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 __
Sol :
180°

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

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

(iv) In figure (i) , if ∠ACD=130° , ∠ABC=48° , then ∠BAC=__
Sol :
82°

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

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°.T
(ii) Sum of the four angles of a quadrilateral is three right angles.F
(iii) In a triangle, there may be two right angles.F
(iv) In a triangle, exterior angle is smaller than each interior opposite angle.F
(v) In a triangle, there may be two obtuse angles.F
(vi) In a triangle, there may be two acute angles.T
(vii) In a triangle, exterior angle is equal to the sum of interior opposite angles.T

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
Sol :
(i) The sum of two remaining angle
=180°-70°
=110°

(ii) ∠A+∠B+∠C=180°(sum of angles of triangle is 180°)
∠A+80°+40°=180°
∠A+120°=180°
∠A=60°

(iii) ∠A+∠B+∠C=180°
∠A+105°+50°=180°
∠A+155°=180°
∠A=25°

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 ?
Sol :

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 :
Four right angles

Question 9

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

Sol :
∠BAC+∠ABC=∠ACM
∠BAC+45°=100°
∠BAC=55°

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

Sol :
We know that exterior angle is equal to the sum of opposite interior angles.
∠MBC=∠BAC+∠BCA
140°=∠BAC+40°
∠BAC=140°-40°
∠BAC=100°

(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 ?
Sol :
θ>β

(iv) If in ΔABC (figure below) , ∠ACD=105° and ∠BAC=35° , find the value of ∠ABC.
<fig to be added>
Sol :
We know that in triangle exterior angle is equal to the sum of opposite interior angles.
∠ACD=∠BAC+∠ABC
105°=35°+∠ABC
∠ABC=105°-35°
∠ABC=70°

Question 10

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

Sol :
∠A+∠B+∠C=180°
4x+2x+3x=180°
9x=180°
$x=\frac{180}{9}$
x=20°

∠A=4x
=4×20°
=80°

∠B=2x
=2×20°
=40°

∠C=3x
=3×20°
=60°

(b)

Sol:

∠A+∠B+∠C+∠D=360°
3x+4x+x+2x=360°

10x=360°

$x=\frac{360}{10}$

x=36°

∠A=3x

=3×36°

=108°

∠B=4x

=4×36°

=144°

∠C=x=36°

∠D=2x

=2×36°

=72°

(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.
<fig to be added>
Sol :
AM and DM are the bisector of ∠A and ∠D

∠MAD=$\frac{1}{2}$∠A
⇒2∠MAD=∠A
$\angle MDA=\frac{1}{2} \angle D$
⇒2∠MDA=∠D

In quadrilateral ABCD

∠A+∠B+∠C+∠D=360°

2∠MAD+50°+100°+2∠MDA=360°

2(∠MAD+∠MDA)+150°=360°

2(∠MAD+∠MDA)=360°-150°
∠MAD+∠MDA$=\frac{210^{\circ}}{2}$
∠MAD+∠MDA=105°

ΔMDA
∠MAD+∠MDA+∠AMD=180°
105°+∠AMD=180°
∠AMD=75°

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 .
Sol :
(i)
Diagram
∠B=x ,∠C=x,∠A=x+30°
∠A+∠B+∠C=180°
x+30+x+x=180°
3x+30°=180°
3x=150°
$x=\frac{150}{3}$
x=50°

∠A=x+30°
=50°+30°
=80°

∠B=x=50°

∠C=x=50°

(ii)
Diagram
∠BAC+∠ABC=∠ACD
∠BAC+35°=115°
∠BAC=80°
∠ACB+∠ACD=180°(linear pair)
∠ACB+115°=180°
∠ACB=65°

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 .
Sol :

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°

[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
∴$y=\dfrac{1}{2}\angle ADC$ )]