Tag: theorems on triangles

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Questions Related to theorems on triangles

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

$\Delta ABC$ 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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In $\triangle  ABC$

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

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

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. $\dfrac{3}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $

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

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. $\dfrac{m(n-1)}{(m+1)(n+1)}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $2$
$\dfrac{mn(n+m-2)}{2}/\dfrac{mn(m+n)}{2}$
$\dfrac{m+n-2}{m+n}$



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

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 $(120^o, 150^o)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let A=(1, -2), B=(-7, 6), C=(11/5, 2/5). Vector AB = (-8, 8), vector AC = (11/5 - 1, 2/5 + 2) = (6/5, 12/5). The dot product AB dot AC = (-8)(6/5) + (8)(12/5) = -48/5 + 96/5 = 48/5. The magnitudes are |AB| = sqrt(64+64) = 8*sqrt(2) and |AC| = sqrt(36/25 + 144/25) = sqrt(180/25) = (6/5)*sqrt(5). Cos(A) = (48/5) / (8*sqrt(2) * (6/5)*sqrt(5)) = 48 / (48*sqrt(10)) = 1/sqrt(10). Since cos(A) is positive and approximately 0.316, A is acute and arccos(0.316) is approximately 71.5 degrees.

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

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. $\dfrac{1}{2}xyz$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\dfrac{-(ac^2+ba^2+cb^2)}{(abc)^2}$

Multiple choice maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle
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. $\underset{r}{\rightarrow}=\widehat{i}+2\widehat{j}+3\widehat{k}+\mu (3\widehat{i}+3\widehat{j}+4\widehat{k})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The internal angle bisector of a triangle divides the opposite side in the ratio of the adjacent sides. The lengths of sides AB and AC can be found to determine the ratio, and then the coordinates of the dividing point on BC are used to find the direction vector of the angle bisector.

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

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. $x + y = - 11$
Reveal answer Fill a bubble to check yourself
B Correct answer
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,
$AC= x+y=11$

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of angles in a triangle is 180 degrees. x + (x + 10) + (x + 20) = 180. 3x + 30 = 180, so 3x = 150, x = 50. The angles are 50, 60, and 70 degrees. The biggest is 70 degrees.