Exercise 8 B
Question 1
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Sol: Sum of angles of a triangle =$180^{\circ}$
Vertical Angles of an isosceles Triangle $=90^{\circ}$
So sum of its base angle $=180^{\circ}-90^{\circ}=90^{\circ}$
Is basic angles are equal to each other
So Each angle will be $=\frac{90^{\circ}}{2}=45^{\circ}$
Hence each base angle =$45^{\circ}$
Question 2
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Sol: Given : An equilateral $\triangle A B C$
To prove :
(i)Each angle is of $60^{\circ}$
(ii)What is each exterior angle.
Proof : $\triangle A B C$
is AB= AC
So $\angle C=\angle B$
Similarly , AC= CB
So $\angle B=\angle A$
From (i) and (ii)
$\angle A=\angle B=\angle C$
But $\angle A+\angle B+\angle C=180^{\circ}$
So $\angle A=\angle B=\angle C=\frac{180^{\circ}}{3}=60^{\circ}$ Hence proved
is Each equilateral triangle has $60^{\circ}$ each
So every equilateral triangle is equilateral
If $\angle A B C=60^{\circ}$
and $\angle A B C+\angle C B Z=180^{\circ}$
$60^{\circ}+\angle C B Z=180^{\circ} \Rightarrow \angle C B Z=180^{\circ}-60^{\circ}=130^{\circ}$
So Each exterior angle of an equilateral triangle measures $120^{\circ}$
Question 3
Sol: ABC is an isosceles triangle in which AB=AC
BC is produced to both sides to O and E forming
Exterior angles ACD and ABE respectively
( Image to be added)
To prove: $\angle A C D=\angle A B E$
Proof : In $\triangle A B C$
if $A B=A C$
so $\angle C=\angle B$
$\angle A C B=\angle A B C$
But $\angle A B E+\angle A B C=180^{\circ}$ Linear pair .....(i)
Similarly $\angle A C D+\angle A C B=180^{\circ}$............(ii)
From (i) and (iii)
$\angle A B E+\angle A B C=\angle A C D+\angle A C B$
But $\angle A B C = \angle A C B$ (Proved)
$\angle A B E=\angle A C D$ or $\angle A C D=\angle A B E$ Hence proved.
Question 4
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Sol: $\triangle A B C, A D$ is the bisector of $\angle B A C$ Which meets BC at AD, AD =DC and $\angle B D A=70^{\circ}$
if $A D$ is the bisector of $\angle B A C$
$\angle 1=\angle 2$
In $\triangle A D C$
If AD=DC
So $\angle Z=\angle C$
But ext. $\angle A D B=\angle 2+\angle C$
=$70^{\circ}=\angle 2+\angle C=\angle C+\angle C=2 \angle C$
So $\triangle C=\frac{70^{\circ}}{2}=35^{\circ}$
In $\triangle A B D$
$\begin{aligned} & \angle 1+\angle B+\angle A D B=180^{\circ} \\ \Rightarrow & \angle 2+\angle B+70^{\circ}=180^{\circ} \\ \Rightarrow & \angle C+\angle B+70^{\circ}=180^{\circ} \\ \Rightarrow & 35^{\circ}+\angle B+70^{\circ}=180^{\circ} \\ \Rightarrow & \angle B+105^{\circ}=180^{\circ} \\ \Rightarrow & B=180^{\circ}-105^{\circ}=75^{\circ} \\ & \text { Hence } \angle A C D=35^{\circ} \\ & \text { and } \angle A B D=75^{\circ} \end{aligned}$
Question 5
Sol:(i) Vertical angle of an isosceles triangle =x
Then each equal angle = 2x
So x+2x+2x =$180^{\circ}$
$5 x=180^{\circ}$
$x=\frac{180^{\circ}}{5}=36^{\circ}$
So vertical angle = $36^{\circ}$
And each of base angles =$36^{\circ} \times 2=72^{\circ}$
Hence angles are $36^{\circ}, 72^{\circ}, 72^{\circ}$
(ii) Vertical angle of an isosceles triangle =x
Then each base angle = 3x
So x+3x+ 3x = $180^{\circ}$
$7 x=180^{\circ}$
$x=\frac{180^{\circ}}{7}$
So vertical angle $=\frac{180^{\circ}}{7}=25 \frac{5^{\circ}}{7}$
and each base angle $=\frac{180^{\circ}}{7} \times 3=\frac{540^{\circ}}{7}$ $=77 \frac{1}{7}$
So angle are $25 \frac{50}{7} 177 \frac{1}{7}^{\circ}$ and $77 \frac{1}{7}$
Question 6
Sol :(i) If $A B \| C D$ and BC is its transversal
So $\angle y=\angle 1$
In $\triangle E C D$,
Ext. $\angle X=\angle C+\angle D=\angle 1+40^{\circ}$
In $\triangle A C E$
AE= AC
So $\angle A C E=\angle A E C=X$
If $A B \| C D$
$\begin{aligned} \angle B A C+\angle A C B &=180^{\circ} \\ 84^{\circ}+x+\angle 1 &=180^{\circ} \end{aligned}$
$\Rightarrow x+\angle 1=180^{\circ}-84^{\circ}=96^{\circ}$
$\Rightarrow \angle{1}+40^{\circ}+\angle 1=96^{\circ}$
$\Rightarrow 2 \angle 1=96-40=56$
So $\angle 1=\frac{56^{\circ}}{2}=28^{\circ}$
$\Rightarrow \angle y=28^{\circ}$
But x= $=\angle 1+40^{\circ}=28^{\circ}+40^{\circ}=68^{\circ}$
Hence x=$63^{\circ}, y=2 s^{\circ}$
(ii) In the figure in $\triangle A B C$
AB = BC
So $\angle B=\angle A C B=2 x$
But $\angle B A C+\angle A B C+\angle A C B=180^{\circ}$
$\Rightarrow x+2 x+2 x=180^{\circ}$
$\Rightarrow 5 x=180^{\circ}$
$x=\frac{180^{\circ}}{5}=36^{\circ}$
If $A B \| C D$
So
$\begin{aligned} & \angle A C D=\angle B A C \\=& x=36^{\circ} \end{aligned}$
Now in $\triangle A C D$
$\angle C A D+\angle A C D+\angle A D C=180^{\circ}$
$\Rightarrow y+x+88^{\circ}=180^{\circ}$
$\Rightarrow y+36^{\circ}+88^{\circ}=180^{\circ}$
$\Rightarrow y+124^{\circ}=180^{\circ} \Rightarrow y=180^{\circ}-124^{\circ}=56^{\circ}$
Hence $x=36^{\circ}, y=56^{\circ}$
(iii) In the figure,
$A B=B D=D C$
$\angle B D C=180^{\circ}$
In $\triangle A B C$
If $B D=D C$
So $\angle D C B=\angle D B C=x$
But $\angle D C B+\angle D B C+\angle B D C=180^{\circ}$
$\Rightarrow x+x+108^{\circ}=180^{\circ}$
$\Rightarrow 2 x=180^{\circ}-108^{\circ}=72^{\circ}$
So $x=\frac{72^{\circ}}{2}=36^{\circ}$
In $\triangle A B D$,
$B A=B D$
But $\angle B D A=180^{\circ}-108^{\circ}=72$
So $\angle B A D=\angle B D A=72^{\circ}$
and Ext. $\angle B D C=\angle B A D+\angle A B D$
$\begin{aligned}&\Rightarrow 108^{\circ}=72^{\circ}+y^{2} \\&\Rightarrow y=108^{\circ}-72^{\circ}=36\end{aligned}$
Hence $x=36^{\circ}$ and $y=36^{\circ}$
(iv)In $\triangle A B C, A B=A C$ And BC is produced to D such that AC=CD
AD is joined
In $\triangle A C D$
if $A C=C D$
So $\angle C A D=\angle C D A=2 x$
and ext. $\angle A C B=\angle C A D+\angle C D A=2 x+2 x=4 x$
if In $\triangle A B C, A B=A C$
So $\angle A B C=\angle A C B=4 x$
So But $\angle B A C+\angle A B C+\angle A C B=180^{\circ}$
So $4 x+4 x+x=180^{\circ} \Rightarrow 9 x=180^{\circ}$
$\Rightarrow x=\frac{180^{\circ}}{9}=20^{\circ}$
Hence x= $20^{\circ}$
Question 7
Sol: $\angle M=L N$
$\angle F N=110^{\circ}, \angle F=\angle Q=90^{\circ}$
In quad. FLNQ
$\angle F+\angle Q+\angle Q N L+\angle F L N=360^{\circ}$
$\Rightarrow 90^{\circ}+90^{\circ}+\angle Q N L+110^{\circ}=360^{\circ}$
$\Rightarrow 290^{\circ}+\angle Q N L=360^{\circ}$
$\triangle Q N L=360^{\circ}-290^{\circ}=70^{\circ}$
In $\triangle LMN
If LM=LN
(i) So $\angle L M N=\angle L N M$ or $\angle L N Q=70^{\circ}$
(ii)$\angle L M N=\angle M L N=180^{\circ}-\angle L M N+\angle L N M$
$=180^{\circ}-70^{\circ}-70^{\circ}=180^{\circ}-140^{\circ}=40$
Hence $\angle Q N L=70^{\circ}$ and $\angle M L N=40^{\circ}$
Question 8
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Sol: OABC is a sector
$\angle A O B=40^{\circ}$ and $\angle B O C=30^{\circ}$ $A B, B C$ and $C A$ are Joined
If OA=OB = OC
So In $\triangle A O B$
$\angle O A B=\angle O B A$ and $\angle O B C=\angle O C B$
But $\angle A O B=40^{\circ}$ and $\angle B O C=30^{\circ}$
So $\angle O A B=\angle O B A=\frac{180^{\circ}-40^{\circ}}{2}=\frac{140^{\circ}}{2}=70^{\circ}$
and $\angle O B C=\angle O C B=\frac{180^{\circ}-30^{\circ}}{2}=\frac{150^{\circ}}{2}=75^{\circ}$
So $\angle A B C=70^{\circ}+75^{\circ}=145$
Arc AB subtends $\angle A O B$ at the center and $\angle A C B$ at the remaining part of the circle
$\angle A C B=\frac{1}{2} \angle A O B=\frac{1}{2} \times 40^{\circ}=20^{\circ}$
Similarly $\angle C A B=\frac{1}{2} \angle B O C=\frac{1}{2} \times 30^{\circ}=15^{\circ}$
Hence angles of $\triangle A B C$ are $15^{\circ}, 20^{\circ}$ and $145^{\circ}$
Question 9
Sol: In the figure $\triangle A B C$ in which AB= AC, $C E \| B A$
and AC is produced to D $\angle A B C=52^{\circ}$
Now We have to find $\angle D C E$ In $\triangle A B C, \quad A B=A C$
$\angle A C B=\angle A B C=52^{\circ}$
If $\angle A=\angle A C E$
$But \angle A=180^{\circ}-(\angle A B C+\angle A C B)$
$=180^{\circ}-\left(52^{\circ}+52^{\circ}\right)=180^{\circ}-104^{\circ}=76^{\circ}$
So $\angle A C E=\angle A=76^{\circ}$
But $\angle A C E+\angle E C D=180^{\circ}$
$\begin{aligned} \Rightarrow & 76^{\circ}+\angle E C D=180^{\circ} \\ \Rightarrow & \angle E C D=180^{\circ}-76^{\circ}=104^{\circ} \\ & \text { Or } \angle D C E=104^{\circ} \end{aligned}$
Question 10
Sol: In the figure In $\triangle A B C, A B=A C$ and $A D \| B C$ is drawn $\angle D A B=70^{\circ}$
We have to find $\angle B A C$
if $D A \| B C$
So $\angle A B C=\angle D A B$
But in $\triangle A B C, A B=A C$
So $\angle A B C=\angle A C B$
So $\angle A C B=\angle A B C=70^{\circ}$
Now $\angle B A C+\angle A B C+\angle A C B=180^{\circ}$
$\Rightarrow \angle B A C+70^{\circ}+70^{\circ}=180^{\circ} \Rightarrow \angle B A C+140^{\circ}=180^{\circ}$
$\Rightarrow \angle B A C=180^{\circ}-140^{\circ}=40^{\circ}$
Hence $\angle RBA C=40^{\circ}$
Question 11
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Sol: $\angle A C B$ is a right angle
$A C=C D, C D E F$ is a rectangle and $\angle B A C=50^{\circ}$
Join DF
We have to find
(i)$\angle B D E$
(ii)The angle between the diagonals CE and DF or rectangle CDEF
In $\triangle A D C, A C=C D$
So , $\angle C D A=\angle D A C or \angle B A C$
$\angle C D A=50^{\circ}$
(i) Now $\angle C D A+\angle C D E+\angle B D E=180^{\circ}$
$\begin{aligned} & 30^{\circ}+90^{\circ}+\angle B D E=180^{\circ} \\ \Rightarrow & 140^{\circ}+\angle B D E=180^{\circ} \\ \Rightarrow & \angle B D E=180^{\circ}-140^{\circ} \\ \Rightarrow & \angle B D E=40^{\circ} \end{aligned}$
(ii) So $D E=E B$
In $\triangle D A C$,
$\angle A C D=180^{\circ}-(\angle A+\angle A D C)$
$=180^{\circ}-\left(30^{\circ}+30^{\circ}\right)=180^{\circ}-100^{\circ}=80^{\circ}$
So $\angle D C O=\angle A C B-\angle A C D=90^{\circ}-80^{\circ}=10^{\circ}$
In $\triangle O C D$,
If OC=OD
$\angle C O D=180^{\circ}-\left(10^{\circ}+10^{\circ}\right)$
$=180^{\circ}-20^{\circ}=160^{\circ}$
Hence $\angle B DE=40^{\circ}$ and $\angle C O D=160^{\circ}$
Question 12
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Sol: In $\triangle A B C$ Bisector of $\angle B C A$ Meets AB in X . A point y lines on. CX
Such that AX=AY
To prove : $\angle C A Y=\angle A B C$
Proof: In $\triangle A B C$
In CX is the bisector of $\angle C$
$\angle 1=\angle 2$
In $\triangle A X Y$
If AX=AY
So $\angle 3=\angle 4$
In $\triangle B C X$
Ext $\angle 4=\angle 2+\angle 5$
$=\angle 1+\angle 5$................(i)
Similarly in $\triangle C A Y$
Ext. $\angle 3=\angle 1+\angle 6$ .................(ii)
From (i) and ( ii)
If \angle 3=\angle 4
So $\angle 1+\angle 5=\angle 1+\angle 6 \Rightarrow \angle 5=\angle 6$
$\Rightarrow \angle C A Y=\angle A B C$
Hence proved
Question 13
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Sol: In the figure
PS=PR, $\angle T P S=\angle Q P R$
To prove: $P T=P Q$
proof : In $\triangle$ PRS,
If PS=PR
So $\angle P S R=\angle P R S$
But $\angle P S R+\angle P S T=180^{\circ}$
similarly $\angle P R S+\angle P R Q=180^{\circ}$
So $\angle P S R+\angle P S T=\angle P R S+\angle P R T$
But $\angle P S R=\angle P R S$
So $\angle P S T=\angle P R Q$
Now in $triangle PST and $\triangle P R Q$
$P S=P R$
$\angle P S T=\angle P R Q$
$\angle T P S=\angle Q P R$
So $\triangle P S T \cong \triangle P R Q$
So $P T=P Q$ Hence proved
Question 14
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Sol: In an isosceles $\triangle A B C$ , AB =AC, BD and CF are the medians of sides AC and AB respectively
To prove : BD=CE
Proof : In $\triangle B D C$ and $\triangle B E C$
$B C=B C$
$\angle C=\angle B$
$C D=B E$
SO $\triangle B D C \cong \triangle B E C$
SO $B D=C E$ PROVED
Question 15
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Sol: In an isosceles $\triangle A B C$ in which AB=AC, D, E and F are the mid point of the sides BC, CA and AB respectively DE, EF and FD are joined.
To prove : $\triangle$ DEF is an isosceles triangle
Proof: If D and F are the mid points of BC and BA respectively
So $D F \| A C$ and $\frac{1}{2} A C$
Similarly D and E are the mid point of BC and CA respectivley
So $D E \| A B$ and $\frac{1}{2} A B$
But $A B=A C$
So $D E=D F$
So $\triangle D E F$ is an isosceles triangle
Hence proved
Question 16
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Sol: In the figure AD = BC , AC= BD prove that PAB is an isosceles triangle
Given : In the figure AD= BC,AC=BD
To prove: $\triangle P A B$ is an isosceles triangle
Proof : In $\triangle A D B$ and and $\triangle A C B$
$A B=A B$
$B D=A C$
$A D=B C$
So $\triangle A D B \cong \triangle A C B$
So $\angle D=\angle C$
Now in $\triangle A P D$ and $\triangle B P C$
$A D=B C$
$\angle A P D=\angle B P C$
$\angle D=\angle C$
So $\triangle A P D \cong \triangle B P C$
So $A P=B P$
So $\triangle P A B$ is an isosceles triangle (Proved)
Question 17
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Sol: In the figure (i) and (ii) AB=AC, BD =DC
To prove : $\angle A B D=\angle A C D$
Proof :$\triangle A B D$ and $\triangle A C D$
$A B=A C$
$D B=D C$
$A D=A D$
So $\triangle A B D \cong \triangle A C D$
So $\angle A B D=\angle A C D$ Proved
Question 18
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Sol: In $\triangle A B C A B=A C$
BA is produced to D such that BA =AD , DC is joined
To prove : $\angle B C D=1$ right angle
Proof : In $\triangle A B C$
If $A B=A C$
$\angle A B C=\angle A C B$
Similarly in $\triangle A C D$
$\begin{aligned} & A B=A D \\ \Rightarrow & A C=A D \\ \text { So } & \angle 1=\angle 2 \end{aligned}$
Now adding
$\angle A B C+\angle 1=\angle A C B+\angle 2$
$\Rightarrow \angle A B C+\angle 1=\angle B C D$
But $\angle A B C+\angle 1+\angle B C D=180^{\circ}$
So $\angle B C D+\angle B C D=180^{\circ}$. $\Rightarrow 2 \angle B C D=180^{\circ} \Rightarrow \angle B C D=\frac{180^{\circ}}{2}=90^{\circ}$ So $\angle B C D$ is a right angle
Question 19
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Sol: In $\triangle A B C, A B=A C$ Bisector of $\angle B$ and $\angle C$ Meets AC and AB is D and E respectively
To prove: BD= CE
Proof : In $\triangle B D C$ and $\triangle B E C$
$B C=B C$
$\angle C=\angle B$
$\angle D B C=\angle E C B$
So $\triangle B D C \cong \triangle B E C$
So $B D=C E$
Question 20
Sol: In $\triangle A B C$ the bisector of vertical Angle A bisector the side BC at D
So $x_{1}=x_{2}$ and $B C=D C$
To prove : $\triangle A B C$ is an isosceles
i.e. $A B=A C$
Construction : Produce AD to E such that AD= DE
Join EC
Proof: In $\triangle A B D$ and $\triangle E C D$
$A D=D E$
$B D=D C$
$\angle A D B=\angle C D E$
SO $\triangle A B D \cong \triangle E C D$
SO $A B=C E$
and $\angle x_{1}=\angle x_{3}$
But $x_{1}=x_{2}$
So $\angle x_{2}=2 x_{3}$
So $A C=C E$
But $A B=C E$
So $A B=A C$
Hence $\triangle A B C$ is an isosceles triangle
Hence proved
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