Maths M.C.Q

2. 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

4. 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)

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

3. 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 $\triangle\text{ABC},$

$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$

$\Rightarrow\angle\text{B}+42^\circ+\angle\text{B}+\angle\text{B}-21^\circ=180^\circ$

$\Rightarrow3\angle\text{B}=159$

$\Rightarrow\angle\text{B}=53^\circ$

3. 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)

$\Rightarrow\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\angle\text{BOC}=180^\circ$

$\Rightarrow65^\circ+\angle\text{BOC}=180^\circ$

$\Rightarrow\angle\text{BOC}=115^\circ$

1. 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)

$\Rightarrow45^\circ+70^\circ+\angle\text{ACB}=180^\circ$

$\Rightarrow\angle\text{ACB}=65^\circ$

2. 25

solution:

$\angle\text{EAF}=\angle\text{CAD}$ (vertically opposite angles)

$\Rightarrow\angle\text{CAD}=30^\circ$

In $\triangle\text{ABD},$ by angle sum property

$\angle\text{A}+\angle\text{B}+\angle\text{D}=180^\circ$

$\Rightarrow(\text{x}+30)^\circ+(\text{x}+10)^\circ+90^\circ=180^\circ$

$\Rightarrow2\text{x}+130^\circ=180^\circ$

$\Rightarrow2\text{x}=50^\circ$

$\Rightarrow\text{x}=25^\circ$

2. 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)

$\Rightarrow70^\circ+\text{y}=180^\circ$

$\Rightarrow\text{y}=110^\circ$

$\therefore\text{x}+\text{y}=120^\circ+110 ^\circ=230^\circ$

4. 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, $\angle\text{A}=75^\circ.$

1. 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)

$\Rightarrow\text{z}=\text{x}+\text{y}-180^\circ$

2. 60º

Solution:

$\angle\text{ACD}=\angle\text{B}+\angle\text{A}$ (Exterior angle property)

$\Rightarrow110^\circ=50^\circ+\angle\text{A}$

$\Rightarrow\angle\text{A}=60^\circ$

1. 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)

$=90^\circ+60^\circ$

$=150^\circ$

$\angle\text{ACE}=\angle\text{ECD}+\angle\text{ACD}$

$\therefore150^\circ=\angle\text{ECD}+90^\circ$

$\therefore\angle\text{ECD}=60^\circ$

1. 60

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)

$\Rightarrow7\text{y}=\text{x}+3\text{y}$

$\Rightarrow7(15^\circ)=\text{x}+3(15^\circ)$

$\Rightarrow105^\circ=\text{x}+45^\circ$

$\Rightarrow\text{x}=60^\circ$