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 congruence introduction to shapes similarity of triangles introduction to similar triangles

If in $\triangle $s $ABC$ and $DEF,$ $\angle A=\angle E=37^{\circ}, AB:ED=AC:EF$ and $\angle F=69^{\circ},$ then what is the value of $\angle B: ?$

  1. $69^{\circ}$
  2. $74^{\circ}$
  3. $84^{\circ}$
  4. $94^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In $\triangle ABC$ and $\triangle DEF$
$\angle A = \angle E =  37^{o}$
$\dfrac{AB}{ED} = \dfrac{AC}{EF}$
Thus, $\triangle ABC \sim \triangle EDF$ ....... (By SAS rule)
Thus, $\angle B = \angle D$

Now, $\triangle DEF$
$\angle D + \angle E + \angle F = 180$
$\angle D + 37 + 69 = 180$
$\angle D = 74^{\circ}$
Hence, $\angle B = \angle D = 74^{\circ}$

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

Two equilateral triangles with side $4 \ cm$ and $6 \ cm$ are _____ triangles.

  1. similar

  2. congruent

  3. both

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Any two equilateral triangles are similar by SSS criteria..
$SSS$ similarity states that if the lengths of the corresponding sides of two triangles are proportional, then the triangles must be similar.
Two equilateral triangles with side $4 \ cm$ and $6 \ cm$ are similar triangles by $SSS$ similarlty. 

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

$\displaystyle \Delta ABC$ and $\displaystyle \Delta DEF$ are two similar triangles such that $\displaystyle \angle A={ 45 }^{ \circ  },\angle E={ 56 }^{ \circ  }$, then $\displaystyle \angle C$ =___.

  1. $\displaystyle { 56 }^{ \circ }$
  2. $\displaystyle { 45 }^{ \circ }$
  3. $\displaystyle { 101 }^{ \circ }$
  4. $\displaystyle { 79 }^{ \circ }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Delta ABC \sim \Delta DEF$        ...Given

$\Rightarrow \angle A = \angle D$                 ...C.A.S.T.
$\Rightarrow \angle B = \angle E$                 ...C.A.S.T.
$\Rightarrow \angle C = \angle F$                 ...C.A.S.T.
$\therefore \angle B = \angle E = 56^o$
In $\Delta ABC$,
$\angle A + \angle B + \angle C = 180^o$        ....Angle sum property of triangles
$\Rightarrow 45^o+56^o+\angle C = 180^o$
$\Rightarrow \angle C = 79^o$

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

In $\Delta ABC$, DE is || to BC, meeting AB and AC at D and E. If AD = 3 cm, DB = 2 cm and AE = 2.7 cm, then AC is equal to:

  1. $6.5$ cm
  2. $4.5$ cm
  3. $3.5$ cm
  4. $5.5$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
In $\triangle$$ ADE$ and $\triangle$$ ABC$,

$\angle$$ADE=$$\angle$$ABC  $  (corresponding angles)

$\angle$$AED=$$\angle$$ACB$    (corresponding angles)

so,$\triangle$$ ADE$ $\sim$$\triangle$$ ABC$

so, $\dfrac{AD}{AB}=\dfrac{AE}{AC}$ 

$AC=2.7$$\times$$\dfrac{5}{3}$$=$$ 4.5$ cm
Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

Point L, M and N lie on the sides AB, BC and CA of the triangle ABC such that $\ell (AL) : \ell (LB) = \ell (BM) : \ell (MC) = \ell (CN) : \ell (NA) = m : n$, then the areas of the triangles LMN and ABC are in the ratio

  1. $\dfrac{m^2}{n^2}$
  2. $\dfrac{m^2 - mn + n^2}{(m + n)^2}$
  3. $\dfrac{m^2 - n^2}{m^2 + n^2}$
  4. $\dfrac{m^2 + n^2}{(m + n)^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

Three of the exterior angles of a hexagon are $40^{\circ}$, $51^{\circ}$ and $86^{\circ}$. If each of the remaining exterior angles is $x^{\circ}$, find the value of $x$.

  1. $58$
  2. $61$
  3. $65$

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Three of the exterior angles of a hexagon are $ 40^o, 51^o$  and  $86^o $. Each of the remaining exterior angles is $ x^o $.
Sum of all exterior angle of any polygon is $ 360^o $
$ 40^o + 51^o + 86^o + 3 \times x^o = 360^o $
$ => 3 \times x^o = 183^o $
$ => x^o = 61^o $

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If in a $\triangle ABC,{a}^{2}+{b}^{2}+{c}^{2}=8{R}^{2},$ where $R=$ circumradius,then the triangle is

  1. equilateral

  2. isosceles

  3. right angled

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since, ${a}^{2}+{b}^{2}+{c}^{2}=8{R}^{2}$
Using sine rule, we have
${\left(2R\sin{A}\right)}^{2}+{\left(2R\sin{B}\right)}^{2}+{\left(2R\sin{C}\right)}^{2}=8{R}^{2}$
$\Rightarrow 4{R}^{2}\left[{\left(\sin{A}\right)}^{2}+{\left(\sin{B}\right)}^{2}+{\left(\sin{C}\right)}^{2}\right]=8{R}^{2}$
$\Rightarrow {\sin}^{2}A+{\sin}^{2}B+{\sin}^{2}C=2$
$\Rightarrow 1-{\cos}^{2}A+{\sin}^{2}B+1-{\cos}^{2}C=2$
$\Rightarrow -{\cos}^{2}A+{\sin}^{2}B-{\cos}^{2}C=2-2=0$
$\Rightarrow {\cos}^{2}A-{\sin}^{2}B+{\cos}^{2}C=0$
$\Rightarrow \cos{\left(A+B\right)}\cos{\left(A-B\right)}+{\cos}^{2}C=0$
Since, $A+B+C=\pi \Rightarrow A+B=\pi-C$
$\Rightarrow \cos{\left(\pi-C\right)}\cos{\left(A-B\right)}+{\cos}^{2}C=0$
$\Rightarrow  -\cos{C}\cos{\left(A-B\right)}+{\cos}^{2}C=0$
$\Rightarrow  \cos{C}\left[-cos{\left(A-B\right)}+\cos{C}\right]=0$
$\Rightarrow \cos{C}=0, -cos{\left(A-B\right)}+\cos{C}=0$
$\Rightarrow C=\dfrac{\pi}{2}$ or $-cos{\left(A-B\right)}+\cos{\left(\pi-\left(A+B\right)\right)}=0$
$\Rightarrow -cos{\left(A-B\right)}-cos{\left(A+B\right)}=0$
Using transformation angle formula, we have
$\Rightarrow 2\cos{A}\cos{B}=0$
$\therefore \cos{A}=0,\cos{B}=0$
Hence $\angle{A}=\dfrac{\pi}{2} $or $\angle{B}=\dfrac{\pi}{2}, $ or  $\angle{C}=\dfrac{\pi}{2}$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

In $\triangle ABC,$ which of the following statements are true:

  1. maximum value of $\sin{2A}+\sin{2B}+\sin{2C}$ is same as the maximum value of $\sin{A}+\sin{B}+\sin{C}$
  2. $R\ge 2r,$ where $R$ is circumradius and $r$ is the inradius.
  3. ${R}^{2}\ge \dfrac{abc}{\left(a+b+c\right)}$
  4. $\triangle ABC$ is right angled if $r+2R=s,$ where $s$ is semi perimeter.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Option$\left(a\right)$
$\because$ Maximum value of $\sin{2A}+\sin{2B}+\sin{2C}$ and  $\sin{A}+\sin{B}+\sin{C}$ is same that is $3$.
Option$\left(b\right)$
$\because \sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}\le \dfrac{1}{8}$
$\Rightarrow \dfrac{r}{4R}\le \dfrac{1}{8}$
$\Rightarrow R\ge 2r$
Option$\left(c\right)$
$\because \dfrac{abc}{a+b+c}=\dfrac{4R\triangle}{2s}$
                 $=2R.r=R\left(2r\right)\le {R}^{2}$
$\therefore {R}^{2}\ge \dfrac{abc}{a+b+c}$   ($\because R\ge 2r$)
Option$\left(d\right)$
$\angle{B}={90}^{0}$
$\therefore r=\left(s-b\right)\tan{\dfrac{B}{2}}=s-b$
$R=\dfrac{b}{2\sin{B}}=\dfrac{b}{2}$
$\Rightarrow 2R=b$
$\therefore r+2R=s-b+b=s$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

There exist a triangle $ABC$ satisfying

  1. $\tan{A}+\tan{B}+\tan{C}=0$
  2. $\dfrac{\sin{A}}{2}=\dfrac{\sin{B}}{3}=\dfrac{\sin{C}}{7}$
  3. ${\left(a+b\right)}^{2}={c}^{2}+ab$ and $\sqrt{2}\left(\sin{A}+\cos{A}\right)=\sqrt{3}$
  4. $\sin{A}+\sin{B}=\left(\dfrac{\sqrt{3}+1}{2}\right), \cos{A}\cos{B}=\dfrac{\sqrt{3}}{4}=\sin{A}\sin{B}$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

Option$\left(a\right)$ SInce
$\tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}$
But here, $\tan{A}+\tan{B}+\tan{C}=0,$ is impossible.
Option$\left(b\right)$
$\dfrac{\sin{A}}{2}=\dfrac{\sin{B}}{3}=\dfrac{\sin{C}}{7}$
or $\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{7}$
$\Rightarrow \dfrac{a+b}{5}=\dfrac{c}{7}$
$\Rightarrow \dfrac{a+b}{c}=\dfrac{5}{7}<1$
$\therefore a+b<c$ is impossible.
Option$\left(c\right)$
${\left(a+b\right)}^{2}={c}^{2}+ab$
$\Rightarrow {a}^{2}+{b}^{2}+2ab={c}^{2}+ab$
$\Rightarrow {a}^{2}+{b}^{2}-{c}^{2}=-ab$
$\Rightarrow \dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}=\dfrac{-1}{2}$
$\Rightarrow \cos{C}=\dfrac{-1}{2}$
$\therefore \angle{C}={120}^{0}$
and $\sqrt{2}\left(\sin{A}+\cos{A}\right)=\sqrt{3}$
$\Rightarrow \sqrt{2}\left[\sqrt{2}\left(\dfrac{1}{\sqrt{2}}\sin{A}+\dfrac{1}{\sqrt{2}}\cos{A}\right)\right]=\sqrt{3}$
We know that $\sin{\dfrac{\pi}{4}}=\cos{\dfrac{\pi}{4}}=\dfrac{1}{\sqrt{2}}$
$\Rightarrow 2\left[\sin{\left(A+\dfrac{\pi}{4}\right)}\right]=\sqrt{3}$
$\Rightarrow \left[\sin{\left(A+\dfrac{\pi}{4}\right)}\right]=\dfrac{\sqrt{3}}{2}$
$\Rightarrow \left[\sin{\left(A+\dfrac{\pi}{4}\right)}\right]=\sin{\dfrac{\pi}{3}}$
$\Rightarrow A+\dfrac{\pi}{4}=\dfrac{\pi}{3}$
$\therefore A=\dfrac{\pi}{3}-\dfrac{\pi}{4}=\dfrac{\pi}{12}$ is possible.
Option$\left(d\right)$
$\because \sin{A}+\sin{B}=\dfrac{\sqrt{3}+1}{2}$                 ................$\left(1\right)$
and $\cos{A}\cos{B}=\dfrac{\sqrt{3}}{4}=\sin{A}\sin{B}$
$\therefore \cos{A}\cos{B}-\sin{A}\sin{B}=\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}=0$
$\Rightarrow \cos{\left(A+B\right)}=0$
$\Rightarrow A+B=\dfrac{\pi}{2}$
$\therefore B=\dfrac{\pi}{2}-A$
From eqn$\left(1\right)$ 
$\sin{A}+\cos{A}=\dfrac{\sqrt{3}+1}{2}$
$\Rightarrow \sqrt{2}\left[\sin{A}\dfrac{1}{\sqrt{2}}+\cos{A}\dfrac{1}{\sqrt{2}}\right]=\dfrac{\sqrt{3}+1}{2}$
We know that $\sin{\dfrac{\pi}{4}}=\cos{\dfrac{\pi}{4}}=\dfrac{1}{\sqrt{2}}$
$\Rightarrow \sqrt{2}\left[\sin{A}\cos{\dfrac{\pi}{4}}+\cos{A}\sin{\dfrac{\pi}{4}}\right]=\dfrac{\sqrt{3}+1}{2}$
$\Rightarrow \sin{\left(A+\dfrac{\pi}{4}\right)}=\dfrac{\sqrt{3}+1}{2\sqrt{2}}$
$\Rightarrow \sin{\left(A+\dfrac{\pi}{4}\right)}=\sin{\dfrac{5\pi}{12}}$
$\Rightarrow A+\dfrac{\pi}{4}=\dfrac{5\pi}{12}$
$\Rightarrow A=\dfrac{5\pi}{12}-\dfrac{\pi}{4}=\dfrac{\pi}{6}$
$\therefore B=\dfrac{\pi}{2}-A=\dfrac{\pi}{2}-\dfrac{\pi}{6}=\dfrac{2\pi}{6}=\dfrac{\pi}{3}$
Thus, $\angle{C}=\dfrac{\pi}{2}$ is possible. 

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If in a $\triangle ABC, \sin{C}+\cos{C}+\sin{\left(2B+C\right)}-\cos{\left(2B+C\right)}=2\sqrt{2}$, then $\triangle ABC$ is

  1. equilateral

  2. isosceles

  3. right-angled

  4. obtuse angled
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$\because \sin{C}+\cos{C}+\sin{\left(2B+C\right)}-\cos{\left(2B+C\right)}=2\sqrt{2}$
$\Rightarrow \left[\sin{C}+\sin{\left(2B+C\right)}\right]+\left[\cos{C}-\cos{\left(2B+C\right)}\right]=2\sqrt{2}$
$\Rightarrow 2\sin{\left(B+C\right)}\cos{B}+2\sin{\left(B+C\right)}\sin{B}=2\sqrt{2}$
$\Rightarrow \sin{\left(\pi-A\right)}\cos{B}+\sin{\left(\pi-A\right)}\sin{B}=\sqrt{2}$
$\Rightarrow \sin{A}\cos{B}+\sin{A}\sin{B}=\sqrt{2}$
$\Rightarrow \sin{A}\left[\cos{B}+\sin{B}\right]=\sqrt{2}$
Divide both sides by $\sqrt{2}$ we get
$\Rightarrow \sin{A}\left[\dfrac{1}{\sqrt{2}}\cos{B}+\dfrac{1}{\sqrt{2}}\sin{B}\right]=1$
We know that $\sin{\dfrac{\pi}{4}}=\cos{\dfrac{\pi}{4}}=\dfrac{1}{\sqrt{2}}$ we get
$\Rightarrow \sin{A}\left[\sin{\dfrac{\pi}{4}}\cos{B}+\cos{\dfrac{\pi}{4}}\sin{B}\right]=1$
$\Rightarrow \sin{A}\sin{\left(B+\dfrac{\pi}{4}\right)}=1$
$\therefore \sin{A}=1$ and $\sin{\left(B+\dfrac{\pi}{4}\right)}=1$
Hence $A={90}^{0}, \dfrac{\pi}{4}+B=\dfrac{\pi}{2}$
$\Rightarrow B=\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}$ (on simplification)
$\therefore A={90}^{0}, B={45}^{0}, C={45}^{0}$

Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

Two triangles are similar, if their corresponding angles are ________.

  1. Proportional

  2. Equal

  3. A & B

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Two triangles are similar, if their corresponding angles are equal.

(Two triangles are similar, if their corresponding angles are equal and corresponding sides are proportional.)

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The angle between the planes
$\vec{r}(\hat{i}+2\hat{j}+\hat{k})=4$ and $\vec{r}(\hat{-i}+\hat{j}+2\hat{k})=9$

  1. $30^{\mathrm{o}}$
  2. $60^{\mathrm{o}}$
  3. $45^{\mathrm{o}}$
  4. $90^{0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Angle between two planes is the angle between their normal vectors.

For the first plane, normal vector is $\vec{n _0}=(1,2,1)$
For second plane, normal vector is $\vec{n _1}=(-1,1,2)$
Let $\theta$ be the angle between the planes, it is also the angle between their normals.
$\implies \cos \theta = \dfrac{{n} _{1}.{n} _{2}}{|n _1||n _2|} $

$\implies \cos \theta = \dfrac{(1,2,1)\cdot (-1,1,2)}{\sqrt{1^2+2^2+1^2}\sqrt{(-1)^2+1^2+2^2}} $

$\implies \cos \theta = \dfrac{-1+2+2}{6}=\dfrac{1}{2}$
$\implies \theta $ = $ {60}^{o}$
Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The tetrahedron has vertices $0\left ( 0,0,0 \right ),A\left ( 1,2,1 \right ),B\left ( 2,1,3 \right )$ and $C\left ( -1,1,2 \right )$, then  the angle between the faces $OAB$ and $ABC$ will be

  1. $\displaystyle \cos ^{-1}\frac{17}{31}$
  2. $30^{0}$
  3. $90^{0}$
  4. $\displaystyle \cos ^{-1}\frac{19}{35}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Concept using the angle between the  phases is equal to their normals.
$\therefore$ vector $\perp$ to the face $OAB$ is $\overline{OA}\times \overline{OB}=5\hat{i}-\hat{j}-3\hat{k}$
and vector $\perp$ to the face $ABC$ is $\overline{AB}\times \overline{AC}=\hat{i}-5\hat{j}-3\hat{k}$
$\therefore$ Let $\theta$ be the angle between the faces $OAB$ and $ABC$ 
$\displaystyle \therefore \cos \theta =\frac{\left ( 5\hat{i}-\hat{j}-3\hat{k} \right )\left ( \hat{i}-5\hat{j}-3\hat{k} \right )}{\left | 5\hat{i}-\hat{j}-3\hat{k} \right |\left | \hat{i}-5\hat{j}-3\hat{k} \right |}$
$\displaystyle \therefore \cos \theta =\frac{19}{35}$