Lines and Angles Test

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Lines and Angles Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

If the straight lines AB and XY intersect at the point O and $$\angle AOX=3 \angle XOB$$, then the four angles formed at O are:

A
$$30^{\circ},30^{\circ},90^{\circ},210^{\circ}$$

B
$$30^{\circ},30^{\circ},150^{\circ},150^{\circ}$$

C
$$45^{\circ},135^{\circ},90^{\circ},90^{\circ}$$

D
$$45^{\circ},45^{\circ},135^{\circ},135^{\circ}$$

Solution

As AB is straight line, let angle AOX = X` then
So,$$3X+X=180°$$
$$x=45°$$
$$3X=135°$$
Opposite angle will be equal,
so, other 2 angles will be 45° and 135°
Answer (D) 45°,45°,135°,135°
Question 2
1 / -0

Find the angle which is $$60^o$$ more than its complement.

A
$$55$$

B
$$75$$

C
$$85$$

D
$$60$$

Solution

Let the required angle be $$ x$$.
Then, its complement $$=(90^o-x)$$.
Given, the angle is $$60^o$$ more than its complement.
Then, $$ x=(90^o-x)+60^o$$
$$\Rightarrow x+x=90^o+60^o$$
$$ \Rightarrow 2x=150^o\\ \Rightarrow x=\dfrac { 150 }{ 2 } ^o\\ \Rightarrow x={ 75 }^{ o }$$.

Therefore, option $$B$$ is correct.
Question 3
1 / -0

In figure, if $$l \parallel m$$, then the value of $$x$$ is:

A
$$30^o$$

B
$$45^o$$

C
$$50^o$$

D
$$60^o$$

Solution

Since $$l \parallel m$$, then
$$3y = 2y + 25^0$$ [Alternate $$\angle s$$]
$$\Rightarrow 3y - 2y = 25^0$$
$$\Rightarrow        y = 25^0$$ ......(1)
Also $$x + 15^0 = 2y + 25^0$$    [Vertically opposite $$\angle s$$]
$$\Rightarrow    x + 15^0 = 2\left(25\right) + 25^0$$ [using (1)]
$$\Rightarrow x + 15^0 = 50^0 + 25^0$$
$$\Rightarrow         x = 75^0 - 15^0$$
$$\Rightarrow        x = 60^0$$.(in degrees)
Question 4
1 / -0

Two supplementary angles differ by $$48^o$$. Then find these angles.

A
$$36^o, 84^o$$

B
$$46^o, 94^o$$

C
$$56^o, 104^o$$

D
$$66^o, 114^o$$

Solution

We know that sum of supplementary angles is $$ 180^o$$.
Let one angle be $$x$$ and other be $$ 180^o - x$$
Hence, $$x - (180^o- x) =48^o$$ ....(Given)
$$ \Rightarrow x- 180^o+ x= 48^o$$
$$ \Rightarrow 2x= 48^o+ 180^o= 228^o$$
$$\Rightarrow x= \dfrac { 228^o }{ 2 } = 114^o$$
Hence, other angle $$= 180^o- x= 180^o- 114^o= 66^o$$
Two angles are $$ 114^o$$ and $$66^o$$.
Question 5
1 / -0

Angles forming a linear pair can both be acute angles.

A
True

B
False

C
Ambiguous

D
Data Insufficient

Solution

Answer is option B (False)
Both Angles forming linear pair cannot be acute as they add up to form 180 degrees
.Hence one angle can be acute and other be obtuse or both the angles can be right angles if they form linear pair.
Hence the above statement is false.
Question 6
1 / -0

Find the angle which is $$80^o$$ more than its complement.

A
$$75$$

B
$$85$$

C
$$95$$

D
$$100$$

Solution

Let the required angle be $$ x$$.
Then, its complement $$=(90^o-x)$$.
Given, the angle is $$80^o$$ more than its complement.
Then, $$ x=(90^o-x)+80^o$$
$$\Rightarrow x+x=90^o+80^o$$
$$ \Rightarrow 2x=170^o\\ \Rightarrow x=\dfrac { 170 }{ 2 } ^o\\ \Rightarrow x={ 85 }^{ o }$$.

Therefore, option $$B$$ is correct.
Question 7
1 / -0

How many pairs of adjacent angles, in all, can you name in figure?

A
8

B
7

C
12

D
10

Solution

The pairs of adjacent angles are:

1. $$\angle AOB$$ and $$\angle BOC$$

2. $$\angle AOB$$ and $$\angle BOD$$

3. $$\angle AOB$$ and $$\angle BOE$$

4. $$\angle BOC$$ and $$\angle COD$$

5. $$\angle BOC$$ and $$\angle COE$$

6. $$\angle COD$$ and $$\angle AOC$$

7. $$\angle COD$$ and $$\angle DOE$$

8. $$\angle DOE$$ and $$\angle BOD$$

9. $$\angle DOE$$ and $$\angle AOD$$

10. $$\angle COE$$ and $$\angle AOC$$

Hence there are 10 pairs of adjacent angles.
Question 8
1 / -0

If two adjacent angles are equal, then each angle measures $$90^o$$.

A
True

B
False

C
Ambiguous

D
Data Insufficient

Solution

The answer is $$B$$
Two adjacent angles measure $$90^o$$, only when the lines are perpendicular to each other.
hence the above statement is False.
Question 9
1 / -0

If $$AB\parallel CD$$ and $$BC\parallel DE$$, then the find the value of $$x$$.

A
$$ { 85 }^{ o } $$

B
$$ { 180 }^{ o } $$

C
$$ { 95 }^{ o } $$

D
$$ { 75 }^{ o } $$

Solution

Given that,
$$AB\parallel CD$$ and $$BC\parallel DE.$$
$$\angle CDE={ 85 }^{ o }.$$

To find out,
$$\angle ABC$$

$$AB\parallel CD$$ and $$BC\parallel DE.$$
$$\therefore \quad \angle EDC+\angle BCD={ 180 }^{ o }\\ \Longrightarrow \angle BCD={ 180 }^{ o }-\angle EDC$$
$$\Longrightarrow \angle BCD={ 180 }^{ o }-{ 85 }^{ o }={ 95 }^{ o }$$ (sum of the internal same side angles $$=180^o$$)
$$\angle ABC=\angle BCD={ 95 }^{ o }$$ (alternate angles)

Hence, the value of $$x$$ is $$95^o$$.
Question 10
1 / -0

In the figure measure of two angles are given. If line ED $$\parallel$$ seg AB and E-C-D then find the values of $$x$$ in degrees.

A
$$35^o$$

B
$$55^o$$

C
$$65^o$$

D
$$95^o$$

Solution

$$In\quad the\quad above\quad figure\quad ED\parallel AB\\ x=\angle BAC=65(alternate\quad interior\quad angle)\\ x=65^0$$