M.C.Q

MCQ 1511 Mark

The measurement of Complete angle is:

A
$180^\circ$
B
$0^\circ$
✓
$360^\circ$
D
$90^\circ$

Answer

Correct option: C.

$360^\circ$

$x + y = 180^\circ $ (Linear Pair)
$w + z = 180^\circ $ (Linear Pair)
On adding above equations
We get, $x + y + w + z = 180^\circ + 180^\circ = 360^\circ $
Sum of all the angles around a point is $360^\circ .$

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MCQ 1521 Mark

In figure, $AB \| CD \| EF$ and $GH \| KL.$ The measure of $\angle\text{HKL},$ is:

A
$85^\circ$
B
$135^\circ$
✓
$145^\circ$
D
$215^\circ$

Answer

Correct option: C.

$145^\circ$

$GH \| KL$
$\Rightarrow\ \angle\text{GHK}=\angle\text{HKL}$ [Interior opposite angles]
Now,
$\angle\text{GHK}=\angle\text{GHD}+\angle\text{DHR}$
$=(180^\circ-\angle\text{GHC})+\angle\text{DHK}$
[$\angle\text{GHC}$ and $\angle\text{GHD}$ are supplementary]
$=180^\circ-60^\circ+25^\circ$
$\Rightarrow\ \angle\text{GHK}=145^\circ$
$\Rightarrow\ \angle\text{HKL}=\angle\text{GHK}=145^\circ$

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MCQ 1531 Mark

In fig if $x = 30^\circ $ then $y =$

A
$210^\circ$
B
$180^\circ$
C
$90^\circ$
✓
$150^\circ$

Answer

Correct option: D.

$150^\circ$

$x + y = 180^\circ $ (linear pair)
$x = 30^\circ$
$30^\circ + y = 180^\circ$
$y = 180^\circ - 30^\circ$
$y = 150^\circ .$

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MCQ 1541 Mark

In the adjoining figure, $\text{AB}\parallel\text{CD}$ and $\text{AB}\parallel\text{EF}$ The value of $x$ is:

A
$70^\circ$
B
$40^\circ$
✓
$60^\circ$
D
$50^\circ$

Answer

Correct option: C.

$60^\circ$

$\angle\text{FEC}+\angle\text{ECD}=180^\circ$ (sum of 2 supplimentary angles is 180°)
$\angle\text{FEC}+\angle180^\circ-150^\circ=30^\circ$
$\angle\text{X}=\angle\text{BCE}=\angle\text{ECD}$
$\angle\text{X}=30^\circ+30^\circ=60^\circ$

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MCQ 1551 Mark

Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is:

A
$45^\circ$
B
$30^\circ$
✓
$36^\circ$
D
None of these

Answer

Correct option: C.

$36^\circ$

Let one angle be $\theta$
Then, its complementary $=90-\theta$
According to question,
$2\theta=3(90-\theta)$
$=5\theta=270$
$\theta=54^\circ$
Then, $90-\theta^\circ=36^\circ$
Hence, the smaller angle is $36^\circ .$

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MCQ 1561 Mark

In the given figure, $\angle\text{OEB}=75^\circ,\angle\text{OBE}=55^\circ$ and $\angle\text{OCD}=100^\circ.$ Then $\angle\text{ODC}=?$

A
$25^\circ$
✓
$30^\circ$
C
$35^\circ$
D
$20^\circ$

Answer

Correct option: B.

$30^\circ$

In $\triangle\text{OEB}$
$\angle\text{OEB}+\angle\text{EBO}+\angle\text{BOE}=180^\circ$ (Angle sum property)
$75^\circ+55^\circ+\angle\text{BOE}=180^\circ$
$\angle\text{BOE}=50^\circ$
$\angle\text{BOE}=\angle\text{COD}=50^\circ$ (Vertically opposite angle)
In $\triangle\text{ODC}$
$\angle\text{ODC}+\angle\text{DOC}+\angle\text{DCO}=180^\circ$
$\angle\text{ODC}=180^\circ-100^\circ-50^\circ$
$\angle\text{ODC}=30^\circ.$

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MCQ 1571 Mark

An exterior angle of a triangle is $105^\circ $ and its two interior opposite angles are equal. Each of these equal angles is:

A
$37\frac{1}{2}^\circ$
✓
$52\frac{1}{2}^\circ$
C
$75^\circ$
D
$72\frac{1}{2}^\circ$

Answer

Correct option: B.

$52\frac{1}{2}^\circ$

Let one of interior angle be $x^\circ $
Sum of two opposite interior angles = Exterior angle
$\therefore x^\circ + x^\circ = 105^\circ $ [ $\therefore$ Exterior angle $= 105^\circ$ (given)]
$\Rightarrow2\text{x}^\circ=105^\circ$
$\therefore\text{x}^\circ=\frac{105^\circ}{2}$
$\Rightarrow\text{x}^\circ=52\frac{1}{2}$
Hence, each angle of a triangle is $52\frac{1}{2}^\circ.$

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MCQ 1581 Mark

In the given figure $L1 \| L2$ and $\angle1=520.$ Find the measure of $\angle2:$

A
$38^\circ$
B
$48^\circ$
✓
$128^\circ$
D
$52^\circ$

Answer

Correct option: C.

$128^\circ$

$\angle1+\angle\text{a}=180^\circ$
$\angle1=52^\circ$
$52^\circ+\angle\text{a}=180^\circ$
$\angle\text{a}=180^\circ-52^\circ$
$\angle\text{a}=128^\circ$
$\angle\text{a}=\angle2$ (Corresponding angles)
Therefore $\angle2=128^\circ.$

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MCQ 1591 Mark

An exterior angle of a triangle is $110^\circ $ and its two interior opposite angles are equal. Each of these equal angles is:

A
$70^\circ$
B
$35^\circ$
✓
$55^\circ$
D
$27\frac{1}{2}^\circ$

Answer

Correct option: C.

$55^\circ$

Let the measure of each of the two equal interior opposite angles of the triangle be $x.$
In a triangle, the exterior angle is equal to the sum of the two interior opposite angles.
$\therefore$ $x + x = 110^\circ$
$\Rightarrow 2x = 110^\circ$
$\Rightarrow x = 55^\circ$
Thus, the measure of each of these equal angles is $55^\circ .$

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MCQ 1601 Mark

In Fig., if lines $l$ and $m$ are parallel lines, then $x =$

A
$70^{\circ}$
✓
$40^{\circ}$
C
$100^{\circ}$
D
$30^{\circ}$

Answer

Correct option: B.

$40^{\circ}$

Given that,
$l \| m$
Let, $l \| m$ and transversal cuts them and
$\angle1=70^\circ$
$\angle3=20^\circ$
$\angle4=30^\circ$
$\angle1+\angle2=180^\circ$ (Interior angle)
$\angle2=110^\circ\text{(i})$
$\angle2=\angle5$ (Vertically opposite angle)
$\angle5=110^\circ\text{(ii)}$
$\angle5+\angle3+\angle4=180^\circ$(Sum of angles of a triangle is $180^{\circ}$)
$110^{\circ} + x + 30^{\circ} = 180^{\circ}$
$x = 40^{\circ}$.

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MCQ 1611 Mark

In the given figure, $AB \| CD$. If $\text{AOC}=30^\circ$ and $\angle\text{OAB}=100^\circ$ then $\angle\text{OCD}=?$

A
$150^\circ$
B
$80^\circ$
✓
$130^\circ$
D
$100^\circ$

Answer

Correct option: C.

$130^\circ$

Draw $OE \| AB \| CD$
Now, $OE \| AB$ and $OA$ is the transversal.
$\therefore\angle\text{OAB}+\angle\text{AOE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow\angle\text{OAB}+\angle\text{AOC}+\angle\text{COE}=180^\circ$
$\Rightarrow100^\circ+30^\circ+\angle\text{COE}=180^\circ$
$\Rightarrow\angle\text{COE}=50^\circ$
Also,
$OE \| CD$ and $OC$ is the transversal.
$\therefore\angle\text{OCD}+\angle\text{COE}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow\angle\text{OCD}+50^\circ=180^\circ$
$\Rightarrow\angle\text{OCD}=130^\circ.$

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MCQ 1621 Mark

Write the correct answer in the following: If one of the angles of a triangle is $130^\circ $, then the angle between the bisectors of the other two angles can be.

A
$50^\circ$
B
$65^\circ$
C
$145^\circ$
✓
$155^\circ$

Answer

Correct option: D.

$155^\circ$

In $\Delta\text{ABC},$ we have $\angle\text{A}=130^\circ$OB and $OC$ are the bisectors of the angles $B$ and $C.$
Let $\angle\text{OBC}=\angle\text{OBA}=\text{x}\ \text{and}\angle\text{OCB}=\angle\text{OCA}=\text{y}$
In $\Delta\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow130^\circ+2\text{x}+2\text{y}=180^\circ$
$\Rightarrow\text{x}+\text{y}=25^\circ$
$\text{i.e}\angle\text{OBC}+\angle\text{OCB}=25^\circ$
Now, In $\Delta\text{BOC}$
$\angle\text{BOC}=180^\circ-\big(\angle\text{OBC}+\angle\text{OCB}\big)$ (Angle sum Property)
$=180^\circ-25^\circ=155^\circ$
Hence, $(d)$ is the correct answer.

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MCQ 1631 Mark

Consider the following statement:When two straight lines intersect:
$i$. Adjacent angles are complementary
$ii$. Adjacent angles are supplementary
$iii$. Opposite angles are equal
$iv$. Opposite angles are supplementary
Of these statements

A
$(i)$ and $(ii)$ are correct
✓
$(ii)$ and $(iii)$ are correct
C
$(i)$ and $(iv)$ are correct
D
$(ii)$ and $(iv)$ are correct

Answer

Correct option: B.

$(ii)$ and $(iii)$ are correct

Let two lines $AB$ and $CD$ intersect each other at $O.$
Now,
We can see from fogure any two adjacent angles
$\angle\text{AOD}$ and $\angle\text{DOB},\angle\text{DOB}$ and $\angle\text{BOC}$ etc are supplementary because their sum is $180^\circ .$
$\angle\text{AOD}+\angle\text{DOB}=180^\circ$
$\angle\text{DOB}+\angle\text{BOC}=180^\circ$
So two adjacent angles are always supplementary.
Now,
Two opposite angle like $\angle\text{AOC}$ and $\angle\text{DOB},\angle\text{AOD}$ and $\angle\text{COB}$ are always equal to each other as they are vertically opposite angles
$\angle\text{AOC}=\angle\text{DOB}$
$\angle\text{AOD}=\angle\text{COB}$
Hence statement $(ii)$ and $(iii)$ are correct.

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MCQ 1641 Mark

In figure, lines $I_1 \| I_2$. The value of $x$ is:

A
$40^{\circ}$
✓
$30^{\circ}$
C
$50^{\circ}$
D
$70^{\circ}$

Answer

Correct option: B.

$30^{\circ}$

40$^{\circ}$ + x = 70$^{\circ}$ (exterior angle)
$\angle\text{x}=70^\circ-40^\circ$
$\angle\text{x}=30^\circ.$

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MCQ 1651 Mark

Write the correct answer in the following: If one angle of a triangle is equal to the sum of the other two angles, then the triangle is.

A
An isosceles triangle.
B
An obtuse triangle.
C
An equilateral triangle.
✓
An equilateral triangle.

Answer

Correct option: D.

An equilateral triangle.

Let the angles of $\Delta\text{ABC}\ \text{be}\ \angle\text{A},\angle\text{B}\ \text{and}\ \angle\text{C}$
Given that $\angle\text{A}=\angle\text{B}+\angle\text{C}.....(1)$
But, in any $\Delta\text{ABC},\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ....(2)$[Angles sum property of triangle]
From equation (1) and (2), we get
$\angle\text{A}+\angle\text{A}=180^\circ\Rightarrow2\angle\text{A}=180^\circ\Rightarrow\angle\text{A}\frac{180^\circ}{2}=90^\circ$
$\Rightarrow\angle\text{A}=90^\circ$
Hence, the triangle is a right triangle and option $(d)$ is correct.

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MCQ 1661 Mark

In the adjoining figure, $BE$ and $CE$ are bisectors of $\angle\text{ABC}$ and $\angle\text{ACD}$ respectively. If $\angle\text{BEC}=25^\circ$ then $\angle\text{BAC}$ is equal to :

✓
$50^\circ$
B
$25\frac{1}{2}^0$
C
$65^\circ$
D
$12\frac{1}{2}^0$

Answer

Correct option: A.

$50^\circ$

In $\triangle\text{BEC}$
$\angle\text{BEC}+\angle\text{EBC}+\angle\text{ECD}$ (Exterior angle property)
$\angle\text{BEC}=\angle\text{ECD}-\angle\text{EBC}$
In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BAC}=\angle\text{ACD}$
$\angle\text{ABC}+2\angle\text{EBC}=2\angle\text{ECD}$
$\angle\text{ABC}=2(\angle\text{ECD}-\angle\text{EBC})$
$\angle\text{ABC}=2(\angle(\text{BEC}))$
$\angle\text{ABC}=50^\circ$

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MCQ 1671 Mark

Two complementary angles are such that twich the measure of the one is equal to three times the measure of the other. The larger of the two measure.

A
$72^\circ$
✓
$54^\circ$
C
$63^\circ$
D
$36^\circ$

Answer

Correct option: B.

$54^\circ$

Let the measure of each angle be $x^\circ $ and $(90 - x)^\circ .$
According to the given condition,
$2x = 3(90 - x)$
$\Rightarrow 2x = 270 - 3x$
$\Rightarrow 5x = 270$
$\Rightarrow x = 54^\circ$
So,$ (90 - x)^\circ = (90 - 54)^\circ = 36^\circ$
So, the larger of the two angles is $54^\circ .$

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MCQ 1681 Mark

In figure, if $I_1 \| I_2$, what is $x + y$ in terms of w and $z?$

✓
$180 - w + z$
B
$180 + w - z$
C
$180 - w - z$
D
$180 + w + z$

Answer

Correct option: A.

$180 - w + z$

Let angle supplement of $w^{\circ}$ be $a^{\circ}$.
$\Rightarrow \mathrm{a}^{\circ}=180^{\circ}-\mathrm{w}^{\circ}$
Now,
$a^{\circ}=x^{\circ} \text { [Alternate opposite angles] }$
$\Rightarrow x^{\circ}=180-w^{\circ} \ldots(1)$
Now,
$y^{\circ}=z^{\circ} \ldots \text { (2) [Alternate angles] }$
Adding $(1)$ and $(2)$, we get
$x^{\circ}+y^{\circ}=180^{\circ}-w^{\circ}+z^{\circ}$

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MCQ 1691 Mark

In the given figure, $AOB$ is a straight line. If $\angle\text{AOC}=(3\text{x}-10)^\circ,\angle\text{COD}=50^\circ$ and $\angle\text{BOD}=(\text{x}+20)^\circ$ then $\angle\text{AOC}=?$

A
$40^\circ$
B
$60^\circ$
✓
$80^\circ$
D
$50^\circ$

Answer

Correct option: C.

$80^\circ$

Since $AOB$ is a straight line,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
$\Rightarrow(3\text{x}-10)+50+(\text{x}+20)=180^\circ$
$\Rightarrow4\text{x}+60=180$
$\Rightarrow4\text{x}=120$
$\Rightarrow\text{x}=30$
So, $\angle\text{AOC}=3\text{x}-10=3(30)-10=80^\circ$

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MCQ 1701 Mark

The sides of a quadrilateral are extended in order to form 4 exterior angles. The sum of these exterior angles is:

✓
$360^\circ$
B
$180^\circ$
C
$720^\circ$
D
$90^\circ$

Answer

Correct option: A.

$360^\circ$

As seen from the above figure:
$\angle1+\angle2=180^\circ$ (Linear Pair)
Similarly $\angle3+\angle4=180=\angle5+\angle6=\angle7+\angle8=180^\circ$
Adding up all of the above
$\angle1+\angle2+\angle3+\angle4+\angle5+\angle6+\angle7+\angle8=180\times4=720^\circ $
But we know that the angles of a quadrilateral add up to $360^\circ$
Therefore $\angle2+\angle3+\angle5+\angle7=360^\circ$
Substituting in the above equation we get:
$\angle1+\angle4+\angle6+\angle8+360^\circ=720^\circ$
Therefore $\angle1+\angle4+\angle6+\angle8=360^\circ$
Hence the sum of all exterior angles $= 360^\circ .$

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MCQ 1711 Mark

In the adjoining figure, $AOB$ is a straight line. If $x : y : z = 4 : 5 : 6$, then $y =?$

A
$72^{\circ}$
✓
$60^{\circ}$
C
$48^{\circ}$
D
$80^{\circ}$

Answer

Correct option: B.

$60^{\circ}$

Let $\angle\text{AOC}=\text{x}^\circ=(4\text{a})^\circ,\angle\text{COD}=\text{y}^\circ=(5\text{a})^\circ$ and $\angle\text{BOD}=\text{z}^\circ=(6\text{a})^\circ$
Then, we have
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$ [Since $AOB$ is a straight line]
$\Rightarrow 4a + 5a + 6a= 180^{\circ}$
$\Rightarrow 15a = 180^{\circ}$
$\Rightarrow a = 12^{\circ}$
$\therefore y = 5 \times a = 5 \times 12^\circ = 60^{\circ}$.

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MCQ 1721 Mark

In the figure, $AB \| CD$. If $\angle\text{APQ}=70 ^\circ$ and $\angle\text{PRD}=120^\circ$ then $\angle\text{QPR}=?$

✓
$55^\circ$
B
$60^\circ$
C
$40^\circ$
D
$35^\circ$

Answer

Correct option: A.

$55^\circ$

Since $AB \| CD,$
$\angle\text{APQ}=\angle\text{PQR}=70^\circ$ (Alternate angles)
$\Rightarrow\angle\text{PRD}=\angle\text{PQR}+\angle\text{QPR}$ (Exterior angle is equal to sum of the remote interior angles)
$\Rightarrow120^\circ=70^\circ+\angle\text{QPR}$
$\Rightarrow\angle\text{QPR}=506^\circ$

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MCQ 1731 Mark

In figure, if $AB \| CD$, then $x =$

✓
$100$
B
$105$
C
$110$
D
$114$

Answer

Correct option: A.

$100$

Extending line $BA$ and $CP$ to meet at point $E.$
$\angle\text{APE}=180^\circ-148^\circ=32^\circ$
$\angle\text{EAP}=180^\circ-132^\circ=48^\circ$
$\angle\text{AEP}=\text{x}^\circ$ [(Correspondence angles) because $AB \| CD$ cut by transverse $EC]$
Now, in $\triangle\text{APE}$
$\angle\text{APE}+\angle\text{EAP}+\angle\text{AEP}=180^\circ$
$\Rightarrow\ 32^\circ+48^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow\ \text{x}^\circ=100$

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MCQ 1741 Mark

One angle is equal to three times its supplement. The measure of the angle is:

A
$130^\circ$
✓
$135^\circ$
C
$90^\circ$
D
$120^\circ$

Answer

Correct option: B.

$135^\circ$

Let the required angle be ${\theta}.$
Then, measure of its supplement $180^\circ-\theta$
According to question, we have
$\theta=3(180-\theta)$
$\Rightarrow\ \theta=540^\circ-3\theta$
$\Rightarrow\ 4\theta=540^\circ $
$\Rightarrow\ \theta=135^\circ$

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MCQ 1751 Mark

In the adjoining figure, the bisectors of $\angle\text{CBD}$ and $\angle\text{BCE}$ meet at the point O. If $\angle\text{BAC}=70^\circ$ then $\angle\text{BOC}$ is equal to:

A
$11^\circ$
✓
$55^\circ$
C
$70^\circ$
D
$35^\circ$

Answer

Correct option: B.

$55^\circ$

$\angle\text{BOC}=90^\circ-\frac{1}{2 }\angle\text{BAC}$
$\angle\text{BOC}=90^\circ-35^\circ=55^\circ$

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MCQ 1761 Mark

Two planes intersect each other to form a:

A
Plane.
B
Point.
C
Angle.
✓
Straight line.

Answer

Correct option: D.

Straight line.

As can be seen from the above diagram, the two planes $"P"$ and $"Q"$ are intersecting in a line, which is $AB.$

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MCQ 1771 Mark

In the given figure, $AB \I CD$. If $\angle\text{AOC}=30^\circ$ and $\angle\text{OAB}=100^\circ.$ then $\angle\text{OCD}=?$

✓
$80^\circ$
B
$100^\circ$
C
$130^\circ$
D
$150^\circ$

Answer

Correct option: A.

$80^\circ$

Extend line $CD$ which intersect $AO$ at $M.$
(Corresponding angle)
$\angle\text{MOC}+\angle\text{CMO}=\angle\text{DCO}$ (exterior angle is equal to the sum of two opposite interior angles)
$\angle\text{DCO}=100^\circ+30^\circ=130^\circ$

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MCQ 1781 Mark

In the given figure, $AB$ is a mirror, $PQ$ is the incident ray and $QR$ is the reflected ray. If $\angle\text{PQR}=108^\circ$ then $\angle\text{AQP}=?$

A
$72^\circ$
B
$18^\circ$
✓
$36^\circ$
D
$54^\circ$

Answer

Correct option: C.

$36^\circ$

We know that, angle of incidance = angle reflection.
that is, $\angle\text{AQP}=\angle\text{BQR}$
Since $AOB$ is a straight line,
$\angle\text{AQP}+\angle\text{BQR}+\angle\text{PQR}=180^\circ$
$\Rightarrow\angle\text{AQP}+\angle\text{AQP}+\angle\text{PQR}=180^\circ$
$\Rightarrow2\angle\text{AQP}=72$
$\Rightarrow \angle\text{AQP}=36^\circ$

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MCQ 1791 Mark

Measurement of reflex angle is:

A
Between $0^\circ $ and $90^\circ $
B
$90^\circ$
✓
Between $180^\circ $ and $360^\circ $
D
Between $90^\circ $ and $180^\circ $

Answer

Correct option: C.

Between $180^\circ $ and $360^\circ $

Let $x$ be the angle
then its reflex angle is $360^\circ - x$
and in any triangle the angle lies between $0$ to $180^\circ $

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MCQ 1801 Mark

One angle is equal to three times its supplement. The measure of the angle is:

A
$130^\circ$
B
$120^\circ$
✓
$135^\circ$
D
$90^\circ$

Answer

Correct option: C.

$135^\circ$

Let the required angle be $x$
Supplement $= 180^\circ - x$
According to question,
$x = 3 (180^\circ - x)$
$x = 540^\circ - 3x$
$x = 135^\circ .$

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MCQ 1811 Mark

In the given figure, $\angle\text{BAC}=40^\circ,\angle\text{ACB}=90^\circ$ and $\angle\text{BED}=100^\circ$ Then $\angle\text{BDE}=?$

A
$25^\circ$
B
$40^\circ$
C
$50^\circ$
✓
$30^\circ$

Answer

Correct option: D.

$30^\circ$

In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BAC}+\angle\text{ACB}=180^\circ$(Angle sum property)
$\angle\text{ABC}=180^\circ-90^\circ-40^\circ$
$\angle\text{ABC}=50^\circ$
In $\triangle\text{BED}$
$\angle\text{BED}+\angle\text{EBD}+\angle\text{BDE}=180^\circ$(Angle sum property)
$\angle\text{EBD}=180^\circ-50^\circ-100^\circ$
$\angle\text{EBD}=30^\circ$

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MCQ 1821 Mark

In the adjoining figure, $\angle\text{a}$ and $\angle\text{g}$ are called:

A
Co-interior angles.
B
Corresponding angles.
✓
Alternate exterior angles.
D
Alternate interior angles.

Answer

Correct option: C.

Alternate exterior angles.

$\angle\text{a}$ and $\angle\text{g}$ are on alternate side and are exterior.

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MCQ 1831 Mark

In figure, if $AB \| CF, CD \| EF$, then the value of $x$ is:

A
$140^\circ$
B
$120^\circ$
C
$100^\circ$
✓
$110^\circ$

Answer

Correct option: D.

$110^\circ$

In the above figure, $\angle\text{B}=\angle\text{BCF}$ (Alternate Interior angles)
Now $\angle\text{FCA}=\angle\text{BCA}+\angle\text{FCB}$
$= 60^\circ + 50^\circ = 110^\circ$
Now $\angle\text{FCA}=\angle\text{x} $ (Alternate interior angles)
Therefore $\angle\text{x}=110^\circ.$

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MCQ 1841 Mark

The sides $BC, CA$ and $AB$ of $\triangle\text{ABC}$ have been produced to $D, E$ and $F$ respectively as shown in the figure, forming exterior angles $\angle\text{ACD},\angle\text{BAE}$ and $\angle\text{CBF}.$ Then, $\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=?$

A
$240^\circ$
✓
$360^\circ$
C
$300^\circ$
D
$320^\circ$

Answer

Correct option: B.

$360^\circ$

We know that the exteior angle is sum of opposite interior angle
$\angle\text{ACD}=\angle1+\angle2\text{ (i)}$
$\angle\text{BAE}=\angle3+\angle2\text{ (ii)}$
$\angle\text{CBF}=\angle1+\angle3\text{ (iii)}$
$ADD$ Equation $(i), (ii)$ and $(iii)$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=2\times(\angle1+\angle2+\angle3)$
$\angle1+\angle2+\angle3=180^\circ\text{(ASP)}$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=2\times180^\circ$
$\angle\text{ACD} +$ We know that the exteior angle is sum of opposite interior angle
$\angle\text{ACD}=\angle1+\angle2\text{ (i)}$
$\angle\text{BAE}=\angle3+\angle2\text{ (ii)}$
$\angle\text{CBF}=\angle1+\angle3\text{ (iii)}$
$ADD$ Equation $(1i), (ii)$ and $(iii)$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=2\times(\angle1+\angle2+\angle3)$
$\angle1+\angle2+\angle3=180^\circ\text{(ASP)}$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=2\times180^\circ$
$\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=360^\circ.$

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MCQ 1851 Mark

In the adjoining figure, $AB \| CD$ and $AB \| EF$. If $\text{EA}\bot\text{BA}$ and $\angle\text{BEF}=55^\circ,$ then the values of $x, y$ and $z:$

A
$35^\circ , 125^\circ , 120^\circ$
✓
$125^\circ , 125^\circ , 35^\circ$
C
$60^\circ , 60^\circ , 60^\circ$
D
$120^\circ , 130^\circ , 25^\circ$

Answer

Correct option: B.

$125^\circ , 125^\circ , 35^\circ$

$x + 55 = 180^\circ $ (Sum of supplementary angles or co-interior angles)$x = 125^\circ$
$x = y = 125^\circ $ (Corresponding angles)
$\text{z}+\angle\text{EAB}=\text{y} $ (Exterior angle property)
$z = 125^\circ - 90^\circ = 35^\circ .$

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MCQ 1861 Mark

The number of line segments determined by three given non$-$collinear points is:

A
Four.
B
Infinitely many.
C
Two.
✓
Three.

Answer

Correct option: D.

Three.

Three because non$-$collinear points means the point does not lies in a same line.

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MCQ 1871 Mark

In the figure, $POQ$ is a line. The value of $x$ is:

A
$30^\circ$
✓
$20^\circ$
C
$35^\circ$
D
$25^\circ$

Answer

Correct option: B.

$20^\circ$

Since, $POQ$ is a line segment.
$\therefore\angle\text{POQ}=180^\circ$
$\Rightarrow\angle\text{POA}+\angle\text{AOB}+\angle\text{BOQ}=180^\circ$
$\Rightarrow40^\circ+4\text{x}+3\text{x}=180^\circ$
[Putting $\angle\text{POA}=40^\circ,\angle\text{AOB}=4\text{x}$ and $\angle\text{BOQ}=3\text{x}$]
$\Rightarrow7\text{x}=180^\circ-40^\circ$
$\Rightarrow7\text{x}=140^\circ$
$\text{x}=\frac{140^\circ}{7}$
$\text{x}=20^\circ.$

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MCQ 1881 Mark

In Fig. the value of $y$ is:

A
$20^{\circ}$
B
$60^{\circ}$
✓
$30^{\circ}$
D
$45^{\circ}$

Answer

Correct option: C.

$30^{\circ}$

$3x + y + 2x = 180^{\circ}$(Linear pair)
$5x + y = 180^{\circ} (i)$
From figure,
$y = x$ (Vertically opposite angles)
Using it in $(i)$, we get
$5x + x = 180^{\circ}$
$6x = 180^{\circ}$
$x = 30^{\circ}$
Thus,
$y = x = 30^{\circ}$.

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MCQ 1891 Mark

An angle which measures more than $180^{\circ}$ but less than $360^{\circ}$, is called.

A
An acute angle.
B
An obuse angle.
C
A straight angle.
✓
A reflex angle.

Answer

Correct option: D.

A reflex angle.

An angle which measures more than $180^{\circ}$ but less than $360^{\circ}$is called a reflex angle.

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MCQ 1901 Mark

In Fig., if $l \| m,$ then $x =$

A
$40^{\circ}$
B
$25^{\circ}$
C
$65^{\circ}$
✓
$105^{\circ}$

Answer

Correct option: D.

$105^{\circ}$

Given that,
$l \| m$ and n cuts them
Let,
$\angle1=65^\circ$
$\angle2=\text{x}$
$\angle3=40^\circ$
$\angle1+\angle4=65^\circ$ (Alternate angle) $(i)$
$\angle3+\angle4+\angle5=180^\circ$(Angle sum property)
$40^\circ+165^\circ+\angle5=180^\circ$
$\angle5=75^\circ$
Now,
$\angle2+\angle5=180^\circ$ (Linear pair)
$x + 75^{\circ}$ = 180$^{\circ}$
$x = 105^{\circ}$.

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MCQ 1911 Mark

Angles of a triangle are in the ratio $2 : 4 : 3$. The smallest angle of the triangle is:

A
$20^\circ$
✓
$40^\circ$
C
$60^\circ$
D
$80^\circ$

Answer

Correct option: B.

$40^\circ$

Given, the ratio of angles of a triangle is $2 : 4 : 3.$
Let the angles of a triangle be $\angle\text{A},\angle\text{B}$ and $\angle\text{C}.$
$\angle\text{A}=2\text{x}, \angle\text{B}=4\text{x}$
$\angle\text{C}=3\text{x},\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
[sum of all the angles of a triangle is 180º]
$2\text{x}+4\text{x}+3\text{x}=180^\circ$
$9\text{x}=180^\circ$
$\text{x}=\frac{180^\circ}{9}=20^\circ$
$\angle\text{A}=2\text{x}=2\times20^\circ=40^\circ$
$\angle\text{B}=4\text{x}=2\times20^\circ=80^\circ$
$\angle\text{C}=3\text{x}=3\times20^\circ=60^\circ$
Hence, the smallest angle of a triangle is $40^\circ .$

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MCQ 1921 Mark

The number of lines that can pass through a given point is:

A
Only one
B
Two
C
One
✓
Infinity

Answer

Correct option: D.

Infinity

As seen from the above image, any number of lines can be drawn through a given point.
Hence, the answer may be given as $"$Infinity$"$.

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MCQ 1931 Mark

If two angles are complements of each other then each angle is:

A
A straight angle.
B
A reflex angle.
✓
An acute angle.
D
An obtuse angle.

Answer

Correct option: C.

An acute angle.

If two angles are complements of each other, that is, the sum of their measures is $90^\circ $, then each angle is an acute angle.

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MCQ 1941 Mark

Write the correct answer in the following: In Fig. if $\text{OP}||\text{RS},$ $\angle\text{OPQ}=110^\circ$ and $\angle\text{QRS}=130^\circ,$ then $\angle\text{PQR}$ is equal to,

A
$40^\circ$
B
$50^\circ$
✓
$60^\circ$
D
$70^\circ$

Answer

Correct option: C.

$60^\circ$

In the given figure, producing $OP$, to interscet $RQ$ at $X.$
Since, $\text{OP}||\text{RS}$ and RX is a transversal.
So, $\angle\text{RXP}=\angle\text{XRS}$

$\Rightarrow\angle\text{RXP}=130^\circ$$\big[\because\angle\text{QRS}=130^\circ(\text{given})\big]....(\text{i})$
Now, $RQ$ is a line segment.
So,$\angle\text{PXQ}+\angle\text{RXP}=180^\circ$ [linear pair axiom]
$\Rightarrow\angle\text{PXQ}=180^\circ-\angle\text{RXP}=180^\circ-130^\circ$ [from eq. $(i)]$
$\Rightarrow\angle\text{PXQ}=50^\circ$
In $\Delta\text{PQX},$ $\angle\text{OPQ}$ is an exterior angle,
$\therefore\angle\text{OPQ}=\angle\text{PXQ}+\angle\text{PQX}$
[$\because\ $exterior asngle = sum of two opposite interior angles]
$110^\circ=50^\circ+\angle\text{PQX}$
$\angle\text{PQX}=110^\circ-50^\circ$
$\angle\text{PQR}=60^\circ$ $[\because\ \angle\text{PQX}=\angle\text{PQR}]$

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MCQ 1951 Mark

In figure, if $AB \| HF$ and $DE \| FG$, then the measure of $\angle\text{FDE}$ is:

A
$108^\circ$
✓
$80^\circ$
C
$100^\circ$
D
$90^\circ$

Answer

Correct option: B.

$80^\circ$

AB || HF and $\angle\text{CFH}=28^\circ$ [Given]
$\angle\text{CFH}=\angle\text{FDA}$ [Correspondence angels are equal]
$\angle\text{FDA}=28^\circ$
Now,
$\angle\text{FDA}+\angle\text{FDE}+\angle\text{EDB}=180^\circ$
$\Rightarrow\ 28^\circ+\angle\text{FDE}+72^\circ=180^\circ$
$\Rightarrow\ \angle\text{FDE}=80^\circ$

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MCQ 1961 Mark

In the given figure, straight lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}+\angle\text{BOD}=130^\circ$ then $\angle\text{AOD}=?$

✓
$115^\circ$
B
$125^\circ$
C
$110^\circ$
D
$65^\circ$

Answer

Correct option: A.

$115^\circ$

We have,
$\angle\text{AOC}=\angle\text{BOD}$ [Vertically-Opposite Angles]
$\therefore\angle\text{AOC}+\angle\text{BOD}=130^\circ$
$\Rightarrow\angle\text{AOC}+\angle\text{AOC}=130^\circ [\therefore\angle\text{AOC}=\angle\text{BOD}]$
$\Rightarrow2\angle\text{AOC}=130^\circ$
$\Rightarrow\angle\text{AOC}=65^\circ$
Now,
$\angle\text{AOC}+\angle\text{AOD}=180^\circ$ [$\because$ COD is a straight line]
$\Rightarrow65^\circ+\angle\text{AOD}=180^\circ$
$\Rightarrow\angle\text{AOC}=115^\circ.$

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MCQ 1971 Mark

In the given figure, straight lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}=\phi,\angle\text{BOC}=\theta$ and $\theta=3\phi,$ then $\phi=?$

✓
$45^\circ$
B
$30^\circ$
C
$60^\circ$
D
$40^\circ$

Answer

Correct option: A.

$45^\circ$

We have.
$\theta+\phi=180^\circ$ [$\because AOD$ is a straight line]
$\Rightarrow3\phi+\phi=180^\circ [\because\theta=3\phi]$
$\Rightarrow4\theta=180^\circ$
$\Rightarrow\phi=45^\circ.$

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MCQ 1981 Mark

If two lines intersect each other then:

A
Corresponding angles are equal.
B
Alternate interior angles are equal.
C
Co-interior angles are equal.
✓
Vertically opposite angles are equal.

Answer

Correct option: D.

Vertically opposite angles are equal.

$\angle\text{A}+\angle\text{B}=180^\circ ($Linear Pair$)$
$\angle\text{B}+\angle\text{C}=180 ($Linear Pair$)$
On equating above equations, we get
$\angle\text{A}+\angle\text{B}=\angle\text{B}+\angle\text{C}$
$\angle\text{A}=\angle\text{C}$
Similarly, $\angle\text{B}=\angle\text{D}$

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MCQ 1991 Mark

In the given figure, $AB \| CD \| EF,$ $\text{EA }\bot\text{ AB}$ and BDE is the transversal such that $\angle\text{DEF}=55^\circ,$ Then $\angle\text{AEB}=?$

✓
$35^\circ$
B
$55^\circ$
C
$45^\circ$
D
$25^\circ$

Answer

Correct option: A.

$35^\circ$

$\text{EA }\bot\text{ AB}$
$\angle\text{AEF}=90^\circ$
$\angle\text{AEF}=\angle\text{BEF}+\angle\text{AEB}$
$\angle\text{BEF}+\angle\text{AEB}=90^\circ$
$\angle\text{BEF}=55^\circ$
$55^\circ+\angle\text{AEB}=90^\circ$
$\text{AEB}=90^\circ-55^\circ$
$\angle\text{AEB}=35^\circ.$

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MCQ 2001 Mark

In the given figure, $BO$ and $CO$ are the bisectors of $\angle\text{B}$ and $\angle\text{C}$ respectively. If $\text{A}=50^\circ,$ then $\angle\text{BOC}=?$

A
$130^\circ$
B
$120^\circ$
C
$100^\circ$
✓
$115^\circ$

Answer

Correct option: D.

$115^\circ$

In $\triangle\text{ABC}$
$\text{2x}+\text{2y}+\angle\text{A}=180^\circ$ (Angle sum property)
$\text{x}+\text{y}+(\frac{\angle\text{A}}{2})=90^\circ$
$\text{x}+\text{y}=90^\circ-(\frac{\text{A}}{2})$(1)
In $\triangle\text{BOC},$ we have
$\text{x}+\text{y}+\angle\text{BOC}=180^\circ$
$90^\circ- (\frac{\angle\text{A}}{2})+\angle\text{BOC}=180^\circ$ [From (1)]
$\text{BOC}=180^\circ-90^\circ+(\frac{\text{A}}{2})$
$\angle\text{BOC}=90^\circ+(\frac{\text{A}}{2})$
$\angle\text{BOC}=90^\circ+25^\circ=115^\circ.$

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Maths M.C.Q