In a triangle ABC, ∠A=90° and AL⊥BC , prove that ∠BAL=∠ACB 

<fig to be added>

[Hint : Draw f‌igure yourself, in ΔABC , ∠ABC ,∠CAB+∠B+∠ACB=180°..(i)

Now, in ΔALB , ∠ALB=90°

∴ ∠ALB+∠B+∠BAL=180°..(ii)

From (i) and (ii) , we get

∠ALB+∠B+∠BAL=∠CAB+∠B+∠ACB

⇒90°+∠B+∠BAL=90°+∠B+∠ACB

⇒∠BAL=∠ACB ]

 


Question 22

If a transversal intersects two parallel lines, prove that bisectors of two interior angles on the same side form 90° with one another.

<fig to be added>

[Hint: AB||CD and they are intersected by transversal EF at M and N (see the given fig)

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

∴ ∠BMN+∠DNM=180°

\dfrac{1}{2}\angle BMN+\dfrac{1}{2}\angle DNM=90^{\circ}

Now MO and NO are bisectors of ∠BMN and ∠DNM respectively.

∴ ∠OMN+∠ONM=90° or ∠1-∠2=90°

Now , ∠1+∠2+∠MON=180°

⇒90°+∠MON=180°

⇒∠MON=90°]

 


Question 23

Prove that the sum of interior angles of a hexagon is 720°

[Hint: We know that the sum of the angles subtended by sides at the centre of a polygon is 360°.

Angles subtended a side of a regular polygon at its centre =\dfrac{360°}{n} where n is the number of sides which is here 6

∵ ∠AOB =\dfrac{360^{\circ}}{6} = 60°; Also AO=BO=AB

∴ ΔAOB is an equilateral triangle.

∴ ΔAOB=∠OBA=∠OAB=60°

Similarly, ΔOAF is also an equilateral triangle.

∴ ∠OAB+∠OAF=60°+60°=120°=∠FAB=an interior angle

∴ Sum of all the interior angles of a hexagon =6×120°=720]

 


Question 24

In a triangle ABC , CD is perpendicular to side AB and BE is perpendicular to side AC. Prove that ∠ABE=∠ACD

 

[Hint: In ΔABE and ΔADC

∠AEB=∠ADC; ∠A=∠A

∴ ∠ABE=∠ACD]

 


Question 25

In ΔABC , internal bisectors of ∠B and ∠C meet at P and external bisectors of these angles meet at Q. Prove that

∠BPC+∠BQC=180°

In ΔABC , internal bisectors of ∠B and ∠C meet at P and external bisectors of these angles meet at Q. Prove that ∠BPC+∠BQC=180°

[Hint: ∠BPC=90°+\dfrac{\angle A}{2} and

∠BQC=90°-\dfrac{\angle A}{2}

∴ ∠BPC+∠BQC

=90°+\dfrac{\angle A}{2}+90°-\dfrac{\angle A}{2}

=180°]


Question 26

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

[Hint: AP and BP are normals to mirrors m and n respectively ]

 

Angle of incidence = Angle of ref‌lection [Law of reflection]

Now in figure, (ii) ∠CAP=∠PAB=x

∠ABP=∠PBD=y

In ΔAPB ,  ∠APB=90°

∴ ∠PAB+∠PBA=90°

∴ x+y=90° or 2(x+y)=2×90°

or 2x+2y=180° or ∠CAB+∠ABD=180°

Since these angles are on the same side of the transversal AB, therefore CA||BD

 


Question 27

In the given f‌igure, PS is bisector of ∠P and PT⊥QR , prove that 

\angle TPS=\dfrac{1}{2}(\angle Q-\angle R)


[Hint: Let ∠QPS=∠RPS=x and ∠TPS=y

In right angled ΔPQT , ∠QPT+∠Q=(x-y)+∠Q=90°..(i)

In right angled ΔPRT , ∠RPT+∠R=90° or

(x+y)+∠R=90°..(iii) [∵ ∠RPT=∠RPS+∠TPS=x+y]

From (i) and (ii) , x-y+∠Q=x+y+∠R

⇒ ∠Q-∠R=2y

∴ y=∠TPS=\dfrac{1}{2}(\angle Q - \angle R)]

 


Question 28

In the given f‌igure, side BC of ΔABC is produced so as to form the ray BD and CE||AB. Without using the theorem related to sum of angles of a triangle, prove that:

∠ACD=∠A+∠B and hence determine ∠A+∠B+∠C=180°.

<fig to be added>

[Hint: ∵ AB||CE and transversal AC meets them

∴ ∠1=∠4 [alternate angles]

and ∠2=∠5 [corresponding angles]

∴ ∠1+∠2=∠4+∠5=∠ACE+∠ECD=∠ACD

∴ ∠A+∠B=∠ACD

Again , ∠A+∠B+∠C=∠ACD+∠C=180° [∵ Line AC stands on ray BCD ∠C+∠ACD=180°]

 


Question 29

In ΔABC  , side BC is produced to D and bisector of ∠A meets BC at L. Prove that ∠ABC+∠ACD=2∠ALC

<fig to be added>

[Hint: ∠BAL=∠CAL=∠1 (say) [∵ AL is bisector of ∠BAC]

In ΔABL , ∠ALC=∠B+∠1

or 2∠ALC=2∠B+2∠1..(i)

In ΔABC, exterior ∠ACD=∠B+∠A=∠B+2∠1

∴ ∠B+∠ACD=∠B+∠B+2∠1=2∠B+2∠1..(ii)

From (i) and (iii) , we get

2∠ALC=∠B+∠ACD=∠ABC+∠ACD ]

 


Question 30

ABCDE is a regular pentagon and bisector of ∠A meets the side CD at point  . Prove that ∠AMC=90°

<fig to be added>

[Hint: We know that a regular pentagon has all its five sides equal. Also angle subtended by any side at the centre O=\dfrac{360^{\circ}}{5}=72^{\circ}

∴ ∠COD=72°

In ΔOCD, OC=OD

∴ ∠OCD=∠ODC=\dfrac{180^{\circ}-72^{\circ}}{2}=54^{\circ}

∴ ∠OCF=180°-54°=126°

Consider ΔAOB and ΔAOE

AB=AE; ∠BAO=∠EAO [∵ AO is bisector of ∠BAE]

AO=AO

∴ ΔAOB≅ΔAOE; BO=OE

Now , AO will bisect ∠COD

∴ ∠COM=∠MOD=\dfrac{72^{\circ}}{2}=36^{\circ}

∴ ∠OMC=∠OCM+∠COM=54°+36°=90°

Hence,  ∠AMC=90°]

 


Question 31

In the given f‌igure, AE bisects ∠CAD and ∠B=∠C . Prove that AE||BC

<fig to be added>

[Hint: AE is bisector of ∠CAD

∴ ∠CAE=∠DAE=∠1 (say)

Now exterior ∠CAD=∠B+∠C⇒2∠1

But ∠B+∠C=2∠C [∵ ∠B=∠C]

∴ 2∠C=2∠1 or ∠C=∠1=∠CAE

∵ These are alternate angles

∴ AE||BC ]

 


Question 32

If arms of an angle are perpendicular to the arms of another angle, then prove that the angles will either be equal or supplementary

[Hint: In figure (i) , ∠BFO=∠EDO=90°

∠BFO=∠EOD [Vertically opposite angles]

∴ ∠BFO+∠BOF=∠EDO+∠EOD

or 180°-∠FBO=180°-∠OED

or ∠FBO=∠OED

i.e., ∠ABC=∠FED

In figure (ii) , ∠B+∠D+∠E+∠F=360°

or ∠B+90°+∠E+90°=360°

or ∠B+∠E=180° ]

 


Question 33

In the given f‌igure, bisectors of ∠B and ∠D meet produced CD and AB at P and Q respectively.

Prove that ∠P+∠Q=\dfrac{1}{2}(\angle ABC+ \angle ADC)


[Hint: Let ∠ABC=2x and ∠ADC=2y, then in quadrilateral PBQD, we have

∠P+∠PDQ+∠Q+∠QBP=360°

or ∠P+(180°-y)+∠Q+(180°-x)=360°

or ∠P+∠Q=x+y=\dfrac{1}{2}(\angle ABC+\angle ADC)

[(∵ 2x=∠ABC ∴ x=\dfrac{1}{2}∠ABC and 2y=∠ADC

y=\dfrac{1}{2}\angle ADC )]


 

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