M.C.Q

MCQ 11 Mark

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

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

Answer

Correct option: C.

$115^\circ$

$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ (Angle sum property)
$\Rightarrow50^\circ+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{C}=130^\circ$
$\Rightarrow\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}=65^\circ$
In $\triangle\text{OBC},$
$\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^\circ$ (Angle sum property)
$\Rightarrow\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\angle\text{BOC}=180^\circ$
$\Rightarrow65^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=115^\circ$

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MCQ 21 Mark

In a $\triangle\text{ABC},$ if $3\angle\text{A}=4\angle\text{B}=6\angle\text{C}$ then $A : B : C =?$

A
$3 : 4 : 6$
✓
$4 : 3 : 2$
C
$2 : 3 : 4$
D
$6 : 4 : 3$

Answer

Correct option: B.

$4 : 3 : 2$

Given that $3\angle\text{A}=4\angle\text{B}=6\angle\text{C}=\text{k}.$
$\Rightarrow\angle\text{A}=\frac{\text{k}}{3},\angle\text{B}=\frac{\text{k}}{4}\text{ and}\ \angle\text{C}=\frac{\text{k}}{6}$
$\Rightarrow\text{A}:\text{B}:\text{C}=\frac{\text{k}}{3}:\frac{\text{k}}{4}:\frac{\text{k}}{6}$
$\Rightarrow\text{A}:\text{B}:\text{C}=\frac{1}{3}:\frac{1}{4}:\frac{1}{3}$
The $LCM$ of $3, 4$ and 6 is $12.$
Multiply by 12 throughout.
$\Rightarrow A : B : C = 4 : 3 : 2$

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MCQ 31 Mark

The side $BC, CA$ and $AB$ of $\triangle\text{ABC}$ have been produced to $D, E$ and $F$ respectively. $\angle\text{BAE}+\angle\text{CBF}+\angle\text{ACD}=?$

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

Answer

Correct option: D.

$360^\circ$

Clearly, $\angle1+\angle\text{BAE}=180^\circ\ ....(\text{i)}$(Supplementary angles)
Also, $\angle2+\angle\text{CBF}=180^\circ\ .....(\text{ii)}$ (Supplementary angles)
And $\angle\text{3}+\angle\text{ACD}=180^\circ\ .....(\text{iii)}$ (Supplementary angles)
$\therefore(\angle1+\angle2+\angle3)+(\angle\text{BAE}\\+\angle\text{CBF}+\angle\text{ACD})=540^\circ$
$\Rightarrow180^\circ+(\angle\text{BAE}+\angle\text{CBF}+\angle\text{ACD})=540^\circ$ (Angle sum property)
$\Rightarrow\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=360^\circ$

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MCQ 41 Mark

In a $\triangle\text{ABC},$ if $\angle\text{A}-\angle\text{B}=42^\circ$ and $\angle\text{B}-\angle\text{C}=21^\circ$ then $\angle\text{B}=?$

A
$32^\circ$
B
$63^\circ$
✓
$53^\circ$
D
$95^\circ$

Answer

Correct option: C.

$53^\circ$

$\angle\text{A}-\angle\text{B}=42^\circ$
$\Rightarrow\angle\text{A}=\angle\text{B}+42^\circ$
$\angle\text{B}-\angle\text{C}=21^\circ$
$\Rightarrow\angle\text{C}=\angle\text{B}-21^\circ$
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{B}+42^\circ+\angle\text{B}+\angle\text{B}-21^\circ=180^\circ$
$\Rightarrow3\angle\text{B}=159$
$\Rightarrow\angle\text{B}=53^\circ$

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MCQ 51 Mark

In the given figure, the side $CB$ and $BA$ of $\triangle\text{ABC}$ have been produced to $D$ and $E$ respectively such that $\angle\text{ABD}=110^\circ$ and $\angle\text{CAE}=135^\circ.$ Then, $\angle\text{ACB}=?$

✓
$65^\circ$
B
$45^\circ$
C
$55^\circ$
D
$35^\circ$

Answer

Correct option: A.

$65^\circ$

Since $BAE$ is a straight line, we have
$\angle\text{BAC}+\angle\text{CAE}=180^\circ$ .....(Supplementary angles)
$\Rightarrow\angle\text{BAC}+135^\circ=180^\circ$
$\Rightarrow\angle\text{BAC}=45^\circ$
Since $CBD$ is a straight line, we have
$\angle\text{ABD}+\angle\text{ABC}=180^\circ$ ....(Supplmenetary angles)
$\Rightarrow110^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=70^\circ$
In $\triangle\text{ABC},$ we have
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ .....(Angle sum property)
$\Rightarrow45^\circ+70^\circ+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=65^\circ$

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MCQ 61 Mark

In the given figure, $\text{EAD}\perp\text{BCD}.$ Ray $FAC$ cuts ray $EAD$ at a point A such that $\angle\text{EAF}=30^\circ.$ Also, in $\triangle\text{BAC},\angle\text{BAC}=\text{x}^\circ$ and $\angle\text{ABC}=(\text{x}+10).$ Then, value of $x$ is:

A
$20$
✓
$25$
C
$30$
D
$35$

Answer

Correct option: B.

$25$

$\angle\text{EAF}=\angle\text{CAD}$ (vertically opposite angles)
$\Rightarrow\angle\text{CAD}=30^\circ$
In $\triangle\text{ABD},$ by angle sum property
$\angle\text{A}+\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow(\text{x}+30)^\circ+(\text{x}+10)^\circ+90^\circ=180^\circ$
$\Rightarrow2\text{x}+130^\circ=180^\circ$
$\Rightarrow2\text{x}=50^\circ$
$\Rightarrow\text{x}=25^\circ$

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MCQ 71 Mark

Side $BC$ of $\triangle\text{ABC}$ has been produced to $D$ on left and to $E$ on right-hand side of $BC$ such that $\angle\text{ABD}=125^\circ$ and $\angle\text{ACE}=130^\circ.$ Then $\angle\text{A}=?$

A
$50^\circ$
B
$55^\circ $
C
$65^\circ$
✓
$75^\circ$

Answer

Correct option: D.

$75^\circ$

Since $DE$ is a straights line,
$\angle\text{ACB}+\angle\text{ACE}=180^\circ$
$\Rightarrow\angle\text{ACB}+130^\circ=180^\circ$
$\Rightarrow\angle\text{ACB}=50^\circ$
In $\triangle\text{ABC},$
$\angle\text{ABD}=\angle\text{BAC}+\angle\text{ACB}$ ......(Exterior angle is equal to sum of the remote interior angles)
$\Rightarrow125^\circ=\angle\text{BAC}+50^\circ$
$\Rightarrow\angle\text{BAC}=75^\circ$
that is, $\angle\text{A}=75^\circ.$

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MCQ 81 Mark

In the given figure, side $BC$ of $\triangle\text{ABC}$ has been produced to a point $D.$ If $\angle\text{A}=3\text{y},\angle\text{B}=\text{x}^\circ,\angle\text{C}=5\text{y}^\circ$ and $\angle\text{CBD}=7\text{y}^\circ.$ Then, the value of $x$ is:

✓
$60$
B
$50$
C
$45$
D
$35$

Answer

Correct option: A.

$60$

$\angle\text{ACB}=\angle\text{ACD}=180^\circ$ (linear pair)
$\Rightarrow5\text{y}+7\text{y}=180^\circ$
$\Rightarrow12\text{y}=180^\circ$
$\Rightarrow\text{y}=15^\circ$
Now, $\angle\text{ACD}=\angle\text{ABC}+\angle\text{BAC}$ (Exterior angle property)
$\Rightarrow7\text{y}=\text{x}+3\text{y}$
$\Rightarrow7(15^\circ)=\text{x}+3(15^\circ)$
$\Rightarrow105^\circ=\text{x}+45^\circ$
$\Rightarrow\text{x}=60^\circ$

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MCQ 91 Mark

In a $\triangle\text{ABC},$ side $BC$ is produced to $D.$ If $\angle\text{ABC}=50^\circ$ and $\angle\text{ACD}=110^\circ$ then $\angle\text{A}=?$

A
$160^\circ $
✓
$60^\circ$
C
$80^\circ$
D
$30^\circ$

Answer

Correct option: B.

$60^\circ$

$\angle\text{ACD}=\angle\text{B}+\angle\text{A}$ (Exterior angle property)
$\Rightarrow110^\circ=50^\circ+\angle\text{A}$
$\Rightarrow\angle\text{A}=60^\circ$

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MCQ 101 Mark

In the given figure, two rays $BD$ and $CE$ intersect at a point A. The side $BC$ of $\triangle\text{ABC}$ have been produced on both sides to points $F$ and $G$ respectively. If $\angle\text{ABF}=\text{x}^\circ,\angle\text{ACG}=\text{y}^\circ$ and $\angle\text{DAE}=\text{z}^\circ$ then $z =?$

✓
$x + y - 180$
B
$x + y + 180$
C
$180 - (x + y)$
D
$x + y + 360^\circ $

Answer

Correct option: A.

$x + y - 180$

$\angle\text{ABF}+\angle\text{ABC}=180^\circ$ (linear pair)
$\Rightarrow\text{x}+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-\text{x}$
$\angle\text{ACG}+\angle\text{ACB}=180^\circ$ (linear pair)
$\Rightarrow\text{y}+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=180^\circ-\text{y}$
In $\triangle\text{ABC},$ by angle sum property
$\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$
$\Rightarrow(180^\circ-\text{x})+(180^\circ-\text{y})+\angle\text{BAC}=180^\circ$
$\Rightarrow\angle\text{BAC}-\text{x}-\text{y}+180^\circ=0$
$\Rightarrow\angle\text{BAC}=\text{x}+\text{y}-180^\circ$
Now, $\angle\text{EAD}=\angle\text{BAC}$ (vertically opposite angles)
$\Rightarrow\text{z}=\text{x}+\text{y}-180^\circ$

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MCQ 111 Mark

In a $\triangle\text{ABC},$ it is given that $\angle\text{A}:\angle\text{B}:\angle\text{C}=3:2:1$ and $\angle\text{ACD}=90^\circ.$ If $BC$ produced to $E$ then $\angle\text{ECD}=?$

✓
$60^\circ$
B
$50^\circ$
C
$40^\circ$
D
$25^\circ$

Answer

Correct option: A.

$60^\circ$

we know that
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ .....(Angle sum property)
$\therefore\angle\text{A}=\Big(180\times\frac{3}{6}\Big)=90^\circ$
$\angle\text{B}=\Big(180\times\frac{2}{6}\Big)=60^\circ\ \text{and}$
$\angle\text{C}=\Big(180\times\frac{1}{6}\Big)=30^\circ$
Now,
$\angle\text{ACE}=\angle\text{A}+\angle\text{B}$ .....(Exterior angle is equal to sum of the remote interior angles)
$=90^\circ+60^\circ$
$=150^\circ$
$\angle\text{ACE}=\angle\text{ECD}+\angle\text{ACD}$
$\therefore150^\circ=\angle\text{ECD}+90^\circ$
$\therefore\angle\text{ECD}=60^\circ$

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MCQ 121 Mark

In the given figure, lines $AB$ and $CD$ intersect at a point $O.$ The sides $CA$ and $OB$ have been produced to $E$ and repectively such that $\angle\text{OAE}=\text{x}^\circ$ and $\angle\text{DBF}=\text{y}^\circ.$

If $\angle\text{OCA}=80^\circ,\angle\text{COA}=40^\circ$ and $\angle\text{BDO}=70^\circ$ then $\text{x} ^\circ+\text{y}^\circ=?$

A
$190^\circ$
✓
$230^\circ $
C
$210^\circ$
D
$270^\circ$

Answer

Correct option: B.

$230^\circ $

In $\triangle\text{OAC},$ by angle sum property
$\angle\text{OCA}+\angle\text{COA}+\angle\text{CAO}=180^\circ $
$\Rightarrow80^\circ+40^\circ+\angle\text{CAO}=180^\circ$
$\Rightarrow\angle\text{CAO}=60^\circ$
$\angle\text{CAO}+\angle\text{OAE}=180^\circ$ (linear pair)
$\Rightarrow60^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=120^\circ$
$\angle\text{COA}=\angle\text{BOD}$ (vertically opposite angles)
$\Rightarrow\angle\text{BOD}=40^\circ$
In $\triangle\text{OBD},$ by angle sum property
$\angle\text{OBD}+\angle\text{BOD}+\angle\text{ODB}=180^\circ$
$\Rightarrow\angle\text{OBD}+40^\circ+70^\circ=180^\circ$
$\Rightarrow\angle\text{OBD}=70^\circ$
$\angle\text{OBD}+\angle\text{DBF}=180^\circ$ (linear pair)
$\Rightarrow70^\circ+\text{y}=180^\circ$
$\Rightarrow\text{y}=110^\circ$
$\therefore\text{x}+\text{y}=120^\circ+110 ^\circ=230^\circ$

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