Lines and Angles Test

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Lines and Angles Test - 23

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Question 1
1 / -0

In the given figure $$AB\parallel CD$$ and $$EF\parallel GH$$. If $$\angle AIF={120}^{o}$$, then $$\angle CJG$$ is

A
$${80}^{o}$$

B
$${100}^{o}$$

C
$$\cfrac{1}{3}$$ of a right angle

D
$$\cfrac{2}{3}$$ of a right angle

Solution

$$\angle CLI={180}^{o}-\angle AIL\\\ \ \ \ \ \ \ \ \ \ ={180}^{o}-{120}^{o}\\ \ \ \ \ \ \ \ \ \ \ ={60}^{o}$$

($$AB\parallel CD$$ interior angles)

Now,
$$\angle CJG=\angle CLI\\\ \ \ \ \ \ \ \ \ \ ={60}^{o}\\\ \ \ \ \ \ \ \ \ \ =\cfrac{2}{3}\times {90}^{o}$$

($$EF\parallel GH$$, corresponding angles)
Question 2
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Supplementary angle of $$108.5^{\circ}$$ is

A
$$70.5^{\circ}$$

B
$$71.5^{\circ}$$

C
$$71^{\circ}$$

D
$$72.5^{\circ}$$

Solution

Given angle is $$= 108.5^{\circ}$$

Let the angle supplementary with the above angle be $$x$$

Now, the sum of two supplementary angles $$= 180^{\circ}$$

$$\Rightarrow x+108.5^{\circ}=180^{\circ}$$

$$\Rightarrow x=71.5^{\circ}$$
Question 3
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The measure of $$\angle ABC$$ is equal to:

A
$${75}^{o}$$

B
$${105}^{o}$$

C
$${115}^{o}$$

D
$${85}^{o}$$

Solution

As $$ AB $$ is a straight line cutting two parallel lines, it is called the transversal.
When a transversal cuts two parallel lines, the alternate exterior angles are equal.
This means $$ 2x + {35}^{o} = 5x - {25}^{o} $$
$$ => 3x = {60}^{o} $$
$$=> x = {20}^{o} $$
Now $$ \angle ABC + 2x + {35}^{o} =

{180}^{o} $$ , since they together form a straight line.
$$ =>\angle ABC ={180}^{o} - 2x - {35}^{o} $$
$$ =>\angle ABC ={180}^{o} - {40}^{o} - {35}^{o} $$
$$ => \angle ABC ={105}^{o} $$
Question 4
1 / -0

Complementary angle of $$72\displaystyle \frac{1}{2}^{\circ}$$ is:

A
$$17^{\circ}$$

B
$$18\displaystyle \frac{1}{2}^{\circ}$$

C
$$21\displaystyle \frac{1}{2}^{\circ}$$

D
$$17\displaystyle \frac{1}{2}^{\circ}$$

Solution

We know, two angles are complementary when they add up to $$90^o.$$

Given, measure of one complementary angle is $$72\dfrac{1}{2}^o=\dfrac{145}{2}^o.$$

$$\Rightarrow$$ Measure of other complementary angle $$=90^o-\dfrac{145}{2}^o$$ $$=\dfrac{35}{2}^o=17\dfrac{1}{2}^o.$$

$$\therefore$$ Measure of a complementary angle of $$72\dfrac{1}{2}^o=$$ $$17\dfrac{1}{2}^o.$$
Therefore, option $$D$$ is correct.
Question 5
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If the line $${l}_{1}$$ is parallel to line $${l}_{2}$$ in the given figure, what is the value of $$y$$?

A
$${141}^{o}$$

B
$$142^o$$

C
$$143^o$$

D
None of the above

Solution

$${l}_{1}\parallel {l}_{2}$$
$${115}^{\circ}+2x+3x={180}^{\circ}$$ (Co-interior angles)
$$5x={180}^{\circ}-{115}^{\circ}={65}^{\circ}$$
$$x={13}^{\circ}$$
Also, $$y+3x={180}^{\circ}$$ (co-interior angles)
$$y+{39}^{\circ}={180}^{\circ}$$
$$y={180}^{\circ}-{39}^{\circ}={141}^{\circ}$$
Question 6
1 / -0

By observing above figure, comment on nature of straight line AB and CD

A
Parallel

B
Not parallel as corresponding angles are not equal

C
Not parallel as interior angles are not equal

D
Not parallel as alternate angles are not equal

Solution

In the figure, alternate angles are not equal.
It is property of parallel lines that when a transversal intersects two parallel lines, its alternate angles are equal.
Question 7
1 / -0

The measure of angle $$y$$ in the given figure is ......... .

A
$$\angle y = 65^{\circ}$$

B
$$\angle y = 85^{\circ}$$

C
$$\angle y = 75^{\circ}$$

D
$$\angle y = 80^{\circ}$$

Solution

$$\textbf{Step 1: Using property of angle of straight line calculate y}$$
$$\text{We know that a straight line extends an angle of }{180^\circ}$$
$$\implies \angle \text{ABC}+\angle \text{CBD}=180^\circ$$
$$\implies \angle \text{CBD
}=180^\circ - \angle \text{ABC
}$$
$$\implies y^\circ=180^\circ - 115^\circ$$
$$\implies y^\circ=65^\circ$$

$$\mathbf{\text{Thus, the measure of angle y}=65^\circ}$$
Question 8
1 / -0

Choose the pair of complementary angles:

A
$$\displaystyle 66^{o} ,24^{o}$$

B
$$\displaystyle 30^{o},120^{o}$$

C
$$\displaystyle 60^{o},90^{o}$$

D
$$\displaystyle 15^{o},60^{o}$$

Solution

We know, two angles whose sum is equal to $$90^o$$ are known as complementary angles.

Consider option $$(A)$$.

The angles are $$66^o$$ and $$24^o$$.

Then their sum $$=66^o+24^o=90^o$$.

Hence, the angles are complementary.

Consider option $$(B)$$.

The angles are $$30^o$$ and $$120^o$$.

Then their sum $$=30^o+120^o=150^o\ne90^o$$.

Hence, the angles are not complementary.

Consider option $$(C)$$.

The angles are $$60^o$$ and $$90^o$$.

Then their sum $$=60^o+90^o=150^o\ne90^o$$.

Hence, the angles are not complementary.

Consider option $$(D)$$.

The angles are $$15^o$$ and $$60^o$$.

Then their sum $$=15^o+60^o=75^o\ne90^o$$.

Hence, the angles are not complementary.

Hence, only option $$A$$ is correct.
Question 9
1 / -0

In fig. lines AB and CD intersects each other at O. If $$\displaystyle \angle AOC={ 30 }^{ o }$$, find $$\displaystyle \angle BOD$$.

A
$$\displaystyle { 30 }^{ o }$$

B
$$\displaystyle { 150 }^{ o }$$

C
$$\displaystyle { 60 }^{ o }$$

D
$$\displaystyle { 75 }^{ o }$$

Solution

$$\displaystyle \angle AOC={ 30 }^{ o }$$ (given)
$$\displaystyle \because \quad \angle AOC$$ and $$\displaystyle \angle BOD$$ are vertically opposite angle
$$\displaystyle \therefore \quad \angle AOC=\angle BOD$$
$$\displaystyle \therefore \quad \angle BOD={ 30 }^{ o }$$
Question 10
1 / -0

In the figure if $$BD \parallel EF$$, then $$\angle CEF$$ is

A
$$100^o$$

B
$$120^o$$

C
$$140^o$$

D
$$160^o$$

Solution

Given $$BD|| EF$$ and $$ECD=40^{0}$$
As $$BD||EF$$ and line $$CE$$ join both line then
We know that $$\angle CEF+\angle ECD=180^{0}$$ {As interior angles are supplementary}
$$\Rightarrow \angle CEF=180-40=140^{0}$$.