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Triangle And Its Angles · Bihar Board (BSEB) STD 9 (BSEB - English Medium). 15 questions.

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Question 14 Marks
In the given figure, if $\text{AB }||\text{ DE}$ and $\text{BD }||\text{ FG}$ such that $\angle\text{FGH}=125^\circ$ and $\angle\text{B}=55^\circ,$ find $x$ and $y.$
 
Answer
In the given figure, if$\text{AB }||\text{ DE},\text{BD }||\text{ FG},\angle\text{FGH}=125^\circ$ and$\angle\text{B}=55^\circ$
We need to find the value of $x$and $y$

 Here, as $AB \| DE$ and $BD$ is the transversal,
so according to the property, "alternate interior angles are equal",
we get $\angle\text{D}=\angle\text{B}$
$\angle\text{D}=55^\circ\dots(1)$ Similarly, as $BD \| FG$ and $DF$ is the transversal $\angle\text{D}=\angle\text{F}$
$\angle\text{F}=55^\circ(\text{Using 1})$ Further, $EGH$ is a straight line.
So, using the property, angles forming a linear pair are supplementary $\angle\text{FGE}=\angle\text{FGH}=180^\circ$
$\text{y}+125^\circ=180^\circ$
$\text{y}=180^\circ-125^\circ$
$\text{y}=55^\circ$ Also, using the property, "an exterior angle of a triangle is equal to the sum of the two opposite interior angles",
we get, In $\triangle\text{EFG}$ with $\angle\text{FGH}$ as its exterior angle $\text{ext.}\angle\text{FGH}=\angle\text{F}+\angle\text{E}$
$125^\circ=55^\circ+\text{x}$
$\text{x}=125^\circ-55^\circ$
$\text{x}=70^\circ$
Thus, $\text{x}=70^\circ$ and $\text{y}=55^\circ$
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Question 24 Marks
If the bisectors of the base angles of a triangle enclose an angle of $135^\circ $, prove that the triangle is a right angle.
Answer
Given the bisectors of the base angles of a triangle enclose an angle of $135^\circ $

 $\text{i.e}.,\angle\text{BOC}=135^\circ$ But,
We know that $\angle\text{BOC}=90^\circ+\frac{1}{2}\angle\text{A}$
$\Rightarrow135^\circ=90^\circ+\frac{1}{2}\angle\text{A}$
$\Rightarrow\frac{1}{2}\angle\text{A}=135^\circ-90^\circ$
$\Rightarrow\angle\text{A}=45^\circ(2)$
$\Rightarrow\angle\text{A}=90^\circ$
​​​​​​​Therefore, $\triangle\text{ABC}$ is a right angle triangle that is right angled at $A$.
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Question 34 Marks
The bisectors of base angles of a triangle cannot enclose a right angle in any case.
Answer
In $\triangle\text{XYZ},$ Sum of all angles of a triangle is $180^\circ $
$\text{i.e}.,\angle\text{X}+\angle\text{Y}+\angle\text{Z}=180^\circ$

viding both sides by $'2'$ $\Rightarrow\frac{1}{2}\angle\text{X}+\frac{1}{2}\angle\text{Y}+\frac{1}{2}\angle\text{Z}=180^\circ$
$\Rightarrow\frac{1}{2}\angle\text{X}+\angle\text{OYZ}+\angle\text{OYZ}=90^\circ$
$\big[\therefore\text{OY},\text{OZ},\angle\text{Y}\text{ and }\angle\text{Z}\big]$
$\Rightarrow\angle\text{OYZ}+\angle\text{OZY}=90^\circ-\frac{1}{2}\angle\text{X}$
Now in $\triangle\text{YOZ}$
$\therefore\angle\text{YOZ}+\angle\text{OYZ}+\angle\text{OZY}=180^\circ$
$\Rightarrow\angle\text{YOZ}+90^\circ-\frac{1}{2}\angle\text{X}=180^\circ$
$\Rightarrow\angle\text{YOZ}=90^\circ-\frac{1}{2}\angle\text{X}$
Therefore, the bisectors of a base angle cannot enclosure right angle.
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Question 44 Marks
In the given figure, side $BC$ of $\triangle\text{ABC}$ is produced to point $D$ such that bisectors of $\angle\text{ACD}$ meet at a point $E$. If $\angle\text{BAC}=68^\circ,$ find $\angle\text{BEC}.$
Answer
In the given figure, bisectors of $\angle\text{ABC}$ and $\angle\text{ACD}$ meet at E and $\angle\text{BAC}=68^\circ$ We need to find $\angle\text{BEC}$

 Here, using the property: an exterior angle of the triangle is equal to the sum of the opposite interior angles. In $\triangle\text{ABC}$ with $\angle\text{ACD}$ as its exterior angle $\text{ext}.\angle\text{ACD}=\angle\text{A}+\angle\text{ABC}\dots(1)$
Similarly, in $\triangle\text{BE}$ with $\angle\text{ECD}$ as its exterior angle $\text{ext}.\angle\text{ECD}=\angle\text{EBC}+\angle\text{BEC}$
$\frac{1}{2}\angle\text{ACD}=\frac{1}{2}\angle\text{ABC}+\angle\text{BEC}$
$($CE and BE are the bisectors of $\angle\text{ACD}$ and $\angle\text{ABC})$
$\angle\text{BEC}=\frac{1}{2}\angle\text{ACD}-\frac{1}{2}\angle\text{ABC}\dots(2)$
Now, multiplying both sides of (1) by $\frac{1}{2}$
We get, $\frac{1}{2}\angle\text{ACD}=\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{ABC}$
$\frac{1}{2}\angle\text{A}=\frac{1}{2}\angle\text{ACD}-\frac{1}{2}\angle\text{ABC}\dots(3)$
From $(2)$ and $(3)$
we get, $\angle\text{BEC}=\frac{1}{2}\angle\text{A}$
$\angle\text{BEC}=\frac{1}{2}(68^\circ)$
$\angle\text{BEC}=34^\circ$ Thus, $\angle\text{BEC}=34^\circ$
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Question 54 Marks
If the side $BC$ of $\triangle\text{ABC}$ is produced on both sides, then write the difference between the sum of the exterior angles so formed and $\angle\text{A}.$
Answer
In the given problem, we need to find the difference between the sum of the exterior angles and $\angle\text{A}.$
Now, according to the exterior angle theorem $\text{ext}.\angle\text{C}=\angle\text{A}+\angle\text{B}\dots(1)$
Also, $\text{ext}.\angle\text{B}=\angle\text{A}+\angle\text{C}\dots(2)$
Further, adding $(1)$ and $(2)$ $\text{ext}.\angle\text{C}+\text{ext}.\angle\text{B}=\angle\text{A}+\angle\text{B}+\angle\text{A}+\angle\text{C}$
$=2\angle\text{A}+\angle\text{B}+\angle\text{C}\dots(3)$
Also, according to the angle sum property of the triangle,
​​​​​​​we get, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ\dots(4)$
Now, we need to find the difference between the sum of the exterior angles and $\angle\text{A}.$ Thus, $(\text{ext}.\angle\text{C}+\text{ext}.\angle\text{B})-\angle\text{A}=(2\angle\text{A}+\angle\text{B}+\angle\text{C})-\angle\text{A}$
$=\angle\text{A}+\angle\text{B}+\angle\text{C}$
$=180^\circ(\text{Using 4})$ Therefore, $(\text{ext}.\angle\text{C}+\text{ext}.\angle\text{B})-\angle\text{A}=180^\circ$
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Question 64 Marks
If the angles $A, B$ and $C$ of $\triangle\text{ABC}$ satisfy the relation $B − A = C − B$, then find the measure of $\angle\text{B}.$
Answer
In the given $\triangle\text{ABC},$ $\angle\text{A},\angle\text{B}$and $\angle\text{C}$ satisfy the relation $\text{B}-\text{A}=\text{C}-\text{B}$
We need to fine the measure of $\angle\text{B}.$

 As, $\text{B}-\text{A}=\text{C}-\text{B}$
$\text{B}+\text{B}=\text{C}+\text{A}$
$2\text{B}=\text{C}+\text{A}$
$2\text{B}-\text{A}=\text{C}\dots(1)$
Now, using the angle sum property of the triangle,
we get, $\text{A}+\text{B}+\text{C}=180^\circ$
$2\text{B}-\text{A}+\text{A}+\angle\text{B}=180^\circ(\text{Using 1})$
$3\text{B}=180^\circ$
$\text{B}=\frac{180^\circ}{3}$
$=60^\circ$ Therefore, $\angle\text{B}=60^\circ$
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Question 74 Marks
ABC is a triangle in which $\angle\text{A}=72^\circ,$ the internal bisectors of angles $B$ and $C$ meet in $O$. Find the magnitude of $\angle\text{BOC}.$
Answer

Given,
$ABC$ is a triangle where $\angle\text{A}=72^\circ$ and the internal bisector of angles $B$ and $C$ meeting $O.$
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow72^\circ+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{C}=180^\circ-72^\circ$
Dividing bith side by $'2'$
$\Rightarrow\frac{\angle\text{B}}{2}+\frac{\angle\text{C}}{2}=\frac{108^\circ}{2}$
$\Rightarrow\angle\text{OBC}+\angle\text{OCB}=54^\circ$
Now, In $\triangle\text{BOC}\Rightarrow\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^\circ$
$\Rightarrow540^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=180^\circ-54^\circ=126^\circ$
$\therefore\angle\text{BOC}=126^\circ.$
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Question 84 Marks
In the given figure, if $\text{AB }||\text{ CD},\text{EF }||\text{ BC},\angle\text{BAC}=65^\circ$ and $\angle\text{DHF}=35^\circ,$ find $\angle\text{AGH}.$
Answer
In the given figure, if $\text{AB }||\text{ CD},\text{EF }||\text{ BC},\angle\text{BAC}=65^\circ$ and $\angle\text{DHF}=35^\circ$ We need to find $\angle\text{AGH}.$

Here, $GF$ and $CD$ are straight lines intersecting at point $H$,
so using the property, "vertically opposite angles are equal",
we get, $\angle\text{DHF}=\angle\text{GHC}$
$\angle\text{GHC}=35^\circ$ Further, as $AB \| CD$ and $AC$ is the transversal Using the property, "alternate interior angles are equal" $\angle\text{BAC}=\angle\text{ACD}$
$\angle\text{BAC}=65^\circ$ Further applying angle sum property of the triangle In $\triangle\text{GHC}$
$\angle\text{DHF}+\angle\text{HCG}+\angle\text{CGH}=180^\circ$
$\angle\text{CGH}+35^\circ+65^\circ=180^\circ$
$100^\circ+\angle\text{CGH}=180^\circ$
$\angle\text{CGH}=180^\circ-100^\circ$
$\angle\text{CGH}=80^\circ$ Hence, applying the property, "angles forming a linear pair are supplementary" As $AGC$ is a straight line $\angle\text{CGH}+\angle\text{AGH}=180^\circ$
$\angle\text{AGH}+80^\circ=180^\circ$
$\angle\text{AGH}=180^\circ-80^\circ$
$\angle\text{AGH}=100^\circ$
​​​​​​​Therefore,$\angle\text{AGH}=100^\circ$
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Question 94 Marks
In $\triangle\text{ABC},$ if $\angle\text{B}=60^\circ,\angle\text{C}=80^\circ$ and the bisectors of angles $\angle\text{ABC}$ and $\angle\text{ACB}$ meet at a point $O$, then find the measure of $\angle\text{BOC}.$
Answer
In $\triangle\text{ABC},$ if $\angle\text{B}=60^\circ,\angle\text{C}=80^\circ$ and the bisectors of $\angle\text{B}$ and $\angle\text{C}$ meet at $O$.
We need to find the measure of $\angle\text{BOC}$

Since, BO is the bisector of $\angle\text{B}$
$\angle\text{OBC}=\frac{1}{2}\angle\text{B}$
$=\frac{1}{2}(60^\circ)$
$=30^\circ$ Similarly, CO is the bisector of $\angle\text{C}$
$\angle\text{OCB}=\frac{1}{2}\angle\text{C}$
$=\frac{1}{2}(80^\circ)$
$=40^\circ$
Now, applying angle sum property of the triangle, in $ \triangle\text{BOC},$
we get, $\angle\text{OCB}+ \angle\text{OBC}+ \angle\text{BOC}=180^\circ$
$30^\circ+40^\circ+ \angle\text{BOC}=180^\circ$
$\angle\text{BOC}=180^\circ-70^\circ$
$=110^\circ$
​​​​​​​Therefore, $\angle\text{BOC}=110^\circ.$
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Question 104 Marks
In figure $AB $ divides $\angle\text{DAC}$ in the ratio $1 : 3$ and$ AB = DB.$ Determine the value of $x.$ ​​​​​​​
Answer
Let $\angle\text{BAD}=\text{Z},\angle\text{BAC}=3\text{Z}$
$\Rightarrow\angle\text{BDA}=\angle\text{BAD}=\text{Z} $
$(\therefore\text{AB}=\text{DB})$
Now $\angle\text{BAD}+\angle\text{BAC}+108^\circ$ [Linear pair]
$\Rightarrow\text{Z} + 3\text{Z} + 108^\circ =180^\circ $
$\Rightarrow 4\text{Z} = 72^\circ $
$\Rightarrow \text{Z} = 18^\circ $ Now, In $\triangle\text{ADC}$
$\angle\text{ADC}+\angle\text{ACD}=180^\circ$ [Exterior angle property]
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Question 114 Marks
The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.
Answer
In the given problem, the sum of two angles of a triangle is equal to its third angle. We need to find the measure of the third angle.
 Thus, it is given, in $\triangle\text{ABC}$
$\text{A}+\text{B}=\text{C}\dots(\text{i})$
Now, according to the angle sum property of the triangle,
we get, $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{C}+\angle\text{C}=180^\circ (\text{Using i})$
$2\angle\text{C}=180^\circ$
$\angle\text{C}=\frac{180^\circ}{2}$
$\angle\text{C}=90^\circ$
Therefore, the measure of the third angle is $\angle\text{C}=90^\circ$
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Question 124 Marks
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is $10^\circ ,$ find the three angles.
Answer
Given that, The difference between two consecutive angles is $10^{\circ}$
Let $x, x+10^{\circ}, x+20^{\circ}$ be the consecutive angles that differ by $10^{\circ}$
We know that, Sum of all angles in a triangle is $180^{\circ} x+x+10^{\circ}+x+20^{\circ}=180^{\circ} 3 x+30^{\circ}=180^{\circ}$
$\Rightarrow 3x = 180^\circ - 30^\circ$
$\Rightarrow 3x = 150^\circ$
$\Rightarrow x = 50^\circ$
Therefore, the required angles are $x=50^{\circ} x+10^{\circ}=50^{\circ}+10^{\circ}=60^{\circ} x+20^{\circ}=50^{\circ}+20^{\circ}=70^{\circ}$
As the difference between two consecutive angles is $10^{\circ}$,
the three angles are $50^{\circ}, 60^{\circ}, 70^{\circ}$.
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Question 134 Marks
If the angles of a triangle are in the ratio $2 : 1 : 3$, then find the measure of smallest angle.
Answer
In the given problem, angles of $\triangle\text{ABC}$ are in the ratio $2:1:3$
We need to find the measure of the smallest angle.

Let, $\angle\text{A}=2\text{x}$
$\angle\text{B}=\text{x}$
$\angle\text{C}=3\text{x}$ According to the angle sum property of the triangle, in $\triangle\text{ABC},$
we get, $2\text{x}+1\text{x}+3\text{x}=180^\circ$
$6\text{x}=180^\circ$
$\text{x}=\frac{180^\circ}{6}$
$\text{x}=30^\circ$ Thus, $\angle\text{A}=2(30^\circ)=60^\circ$
$\angle\text{B}=1(30^\circ)=30^\circ$
$\angle\text{C}=3(30^\circ)=90^\circ$ Since, the measure of $\angle\text{B}$ is the smallest of all the three angles.
Therefore, the measure of the smallest angle is $30^\circ .$
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Question 144 Marks
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right angle triangle.
Answer
If one angle of a triangle is equal to the sum of the other two angles
$\Rightarrow\angle\text{B}=\angle\text{A}+\angle\text{C}$ In $\triangle\text{ABC},$
Sum of all angles of a triangle is $180^\circ $
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{B}=180^\circ[\angle\text{B}=\angle\text{A}+\angle\text{C}]$
$\Rightarrow2\angle\text{B}=180^\circ$
$\Rightarrow\angle\text{B}=\frac{180^\circ}{2}$
$\Rightarrow\angle\text{B}=90^\circ$
Therefore, $ABC$ is a right angled triangle
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Question 154 Marks
In a triangle $ABC$, if $AB = AC$ and $AB$ is produced to $D$ such that $BD = BC$, find $\angle\text{ACD}:\angle\text{ADC}.$
Answer
In the given $\triangle\text{ABC},\text{AB}=\text{AC}$ and $\text{AB}$ is produced to $D$ such that $BD = BC$
 We need to find $\angle\text{ACD}:\angle\text{ADC}$

Now, using the property, "angles opposite to equal sides are equal" As $AB = AC$
 $\angle6=\angle4\dots({1})$ Similarly, $As AB = AC$
$\angle1=\angle4\dots({2})$
Also, using the property, "an exterior angle of the triangle is equal to the sum of the two opposite interior angle" In $\triangle\text{BDC}$
$\text{ext}.\angle\text{6}=\angle\text{1}+\angle\text{2}$
$\text{ext}.\angle\text{6}=\angle\text{1}+\angle\text{1}(\text{Using 2})$
$\text{ext}.\angle\text{6}=2\angle\text{1}$ From (1), we get $\angle\text{4}=2\angle\text{1}\dots(3)$
Now, we need to find $\angle\text{ACD}:\angle\text{ADC}$
That is, $(\angle\text{4}+\angle\text{2}):\angle\text{1}$
$(2\angle\text{1}+\angle\text{2}):\angle\text{1}(\text{Using 3})$
$(2\angle\text{1}+\angle\text{1}):\angle\text{1}(\text{Using 2})$
$3\angle\text{1}:\angle\text{1}$ Eliminating $\angle1$ from both the sides,
we get $3:1$
Thus, the ratio of $\angle\text{ACD}:\angle\text{ADC} $ is $3 : 1$
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