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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $tan 60^{0}=\frac{AB}{AD}\Rightarrow \sqrt{3}=\frac{5}{AD}\Rightarrow AD=\frac{5}{\sqrt{3}}cm$

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

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.

  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$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $30^o,75^o,75^o$.

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