M.C.Q

MCQ 2011 Mark

In Fig. $AOB$ is a straight line. If $\angle\text{AOC}+\angle\text{BOD}=85^\circ,$ then $\angle\text{COD}=$

A
$90^\circ$
B
$85^\circ$
✓
$95^\circ$
D
$100^\circ$

Answer

Correct option: C.

$95^\circ$

Given,
$AOB =$ Straight line
$\angle\text{AOC}+\angle\text{BOD}=85^\circ$
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ $(Linear pair)
$85^\circ+\angle\text{COD}=180^\circ$
$\angle\text{COD}=95^\circ.$

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MCQ 2021 Mark

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
$145^{\circ}$
B
$155^{\circ}$
✓
$50^{\circ}$
D
$65^{\circ}$

Answer

Correct option: C.

$50^{\circ}$

Let angles of a triangle be $\angle\text{A},\angle\text{B}$ and $\angle\text{C}.$

In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ $[sum of all interior angles of a triangle is $180^{\circ}$]
$\Rightarrow12\angle\text{A}+12\angle\text{B}+12\angle\text{C}=180^\circ2=90^\circ$[dividing both sides by $2]$
$\Rightarrow12\angle\text{B}+12\angle\text{C}=90^\circ−12\angle\text{A} $
$[\because\text{In}\triangle\text{OBC},\angle\text{OBC}+\angle\text{BCO}+\angle\text{COB}=180^\circ]$
⇒ Since, $\angle\text{B}2+\angle\text{C}2+\angle\text{BOC}=180^\circ$ as $BO$ and $OC$ are the angle bisectors of $\angle\text{ABC} $ and $\angle\text{BCA},$ respectively
$\Rightarrow180^\circ−\angle\text{BOC}=90^\circ−12\angle\text{A}$
$\therefore\angle\text{BOC}=180^\circ-90^\circ+12\angle\text{A}$
= $90^{\circ}+12 \times 130^{\circ}=90^{\circ}+65^{\circ}[\therefore\angle\text{A}=130^\circ(\text{given)}]$
$= 155^{\circ}$
Hence, the required angle is 155$^{\circ}$.

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MCQ 2031 Mark

If two angles are supplementary and the larger is $20^{\circ}$ less then three times the smaller, then the angles are:

A
$72\frac{1}{2}^0,17\frac{1}{2}^0$
B
$140^0,40^0$
✓
$130^0,50^0$
D
$62\frac{1}{2}^0,27\frac{1}{2}^0$

Answer

Correct option: C.

$130^0,50^0$

Let the two supplimentary angles be $\mathrm{x}^0$ and $180^{\circ}-\mathrm{x}^0$
Let $180^{\circ}-x$ be the larger angle
$180^{\circ}-x=3 x-20^0$
$4 x=200^{\circ}$
$x=50^{\circ}$
So the angles are $50^{\circ}$ and $130^{\circ}$

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MCQ 2041 Mark

In $\triangle\text{ABC, }\text{BD}\perp\text{AC, }\angle\text{CAE} = 30^\circ$ and $\angle\text{CBD}=40^\circ.$ Then $\angle\text{AEB}=?$

A
$70^\circ$
B
$50^\circ$
C
$60^\circ$
✓
$80^\circ$

Answer

Correct option: D.

$80^\circ$

In $BDC$
$\angle\text{BDC}+\angle\text{BCD}+\angle\text{DBC}=180^\circ$
$\text{BD}\perp\text{AC}$
$\angle\text{BCD}=90^\circ,\angle\text{DBC}=40^\circ$
$90^\circ+\angle\text{BCD}+40^\circ=180^\circ$
$\angle\text{BCD}+130^\circ=180^\circ$
$\angle\text{BCD}=180^\circ-130^\circ$
$\angle\text{BCD}=50^\circ$
$\angle\text{AEB}=\angle\text{CAE}+\angle\text{C}$ (exterior angle)
$\angle\text{CAE}=30^\circ$
$\angle\text{C}=50^\circ$
$\angle\text{AEB}=30^\circ+50^\circ$
$\angle\text{AEB}=80^\circ.$

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MCQ 2051 Mark

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

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

Answer

Correct option: C.

$54^\circ$

Let the measure of the required angle be $x^\circ$
Then, the measure of its complement will be $(90 - x)^\circ$
Therefore, $2x = 3 (90 - x)$
$\Rightarrow 2x = 270 - 3x$
$\Rightarrow 5x = 270$
$\Rightarrow x = 54^\circ .$

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MCQ 2061 Mark

In figure, if $l \| m$, then $x =$

✓
$105^\circ$
B
$65^\circ$
C
$40^\circ$
D
$25^\circ$

Answer

Correct option: A.

$105^\circ$

From figure,
$\angle\text{AGE}=\angle\text{FGB}$ [Opposite angles]
$\Rightarrow\ \angle\text{FGB}=65^\circ$
Also,
$\angle\text{FGB}=\angle\text{HJI}$ [Corresponding angle]
$\Rightarrow\ \angle\text{HJI}=65^\circ$
Now, in $\angle\text{HJI},$
$\angle\text{HJI}+\angle\text{JIH}+\angle\text{IHJ}=180^\circ$
$\Rightarrow\ 65^\circ+40^\circ+\angle\text{IHJ}=180^\circ$
$\Rightarrow\ \angle\text{IHJ}=180^\circ-65^\circ-40^\circ=75^\circ$
Now,
$\text{x}=180^\circ-\angle\text{IHJ}=180^\circ-75^\circ$
$=105^\circ$

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MCQ 2071 Mark

In Fig. if $\frac{\text{y}}{\text{x}}=5$ and $\frac{\text{z}}{\text{x}}=4,$ then the value of $x$ is:

A
$15^\circ$
✓
$18^\circ$
C
$12^\circ$
D
$8^\circ$

Answer

Correct option: B.

$18^\circ$

In the given figure, we have $x^\circ , y^\circ $ and $z^\circ $ forming a linear pair, therefore these must be supplementary.
That is,
$x + y + z = 180^\circ (1)$
Also,
$\frac{\text{y}}{\text{x}}=5$
$y = 5x (2)$
And
$\frac{\text{z}}{\text{x}}=4$
$z = 4x (3)$
Substituting $(ii)$ and $(iii)$ in $(i),$ we get:
$x + 5x + 4x = 180^\circ$
$10x = 180^\circ$
$\text{x}=\frac{180^\circ}{10}$
$x = 18^\circ .$

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MCQ 2081 Mark

Two straight lines $AB$ and $CD$ cut each other at $O$. If $\angle\text{BOD}=63^\circ,$ then $\angle\text{BOC}=$

A
$153^\circ$
B
$17^\circ$
✓
$117^\circ$
D
$63^\circ$

Answer

Correct option: C.

$117^\circ$

$\angle\text{BOD}+\angle\text{BOC}=180^\circ$ (Linear pair)
$63^\circ+\angle\text{BOC}=180^\circ$
$\text{BOC}=117^\circ.$

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MCQ 2091 Mark

In Fig, $PQ \| RS$, $\angle\text{AEF}=95^\circ,\angle\text{BHS}=110^\circ,$ and $\angle\text{ABC}=\text{x}^\circ.$ Then the value of $x$ is:

A
$70^\circ$
B
$15^\circ$
✓
$25^\circ$
D
$35^\circ$

Answer

Correct option: C.

$25^\circ$

Given that,
$PQ \| RS$
$\angle\text{AEF}=95^\circ$
$\angle\text{BHS}=110^\circ$
$\angle\text{ABC}=\text{x}^\circ$
$\angle\text{AEF}=\angle\text{AGH}=95^\circ$ (Corresponding angles)
$\angle\text{AGH}+\angle\text{HGB}=180^\circ$ (Linear pair)
$95^\circ+\angle\text{HGB}=180^\circ$
$\angle\text{HGB}=85^\circ$
$\angle\text{BHS}+\angle\text{BHG}=180^\circ $(Linear pair)
$110^\circ+\angle\text{BHG}=180^\circ$
$\angle\text{BHG}=70^\circ$
In $\triangle\text{BHG},$
$\angle\text{BHG}+\angle\text{HGB}+\angle\text{GBH}=180^\circ$
$70^\circ+85^\circ+\angle\text{GBH}=180^\circ$
$\angle\text{GBH}=25^\circ$
Thus,
$\angle\text{ABC}=\angle\text{GBH}=25^\circ.$

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MCQ 2101 Mark

Two lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ,$then $\angle\text{AOC}=$

A
$70^\circ$
B
$80^\circ$
✓
$90^\circ$
D
$180^\circ$

Answer

Correct option: C.

$90^\circ$

$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=270^\circ$ [Given]
From figure,
$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}+\angle\text{DOA}=360^\circ$
$\Rightarrow\ 270^\circ+\angle\text{DOA}=360^\circ$
$\Rightarrow\ \angle\text{DOA}=360^\circ-270^\circ=90^\circ$
Now,
$\angle\text{DOA}+\angle\text{AOC}=180^\circ$
$\Rightarrow\ \angle\text{AOC}=180^\circ-90^\circ=90^\circ$

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MCQ 2111 Mark

In the given figure, $\angle\text{OAB}=75^\circ, \angle\text{OBA}=55^\circ$ and $\angle\text{OCD}=100^\circ.$ then, $\angle\text{ODC}=?$

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

Answer

Correct option: C.

$30^\circ$

In $\triangle\text{OAB},$ we have
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$ (Angle sum property)
$\Rightarrow55^\circ+75^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=50^\circ$
$\Rightarrow\angle\text{COD}=\angle\text{AOB}=50^\circ$ (Vertivcally opposite angles)
In $\triangle\text{OCD},$ we have
$\angle\text{COD}+\angle\text{OCD}+\angle\text{ODC}=180^\circ$ (Angle sum property)
$\Rightarrow50^\circ+100^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=30^\circ$

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MCQ 2121 Mark

In the adjoining figure, if $A = 40^\circ , B = 95^\circ $ and $D = 60^\circ $, then $E$ is equal to:

✓
$75^\circ$
B
$55^\circ$
C
$65^\circ$
D
$85^\circ$

Answer

Correct option: A.

$75^\circ$

in $\triangle\text{ABC}$$\angle\text{ABC}+\angle\text{BCA}+\angle\text{CAB}=180^\circ$ (Angle sum property)
$\angle\text{BCA}=180^\circ-95^\circ-40^\circ=45^\circ$
$\angle\text{BCA}=\angle\text{ECD}=45^\circ$ (Vertically opposite angle)
In $\triangle\text{ECD}$
$\angle\text{ECD}+\angle\text{DEC}+\angle\text{CDE}=180^\circ$ (Angle sum property)
$\angle\text{ECD}=180^\circ-45^\circ-60^\circ=75^\circ.$

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MCQ 2131 Mark

In the given figure, the value of x which makes $POQ$ a straight line is:

A
$30^\circ$
✓
$25^\circ$
C
$35^\circ$
D
$40^\circ$

Answer

Correct option: B.

$25^\circ$

We know that he measure of a straight angle is $180^\circ$
$(2x + 30^\circ ) + 4x = 180^\circ$
$2x + 30^\circ + 4x = 180^\circ$
$6x = 180^\circ - 30^\circ$
$6x = 150^\circ$
$\text{x}=\frac{150^\circ}{6}=25^\circ.$

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MCQ 2141 Mark

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

A
$20^\circ$
✓
$30^\circ$
C
$45^\circ$
D
$60^\circ$

Answer

Correct option: B.

$30^\circ$

From figure,
$\angle\text{DPQ}+\angle\text{x}^\circ=180^\circ\dots(1)$ [linear pair]
Also,
$\angle\text{DPQ}=\angle\text{AQP}$ [Interior opposite angles]
$\Rightarrow\ \angle\text{DPQ}=120^\circ+\text{x}$
From $(1),$
$120^\circ+\text{x}+\text{x}=180^\circ$
$\Rightarrow\ 2\text{x}=60^\circ$
$\Rightarrow\text{x}=30^\circ$

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MCQ 2151 Mark

In the given figure, $AOB$ is a straight line. If $\angle\text{AOC}=4\text{x}^\circ$ and $\angle\text{BOC}=5\text{x}^\circ$ then $\angle\text{AOC}=?$

A
$40^\circ$
B
$60^\circ$
✓
$80^\circ$
D
$100^\circ$

Answer

Correct option: C.

$80^\circ$

Since $AOB$ is a straight line,
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
$\Rightarrow4\text{x}+5\text{x}=180^\circ$
$\Rightarrow9\text{x}=180^\circ$
$\Rightarrow\text{x}=206^\circ$
So, $\angle\text{AOC}=4\text{x}=4(20)=80^\circ$

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MCQ 2161 Mark

In a figure, if $OP \| RS$, $\angle\text{OPQ}=110^\circ$ and $\angle\text{QRS}=130^\circ,$ then $\angle\text{PQR}$ is equal to:

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

Answer

Correct option: A.

$60^\circ$

Produce $OP$ to intersect $RQ$ at point $N.$
Now, $OP \| RS$ and transversal RN intersects them at N and R respectively.
$\therefore\angle\text{RNP}=\angle\text{SRN}$ (Alternate interior angles)
$\Rightarrow\angle\text{RNP}=130^\circ$
$\therefore\angle\text{PNQ}=180^\circ−130^\circ=50^\circ$ (Linear pair)
$\angle\text{OPQ}=\angle\text{PNQ}+\angle\text{PQN}$ (Exterior angle property)
$\Rightarrow110^\circ= 50^\circ+\angle\text{PQN}$
$\Rightarrow\angle\text{PQN}=110^\circ-50^\circ=60^\circ=\angle\text{PQR}.$

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MCQ 2171 Mark

In the below figure, the value of $x$ is:

A
$30^\circ$
✓
$10^\circ$
C
$15^\circ$
D
$25^\circ$

Answer

Correct option: B.

$10^\circ$

$2x - 20^\circ + x + 100^\circ = 110^\circ $ (Exterior angle is equal to sum of its interior opposite angles)$3x = 110^\circ - 100^\circ + 20^\circ$
$x = 10^\circ .$

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MCQ 2181 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
$50^\circ$
B
$40^\circ$
C
$25^\circ$
✓
$30^\circ$

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 $\angle\text{BED}$
$\angle\text{BED}+\angle\text{EBD}+\angle\text{BDE}=180^\circ$ (Angle sum property)
$\angle\text{BDE}=180^\circ-50^\circ-100^\circ$
$\angle\text{BDE}=30^\circ.$

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MCQ 2191 Mark

In figure, if $l \| m$, what is the value of $x?$

✓
$60$
B
$50$
C
$45$
D
$30$

Answer

Correct option: A.

$60$

$3y^\circ = 2y^\circ + 25^\circ $ [Alternate angles]
$\Rightarrow y^\circ = 25^\circ $
Now,
$x^\circ + 15^\circ = 2y^\circ + 25^\circ $ [Opposite angles]
$\Rightarrow x = 2y^\circ + 25^\circ - 15^\circ $
$\Rightarrow x = 2y^\circ + 10^\circ $
$\Rightarrow x = 2 \times 25^\circ + 10^\circ $
$\Rightarrow x = 60^\circ $

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MCQ 2201 Mark

Which of the following statements is false?

✓
Through a given point, only one straight line can be drawn.
B
Through two given points, it is possible to draw one and only one straight line.
C
Two straight lines can intersect only at one point.
D
A line segment can be produced to any desired length.

Answer

Correct option: A.

Through a given point, only one straight line can be drawn.

Option $(a)$ is false, since through a given point we can draw an infinite number of straight lines.

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MCQ 2211 Mark

In figure, $\text{PQ }||\text{ RS},\angle\text{QPR}=70^\circ,\angle\text{ROT}=20^\circ$ find the value of $x.$

A
$20^\circ$
✓
$50^\circ$
C
$70^\circ$
D
$110^\circ$

Answer

Correct option: B.

$50^\circ$

$\text{PQ }||\text{ RS},\angle\text{QPR}=\angle\text{SRO}=70^\circ$ (Corresponding, Angle)
Now in $\triangle\text{RTO}$
$x + 20^\circ = 70^\circ $ (exterior angle)
$x = 70^\circ - 20^\circ$
$x = 50^\circ .$

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MCQ 2221 Mark

The complement of $(90-a)^0$ is:

A
$(90+2 a)^{\circ}$
B
$(90-a)^0$
✓
$a^{\circ}$
D
$-a^{\circ}$

Answer

Correct option: C.

$a^{\circ}$

We know that the sum of complementary angles are $90^{\circ}$
Let the complementary angle of $(90-\mathrm{a})^{\circ}$ be $x$
$x+(90-a)^{\circ}=90^{\circ}$
$x=90^{\circ}-(90-a)^{\circ}$
$x=90^{\circ}-90^{\circ}+a^{\circ}$
$x=a^{\circ}$.

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MCQ 2231 Mark

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

A
$85^{\circ}$
B
$65^{\circ}$
C
$20^{\circ}$
✓
$45^{\circ}$

Answer

Correct option: D.

$45^{\circ}$

$l \| m$ Let transversal be n and $\angle1=65^\circ$
$\angle2=20^\circ$
$\angle3=\text{x}$
Since,
$l \| m$ and n cuts them so,
$\angle1+\angle4=180^\circ$ (Co. interior angle)
$65^\circ+\angle4=80^\circ$
$\angle4=115^\circ\text{(i)}$
$\angle4=\angle5=115^\circ$ (Vertically opposite angle)
$\angle2+\angle5+\angle3=180^\circ$
$20^{\circ} + 115^{\circ}+ x = 180^{\circ}$
$x = 45^{\circ}$.

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MCQ 2241 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
$18^\circ$
B
$54^\circ$
C
$72^\circ$
✓
$36^\circ$

Answer

Correct option: D.

$36^\circ$

According to question,
$\angle\text{AQP}=\angle\text{BQR}=\text{x}$
$\angle\text{AQP}+\angle\text{BQR}+\angle\text{PQR}=180^\circ$ (Linear Pair)
$2x + 108^\circ = 180^\circ$
$x = 36^\circ $.

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MCQ 2251 Mark

In the adjoining figure, if $m \| n$, then $\angle4+\angle7$ is equal to:

✓
$180^\circ$
B
$120^\circ$
C
$90^\circ$
D
$150^\circ$

Answer

Correct option: A.

$180^\circ$

$\angle1=\angle5$ (Corresponding angle)
$\angle7=\angle5$ (Vertically Opposite angle)
$\angle4+\angle1=180^\circ$ (Linear Pair)
$\angle4+\angle7=180^\circ$ (From above equations).

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MCQ 2261 Mark

In the given figure, $AB \| CD$. If $\angle\text{EAB}=50^\circ$ and $\angle\text{ECD}=60^\circ$ then $\angle\text{AEB}=?$

A
$50^\circ$
B
$60^\circ$
✓
$70^\circ$
D
$55^\circ$

Answer

Correct option: C.

$70^\circ$

Let $\angle\text{AEB}=\text{x}^\circ$
Now, $AB \| CD$ and $BC$ is the transversal
$\therefore\angle\text{ABE}=\angle\text{BCD}=60^\circ$ (Alterante angles)
In $\triangle\text{ABE},$
$\angle\text{BAE}+\angle\text{AEB}+\angle\text{ABE}=180^\circ$ (Angle sum property)
$\Rightarrow50^\circ+\text{x}^\circ+60^\circ=180^\circ$
$\Rightarrow\text{x}=70^\circ$
$\therefore\angle\text{AEB}=70^\circ$

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MCQ 2271 Mark

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

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

Answer

Correct option: A.

$130^\circ$

Construction: Through $O$, draw $OE \| AB \| CD$
$\Rightarrow\angle\text{BAO}+\angle\text{EOA}=180^\circ$
$\Rightarrow100^\circ+\angle\text{EOA}=180^\circ$
$\Rightarrow\angle\text{EOA}=80^\circ$
So, $\angle\text{EOC}=\angle\text{EAO}-\angle\text{COA}=80^\circ-30^\circ=50^\circ$
Since $CD \| EO$
$\angle\text{OCD}+\angle\text{EOC}=1806^\circ$
$\Rightarrow\angle\text{OCD}+50^\circ=180^\circ$
$\Rightarrow\angle\text{OCD}=130^\circ$

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Maths M.C.Q