Maths M.C.Q

4. 155º

Solution:

Let $\angle\text{A}=130^\circ$

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

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

$\Rightarrow\angle\text{B}+\angle\text{C}+130^\circ=180^\circ$

$\Rightarrow\angle\text{B}+\angle\text{C}=50^\circ$

$\Rightarrow\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}=25^\circ$

Now, in $\triangle\text{BOC},$

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

$\Rightarrow25^\circ+\angle\text{BOC}=180^\circ$

$\Rightarrow\angle\text{BOC}=155^\circ$

2. 54

Solution:

AOB is a straight line.

$\therefore$ xº + yº 90º = 180º

⇒ x + y = 90 .....(i)

Since the angles around a point sum up to 360º,

⇒ xº + 90º + yº + 72º + 3xº = 360º

⇒ 4x + y = 198 .....(ii)

Subtracting (i) from (ii), we get

3x = 108 ⇒ x = 36º

Substituting in (i), we get

y = 54º

2. 30º

Solution:

Let the measure of the angle be xº.

So, its supplement = (180º - x)

According to the given condition,

$\text{x}=\frac{1}{5}(180^\circ-\text{x)}$

$\Rightarrow5\text{x}=180-\text{x}$

$\Rightarrow6\text{x}=180$

$\Rightarrow\text{x}=30^\circ$

2. 86º

Solution:

Since AOB is a straight line,

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

$\Rightarrow(3\text{x}+10)+(4\text{x}-26)=180^\circ$

$\Rightarrow7\text{x}-16=180^\circ$

$\Rightarrow7\text{x}=196$

$\Rightarrow\text{x}=28$

So, $\angle\text{BOC}=4 \text{x}-26=4(28)-26=86^\circ$

3. 50º

Solution:

Let $\angle\text{AOC}=\text{x}^\circ$

Draw YOZ || CD || AB.

Now, YO || AB and OA is the transversal.

$\Rightarrow\angle\text{YOA}=\angle\text{OAB}=60^\circ$ (alternate angles)

Again, OZ || CD and OC is the transversal.

$\Rightarrow\angle\text{COZ}+\angle\text{OCD}=180^\circ$ (interior angles)

$\Rightarrow\angle\text{COZ}+110^\circ=180^\circ$

$\Rightarrow\angle\text{COZ}=70^\circ$

Now, $\angle\text{YOZ}=180^\circ$ (straight angle)

$\Rightarrow\angle\text{YOA}+\angle\text{AOC}+\angle\text{COZ}=180^\circ$

$\Rightarrow60^\circ+\text{x}+70^\circ=1806^\circ$

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

$\Rightarrow\angle\text{AOC}=50^\circ$

2. $55^\circ$

Solution:

Let each interior opposite angle be x.

Then, x + x = 110° (Exterior angle property of a triangle)

⇒ 2x = 110°

⇒ x = 55°

3. 60º

Solution:

Through B draw YBZ || OA || CD.

Now, OA || YB and AB is the transversal.

$\Rightarrow\angle\text{OAB}+\angle\text{YBA}=180^\circ$ (interior angles are supplementary)

$\Rightarrow\angle\text{YBA}=70^\circ$

Also, CD || BZ and BC is the transversal.

$\Rightarrow\angle\text{DCB}+\angle\text{CBZ}=180^\circ$ (interior angles are supplementary)

2. 55º

Solution:

Since AB || CD,

$\Rightarrow\angle\text{ACE}=\angle\text{BAC}=80^\circ$

In $\triangle\text{CEF},$

$\angle\text{ACE}=\angle\text{CEF}+\angle\text{CFE}$ (Exterior angle is equal to sum of the remote interior angles)

$\Rightarrow80^\circ=\angle\text{CEF}+25^\circ$

$\Rightarrow\angle\text{CEF}=55^\circ$

3. 45º

Solution:

$\angle\text{AOD}=\angle\text{COB}=\theta$

$\angle\text{AOC}=\angle\text{BOD}=\phi$

Since the sum of the measures of the angles around a point is 360º,

$\angle\text{AOD}+\angle\text{COB}+\angle\text{AOC}+\angle\text{BOD}=360^\circ$

$\Rightarrow\theta+\theta+\phi+\phi=360$

$\Rightarrow2(\theta+\phi)=360$

$\Rightarrow\theta+\phi=180$

Given that $\theta=3\phi.$

So, $3\phi+\phi=180$

$\Rightarrow4\phi=180$

$\Rightarrow\phi=45$

1. 60º

Solution:

The ratio of the angles is given to be 4 : 5 : 6.

So, let the measure of the angles be 4m, 5m and 6m.

Since AOB is a straight line,

$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$

$\Rightarrow4\text{m}+5\text{m}+6\text{m}=180^\circ$

$\Rightarrow15\text{m}=180$

$\Rightarrow\text{m}=12$

So, y = 5m = 5(12) = 60º.

3. 80º

Solution:

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, $\angle\text{AOC}=3\text{x}-10=3(30)-10=80^\circ$

2. 115º

Solution:

$\angle\text{AOC}+\angle\text{BOD}=1306^\circ$ (given)

But $\angle\text{AOC}=\angle\text{BOD}$ (Vartically Opposite angles)

$\Rightarrow2\angle\text{AOC}=130^\circ$

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

Since COD is a straight line,

$\angle\text{AOC}+\angle\text{AOD}=180^\circ$

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

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

1. 55º

Solution:

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)

$\Rightarrow120^\circ=70^\circ+\angle\text{QPR}$

$\Rightarrow\angle\text{QPR}=506^\circ$

3. 36º

Solution:

We know that, angle of incidance = angle reflection.

that is, $\angle\text{AQP}=\angle\text{BQR}$

Since AOB is a straight line,

$\angle\text{AQP}+\angle\text{BQR}+\angle\text{PQR}=180^\circ$

$\Rightarrow\angle\text{AQP}+\angle\text{AQP}+\angle\text{PQR}=180^\circ$

$\Rightarrow2\angle\text{AQP}=72$

$\Rightarrow \angle\text{AQP}=36^\circ$

2. 126º

Solution:

AB || CD

x + y = 180º and y = p (Vertically opposite angles)

Also, CD || EF and t is the transversal.

$\therefore$ p + z = 180º

⇒ y + z = 180º $(\because\text{p}=\text{y})$

$\therefore$ x + y = y + z

⇒ x = z

But y : z = 3 : 7

$\therefore\text{y}=\Big(180\times\frac{3}{10}\Big)=54^\circ$ and $\text{z}=\Big(180\times\frac{7}{10}\Big)=126^\circ$

x = 126º $(\because\text{x}=\text{z})$

3. 30º

Solution:

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)

$\Rightarrow50^\circ+100^\circ+\text{x}=180^\circ$

$\Rightarrow\text{x}=30^\circ$

3. 80º

Solution:

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, $\angle\text{AOC}=4\text{x}=4(20)=80^\circ$

2. 15

Solution:

AOB is a straight line.

$\Rightarrow\angle\text{AOB}=180^\circ$

⇒ 60° + 5x° + 3x° = 180° 

⇒ 60° + 8x° = 180° 

⇒ 8x° = 120° 

⇒ x = 15° 

3. 70º

Solution:

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)

$\Rightarrow50^\circ+\text{x}^\circ+60^\circ=180^\circ$

$\Rightarrow\text{x}=70^\circ$

$\therefore\angle\text{AEB}=70^\circ$

1. 130º

Solution:

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

$\angle\text{OCD}+\angle\text{EOC}=1806^\circ$

$\Rightarrow\angle\text{OCD}+50^\circ=180^\circ$

$\Rightarrow\angle\text{OCD}=130^\circ$