M.C.Q

Question 11 Mark

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

3 : 4 : 6
4 : 3 : 2
2 : 3 : 4
6 : 4 : 3

Answer

4 : 3 : 2

Solution:
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.
⇒ A : B : C = 4 : 3 : 2

View full question & answer→

Question 21 Mark

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

240º
300º
320º
360º

Answer

360º

solution:
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)
$\Rightarrow\angle\text{ACD}+\angle\text{BAE}+\angle\text{CBF}=360^\circ$

View full question & answer→

Question 31 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 $\angle\text{B}=?$

32º
63º
53º
95º

Answer

53º

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

View full question & answer→

Question 41 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 $\angle\text{BOC}=?$

130º
100º
115º
120º

Answer

115º

Solution:
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ (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)
$\Rightarrow\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\angle\text{BOC}=180^\circ$
$\Rightarrow65^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=115^\circ$

View full question & answer→

Question 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, $\angle\text{ACB}=?$

65º
45º
55º
35º

Answer

65º

Solution:
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)
$\Rightarrow45^\circ+70^\circ+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=65^\circ$

View full question & answer→

Question 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:

Answer

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

View full question & answer→

Question 71 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 $\angle\text{A}=?$

160º
60º
80º
30º

Answer

60º

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

View full question & answer→

Question 81 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 $\text{x} ^\circ+\text{y}^\circ=?$

190º
230º
210º
270º

Answer

230º

Solution:
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)
$\Rightarrow70^\circ+\text{y}=180^\circ$
$\Rightarrow\text{y}=110^\circ$
$\therefore\text{x}+\text{y}=120^\circ+110 ^\circ=230^\circ$

View full question & answer→

Question 91 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 $\angle\text{A}=?$

50º
55º
65º
75º

Answer

75º

Solution:
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, $\angle\text{A}=75^\circ.$

View full question & answer→

Question 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
x + y + 180
180 - (x + y)
x + y + 360º

Answer

x + y - 180

Solution:
$\angle\text{ABF}+\angle\text{ABC}=180^\circ$ (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)
$\Rightarrow\text{z}=\text{x}+\text{y}-180^\circ$

View full question & answer→

Question 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 $\angle\text{ECD}=?$

60º
50º
40º
25º

Answer

60º

Solution:
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)
$=90^\circ+60^\circ$
$=150^\circ$
$\angle\text{ACE}=\angle\text{ECD}+\angle\text{ACD}$
$\therefore150^\circ=\angle\text{ECD}+90^\circ$
$\therefore\angle\text{ECD}=60^\circ$

View full question & answer→

Question 121 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:

Answer

Solution:
$\angle\text{ACB}=\angle\text{ACD}=180^\circ$ (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)
$\Rightarrow7\text{y}=\text{x}+3\text{y}$
$\Rightarrow7(15^\circ)=\text{x}+3(15^\circ)$
$\Rightarrow105^\circ=\text{x}+45^\circ$
$\Rightarrow\text{x}=60^\circ$

View full question & answer→

MATHS M.C.Q

Test yourself on this topic