Mathematics · Quantitative Aptitude

Geometry of Triangles and Angles

846 Questions

Triangle and angle geometry covers angle sums, properties of equilateral shapes, and right triangles. These principles form the basis of advanced quantitative aptitude sections. Practicing spatial problems helps secure points in exams.

Triangle angle sumsPythagorean theoremProperties of equilateral trianglesComplementary and supplementary anglesRight triangle formulas

Geometry of Triangles and Angles Questions

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

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
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
B Correct answer
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.

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

  1. $36^o$
  2. $56^o$
  3. $46^o$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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)
$\angle A + \angle C = 124$
$88 + \angle C = 124$
$\angle C = 36$

Multiple choice mathematical modelling proof by contradiction similar triangles

If a triangle is equiangular, then it is an obtuse angled triangle. Which of the following statements doesn't convey the same meaning as of this mentioned sentence.

  1. A triangle is equiangular only if it is an obtuse angled triangle

  2. If a triangle is not obtuse angled triangle then it is not an equiangular triangle.

  3. Equiangularity is a sufficient condition for triangle to be obtuse angled.

  4. A triangle is only obtuse is obtuse angled if it is equiangular
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider the given statements to be in the form of $p\rightarrow{q}$

Options A, B and C represents $q\rightarrow{p}, \sim{p}\rightarrow\sim{q}$ and $\sim{q}\rightarrow\sim{p}$ respectively.

Hence, option C, which is in the form of $p\rightarrow{q}$ is the correct answer.