Tag: maths

Questions Related to maths

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

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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

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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

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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. $\dfrac{3}{5}$

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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 $(120^o, 150^o)$

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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. $1:4$

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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 $\dfrac{-(ac^2+ba^2+cb^2)}{(abc)^2}$