• Angle sum property of triangle - class-VIII
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Angle sum property of triangle - class-VIII

Description: angle sum property of triangle
Number of Questions: 23
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Tags: maths theorems on triangles straight line and angles geometry simple two dimensional shapes triangles triangle and quadrilateral lines and angles parallel lines
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In $\triangle ABC$, if $b\cos A=a\cos B$ then the triangle is 

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. right angled

  2. isosceless

  3. equilateral

  4. scalene

Show answer
Correct Option: B
Explanation:

Given 

In $\triangle ABC,b\cos A=a\cos B$
$2{R}\sin B\cos A=2{R}\sin A \cos B$
$\sin (B-A)=0\implies B=A$

State true or false.
The sum of interior angles of a triangle is ${ 180 }^{ \circ  }$.

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

It is true that the sum of interior angles of a triangle is ${ 180 }^{ \circ  }$.

The angles of a triangle are in the ratio 2: 1: 3. Is the triangle right-angled triangle,

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

The angles of the triangle are in the ratio, 2:1: 3
Let the angles be $2x, x and 3x$
Thus, sum of the angles = 180
$2x + x+ 3x = 180$
$6x = 180$
$x = 30$ 
Hence, the angles will be 30, 60 and 90
Since, one of the angles is 90, the triangle is a right angled triangle.

In a $\triangle ABC$, $\angle A - \angle B = 30^{\circ}$ and $ \angle B -\angle C = 42^{\circ}$; find $\angle A$.

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $84^o$

  2. $94^o$

  3. $32^o$

  4. none of the above

Show answer
Correct Option: B
Explanation:

In $\triangle ABC$,
$\angle A - \angle B = 30$    ....(I)
$\angle B - \angle C = 42$     .....(II)
Also, sum of angles of the triangle $= 180$
$\angle A + \angle B + \angle C = 180$     ....(III)
On subtracting (I) and (II), we get

$\angle A-2\angle B+\angle C=-12^o$     ....(IV)
On subtracting (III) and (IV), we have
$3\angle B=192$
$\angle B=64^o$
From (I), we get
$\angle A=94^o$

If the angles of a triangle are in the ratio 2:3:4, find the three angles.

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $80^o, 120^o, 160^o$

  2. $20^o, 30^o, 40^o$

  3. $40^o, 60^o, 80^o$

  4. None of these

Show answer
Correct Option: C
Explanation:

The angles of a triangle are in the ratio 2:3:4 
Let $x:y:z=2:3:4$
Then $x= 2t; y= 3t; z= 4t$

Sum of all angles of a triangles is $ 180^0$.
$ 2t + 3t + 4t = 180^0 $
$ 9t = 180^0 $
$  t  = 20^0 $
$ x= 2\times 20^0= 40^0; y= 3\times 20^0= 60^0; z = 4\times 20^0= 80^0 $

In a $\triangle ABC$, the sides AB and AC have been produced to D and E. Bisectors of $\angle CBD$ and $\angle BCE$ meet at O. If $\angle A={ 64 }^{ 0 }$, then $\angle BOC$ is 

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. ${ 52 }^{ 0 }$

  2. ${ 58 }^{ 0 }$

  3. ${ 26 }^{ 0 }$

  4. ${ 112 }^{ 0 }$

Show answer
Correct Option: B
Explanation:

Given: OB and OC bisect $ext. \angle B$ and $ext. \angle C$, $\angle A = 64^{\circ}$

Now, In $\triangle OBC$,
Sum of angles = 180
$\angle OBC + \angle OCB + \angle BOC = 180$
$\frac{1}{2} (ext. \angle B + ext. \angle C) + \angle BOC = 180$ (OB and OC bisect exterior angles)
$\frac{1}{2} (180 - \angle ABC + 180 - \angle ACB) + \angle BOC = 180$
$\frac{1}{2} (360 - (\angle ABC + \angle ACB)) + \angle BOC = 180$
$\frac{1}{2} (360 - (180 - \angle A)) + \angle BOC = 180$ (Angle sum property)
$\frac{1}{2} (180 + \angle A) + \angle BOC = 180$
$\angle BOC = 180 - 90 -\frac{1}{2} (64)$
$\angle BOC = 58^{\circ}$

An exterior angle of a triangle is equal to the sum of two ______ opposite angles.

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. interior

  2. exterior

  3. vertical

  4. none of these

Show answer
Correct Option: A
Explanation:

An exterior angle of a triangle is equal to the sum of two interior opposite angles

In $\displaystyle \triangle ABC,\angle C=30^{\circ},\angle B=90^{\circ},BC=10 cm,BD\perp AC$ then the length of AD is

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $\displaystyle \frac{5}{\sqrt{3}}$ cm

  2. $\displaystyle \frac{6}{\sqrt{3}}$ cm

  3. $\displaystyle \frac{7}{\sqrt{3}}$ cm

  4. $\displaystyle \frac{8}{\sqrt{3}}$ cm

Show answer
Correct Option: A
Explanation:

In triangle ABC $\angle C=30^{0}and \angle B =90^{0}$ and BC=10 cm and $BD\perp AC$ 

Then $Sin 30^{0}=\frac{AB}{BC}\Rightarrow \frac{1}{2}=\frac{AB}{10}\Rightarrow AB=5 cm$
And $\angle ABD =60^{0}$
Then $tan 60^{0}=\frac{AB}{AD}\Rightarrow \sqrt{3}=\frac{5}{AD}\Rightarrow AD=\frac{5}{\sqrt{3}}cm$

The interior and boundary of a triangle is called

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. exterior

  2. interior

  3. triangular region

  4. plane

Show answer
Correct Option: C
Explanation:

Remember this..

The interior and boundary of a triangle is called triangular region..

One of the exterior angle of a triangle is $ 105^0$ and the interior opposite angles are in the ratio 2 : 5 . Find the angles of the triangle.

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $ 30^o ; 45^o ; 105^o$ 

  2. $ 45^o ; 45^o ; 90^o$ 

  3. $ 30^o ; 75^o ; 75^o$ 

  4. $ 60^o ; 30^o ; 90^o$ 

Show answer
Correct Option: C
Explanation:

We have the property that, in a triangle exterior angle is equal to the sum of interior opposite angles.


Given, the interior opposite angles to the exterior angle $105^\circ$ are in the ration $2:5$

$\therefore 2x+5x=105^o$

$7x=105^o$ $\implies x=15^o$

Therefore the interior opposite angles to the angle $105^o$ are $2x=2(15)=30^o$ and $5x=5(15)=75^o$

Let the third angle of the triangle be $C$
We have the sum of interior angles of a triangle is $180^o$

$\therefore 30^o+75^o+C=180^o$
$C=180^o-75^o-30^o=180^o-105^o$
$C=75^o$

Hence, the angles are $30^o,75^o,75^o$.

$\Delta ABC$ is a right angled at A, the value of tan B $\times$ tan C is:

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. 0

  2. 1

  3. $- 1$

  4. None of the above

Show answer
Correct Option: B
Explanation:

In $\triangle  ABC$

$\angle A+\angle B+\angle C={ 180 }^{ \circ  }\ \angle B+\angle C={ 180 }^{ \circ  }-{ 90 }^{ \circ  }={ 90 }^{ \circ  }\ \angle C={ 90 }^{ \circ  }-\angle B$
$\tan { B } \times \tan { C } \ =\tan { B } \times \tan { \left( { 90 }^{ \circ  }-\angle B \right)  } \ =\tan { B } \times \cot { B } \ =\tan { B } \times \dfrac { 1 }{ \tan { B }  } \ =1$

In $\Delta ABC$, if $\angle A+\angle B=90^{\circ}$, cot $B=\dfrac{3}{4}$, then the value of tan A is :

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $\dfrac{4}{5}$

  2. $\dfrac{3}{4}$

  3. $\dfrac{4}{3}$

  4. $\dfrac{3}{5}$

Show answer
Correct Option: B
Explanation:

$\angle A+\angle B={ 90 }^{ \circ  }\ \angle B={ 90 }^{ \circ  }-\angle A$

$\cot { B } =\dfrac { 3 }{ 4 } \ \cot { \left( { 90 }^{ \circ  }-\angle A \right)  } =\dfrac { 3 }{ 4 } \ \tan { A } =\dfrac { 3 }{ 4 } $

There are m points on a straight line AB & n points on the line AC none of them being the point A. Triangles are formed with these points as vertices, when (i) A is excluded (ii) A is included.

The ratio of number of triangles in the two cases is?

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $\dfrac{m+n-2}{m+n}$

  2. $\dfrac{m+n-2}{m+n-1}$

  3. $\dfrac{m+n-2}{m+n+2}$

  4. $\dfrac{m(n-1)}{(m+1)(n+1)}$

Show answer
Correct Option: A
Explanation:
Consider triangle without vertex
we can choose $2$ vertices from line $AB$ and one vertex from $A$ the possibilities are 
$\ ^{m}C _{2}\times n$
We can choose $2$ vertices from line $AC$ and one vertex from $AB$ the possibilities are:
$\ ^{n}C _{2}\times m$
As anyone of the above can be done so number of possibilities is 
$\ ^{m}C _{2}\times n+\ ^{n}C _{2}\times m$
Solving 
$\ ^{m}C _{2}\times \ ^{n}C _{2}\times m$
$=\dfrac{m!}{2!(m-2)!}\times n+\dfrac{n!}{2!(n-2)!}\times m$
$=\dfrac{m(m-1)}{2}\times n+\dfrac{n(n-1)}{2}\times m$
$=\dfrac{mn(m+n-2)}{2}$
Consider triangles with vertex $A$
As one vertex is $A$, we can choose one vertex from $AC$ and one from $AB$ the possibilities are 
$l\times m\times n$
$=mn$
Number of triangle is mn(m+n)/2$
Taking the ratio of $1$ and $2$
$\dfrac{mn(n+m-2)}{2}/\dfrac{mn(m+n)}{2}$
$\dfrac{m+n-2}{m+n}$



The position vectors of vertices of $\Delta ABC$ are $(1, -2), (-7, 6)$ and $\left(\dfrac{11}{5}, \dfrac{2}{5}\right)$ respectively. The measure of the interior angle $A$ of the $\Delta ABC$, is

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. acute and lies in $(75^o, 90^o)$

  2. acute and lies in $(60^o, 75^o)$

  3. acute and lies in $(45^o, 60^o)$

  4. obtuse and lies in $(120^o, 150^o)$

Show answer
Correct Option: B

In the above figure, $ABC$ is a right-angled triangle, right angled at $B$ and $AD$ is the external bisector of angle $A$ of triangle $ABC$.

Find $AD$, if $AC= 17 \,cm$ and $BC = 8 \,cm$.

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $15 \,cm$

  2. $23 \,cm$

  3. $60 \,cm$

  4. $15\sqrt{17} cm$

Show answer
Correct Option: D

In a triangle $ABC$, three force of magnitudes $3\overline {AB}\cdot\ 2\overline {AC}$ and $2\overline {CB}$ are acting along the sides $AB,AC$ and $CB$ respectively. If the resultant meets $AC$ at $D$, then the ratio $DC:AD$ will be equal to :

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $1:1$

  2. $1:2$

  3. $1:3$

  4. $1:4$

Show answer
Correct Option: A

In $\Delta ABC$. If $x=\tan\left(\dfrac{B-C}{2}\right)\tan\dfrac{A}{2}, y=\tan\left(\dfrac{C-A}{2}\right)\tan\dfrac{B}{2}, z=\tan\left(\dfrac{A-B}{2}\right)\tan\dfrac{C}{2}$, then $x+y+z$ (in terms of $x,y,z$ only) is 

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $xyz$

  2. $2xyz$

  3. $-xyz$

  4. $\dfrac{1}{2}xyz$

Show answer
Correct Option: A
Explanation:

In $ \triangle ABC$


$\tan\left(\dfrac{B-C}{2}\right)=\dfrac{b-c}{b+c}\cot\dfrac{A}{2}$


$\implies x=\dfrac{b-c}{b+c}\implies\dfrac{-c}{b}$


Similarly $y=\dfrac{-a}{c},z=\dfrac{-b}{a}$

These on adding gives $\dfrac{-(ac^2+ba^2+cb^2)}{(abc)^2}$

Let $ A(1,2,3), B(0,0,1), C(-1,1,1)$ are the vertices of a $\triangle ABC$. Then, the equation of internal angle bisector through A to side BC is 
maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $\underset{r}{\rightarrow}=\widehat{i}+2\widehat{j}+3\widehat{k}+\mu (3\widehat{i}+2\widehat{j}+3\widehat{k})$

  2. $\underset{r}{\rightarrow}=\widehat{i}+2\widehat{j}+3\widehat{k}+\mu (3\widehat{i}+4\widehat{j}+3\widehat{k})$

  3. $\underset{r}{\rightarrow}=\widehat{i}+2\widehat{j}+3\widehat{k}+\mu (3\widehat{i}+3\widehat{j}+2\widehat{k})$

  4. $\underset{r}{\rightarrow}=\widehat{i}+2\widehat{j}+3\widehat{k}+\mu (3\widehat{i}+3\widehat{j}+4\widehat{k})$

Show answer
Correct Option: B

In a  $\triangle A B C,$  side  $A B$  has the equation  $2 x + 3 y = 29$  and the side  $A C$  has the equation  $x + 2 y = 16.$  If the mid point of  $B C$  is  $( 5,6 ) ,$  then the equation of  $B C$  is

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $2 x + y = 16$

  2. $x + y = 11$

  3. $2 x - y = 4$

  4. $x + y = - 11$

Show answer
Correct Option: B
Explanation:

$\cfrac { x _{ 1 }+x _{ 2 } }{ 2 } =5\Rightarrow x _{ 1 }+x _{ 2 }=10.....(1)\quad and\quad y _{ 1 }+y _{ 2 }=12.....(2)$

$Point(x _1,y _1)$ lie on line AC
then
$x _1+2y _1=16...(3)$
Similarly $2x _2+3y _2=29....(4)$
$\Rightarrow 2(x _1+x _2)+4y _1+3y _2=32+29\2\times 10+4y _1+36-3y _1=61\y _1=5\Rightarrow x _1=6\Rightarrow x _2=4\ \Rightarrow y _2=7$
now,
we take these two points and make equation,
$AC= x+y=11$

In triangle, three angles are  $x , x + 10 ^ { \circ } + x + 20 ^ { \circ }$  then the biggest is

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $70 ^ { \circ }$

  2. $80 ^ { \circ }$

  3. $90 ^ { \circ }$

  4. none

Show answer
Correct Option: A

In. triangle ABC,$\angle A$ + $\angle B$ = 144 and$\angle A$ + $\angle C$ = 124.
Calculate smallest angle of the triangle.

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $36^o$

  2. $56^o$

  3. $46^o$

  4. none of these

Show answer
Correct Option: A
Explanation:

$\angle A + \angle B = 144$...(I)
$\angle A + \angle C = 124$...(II)
In triangle ABC,
$\angle A  + \angle B + \angle C = 180 $
Add, I and II,
$\angle A + \angle B + \angle A + \angle C = 144+ 124$
$180 + \angle A = 268 $
$\angle A = 268 - 180 $
$\angle A = 88$
Put this value in (I)
$\angle A + \angle B = 144$
$88 + \angle B = 144$
$\angle B = 56$
Put this value in (II)
$\angle A + \angle C = 124$
$88 + \angle C = 124$
$\angle C = 36$

If every side of a triangle is doubled, then the area of the new triangle is 'K' times the area of the old one. The value of K is

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. 2

  2. 3

  3. $\sqrt 2$

  4. 4

Show answer
Correct Option: D
Explanation:
Let the area of the triangle be $x$.

We know that the area of the triangle
$=\dfrac{1}{2}\times Height \times Base$
$x=\dfrac{1}{2}\times Height \times Base$              $........ (1)$

According to the question,
$Kx=\dfrac{1}{2}\times 2 \times Height \times 2 \times Base$
$Kx=4\times x$
$K=4$

Hence, this is the answer.

The ratio of the areas of two similar triangles is equal to the

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. ratio ofcorresponding medians

  2. ratio ofcorresponding sides

  3. ratio of the squares ofcorresponding sides

  4. none of these

Show answer
Correct Option: C
Explanation:

The ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides.

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