Tag: use of properties of parallel lines

Questions Related to use of properties of parallel lines

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

  1. 0

  2. 1

  3. $- 1$

  4. None of the above


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 :

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

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

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

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


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?

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

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

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

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


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

  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 $(120^o, 150^o)$


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 $BC = 8 \,cm$.

  1. $15 \,cm$

  2. $23 \,cm$

  3. $60 \,cm$

  4. $15\sqrt{17} cm$


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 :

  1. $1:1$

  2. $1:2$

  3. $1:3$

  4. $1:4$


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 

  1. $xyz$

  2. $2xyz$

  3. $-xyz$

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


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


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

  1. $2 x + y = 16$

  2. $x + y = 11$

  3. $2 x - y = 4$

  4. $x + y = - 11$


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

  1. $70 ^ { \circ }$

  2. $80 ^ { \circ }$

  3. $90 ^ { \circ }$

  4. none


Correct Option: A