Lines and Angles Test

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Lines and Angles Test - 38

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Question 1
1 / -0

The supplement of $$45^{o}$$ is

A
$$45^{o}$$

B
$$75^{o}$$

C
$$135^{o}$$

D
$$155^{o}$$

Solution

Two angles are said to be supplementary if the sum of their measures is $$180^{o}$$.
The given angle is $$45^{o}$$
Let the measure of its supplement be $$X^{o}$$.
Then,
$$\implies X+45=180$$
$$\implies X=180-45$$
$$\implies X=135$$
Hence, the supplement of the given angle measures $$135^{o}$$.
Question 2
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In the adjoining figure, what value of $$x$$ will make $$AOB$$ a straight line?

A
$$x$$ = $$30$$

B
$$x$$ = $$35$$

C
$$x$$ = $$25$$

D
$$x$$ = $$40$$

Solution

(b) $$x$$ = $$35$$
Because,
It is given that the angles of the side be $$(2x-10)^{o}$$ and $$(3x+15)^{o}$$.
= $$(2x-10) + (3x+15)$$ = $$180$$...$$[\because Linear\,pair]$$
= $$2x-10+3x + 15$$ = $$180$$
= $$5x + 5$$ = $$180$$
= $$5x$$ = $$180-5$$
= $$x$$ = $$175/5$$
=$$x$$ = $$35$$
Question 3
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In the given figure, what value of $$x$$ will make $$AOB$$ a straight line?

A
$$x$$ = $$50$$

B
$$x$$ = $$100$$

C
$$x$$ = $$60$$

D
$$x$$ = $$80$$

Solution

(d) $$x$$ = $$80$$
Because,
It is given that the angles of the side be $$55^{o}$$, $$45^{o}$$ and $$(x)^{o}$$.
= $$x^{o} + 55^{o} + 45^{o}$$ = $$180^{o}$$...$$[\because Linear\,pair]$$
= $$x$$ = $$180-100$$
= $$x$$ = $$80$$
Question 4
1 / -0

In the given figure, $$AOB$$ is a straight line and the ray $$OC$$ stands on it. If $$\angle BOC=132^{o}$$, then $$\angle AOC=?$$

A
$$68^{o}$$

B
$$48^{o}$$

C
$$42^{o}$$

D
None of these

Solution

Given, $$\angle BOC=132^{o}$$
$$\angle AOC$$ = ?

We can write,
$$\angle AOC + \angle BOC$$ = $$180^{o}$$....$$[\because Linear\,pair]$$
$$\Rightarrow \angle AOC + 132^{o}$$ = $$180^{o}$$
$$\Rightarrow \angle AOC$$ = $$180^{o}-132^{o}$$
$$\Rightarrow \angle AOC$$ = $$48^{o}$$

Hence, the required answer is $$48^o$$
Question 5
1 / -0

In the given figure, $$AOB$$ is a straight line, $$\angle AOC$$ = $$68^{o}$$ and $$\angle BOC$$ = $$x^{o}$$.Find x

A
$$32$$

B
$$22$$

C
$$112$$

D
$$132$$

Solution

Because,
= $$\angle BOC + \angle AOC$$ = $$180^{o}$$....$$[\because Linear\,pair]$$
=$$x^{o} + 68^{o}$$ = $$180^{o}$$
=$$x$$ = $$180-68$$
= $$x$$ = $$112$$
Question 6
1 / -0

The complement of $$80^{o}$$ is:

A
$$100^{o}$$

B
$$10^{o}$$

C
$$20^{o}$$

D
$$280^{o}$$

Solution

Two angles are said to be complementary, if the sum of their measures is $$90^{o}$$.
The given angle is $$80^{o}$$.
Let the measure of its supplement be $$x^{o}$$.
Then,
$$\implies x^{o}+80^{o}=90^{o}$$
$$\implies x=90^o-80^o$$
$$\implies x=10^o$$.
Hence, the supplement of the given angle measures $$10^{o}$$.
Therefore, option $$B$$ is correct.
Question 7
1 / -0

Triangle DEF in figure is right triangle with $$\angle E=90^{o}$$.
What type of angles are $$\angle D$$ and $$\angle F$$?

A
They are equal angles

B
They form a pair of adjacent angles

C
They are complementary angles

D
They are supplementary angles

Solution

$$\angle D+\angle E+\angle F=180^{o}$$ ....Angle sum property
$$\because \angle E=90^{o}$$
$$\Rightarrow \angle D+\angle F=180^{o}-90^{o}$$
$$\Rightarrow \angle D+\angle F=90^{o}$$
Therefore, angle $$D$$ and $$F$$ are complementary angles.
Question 8
1 / -0

Angles between South & West and South & East are

A
vertically opposite angles

B
complementary angles

C
making a linear pair

D
adjacent but not supplementary

Solution

We have,
Angle between South and West $$=90^o$$
Angle between South and East $$=90^o$$
Sum of the angles $$=180^o$$

$$\therefore$$ The angles between South and West and South and East are making a linear pair.

Hence, option (C) is correct.
Question 9
1 / -0

The angles between North & West and South & East are

A
complementary

B
supplementary

C
both are acute

D
both are obtuse

Solution

Option (b) is correct.
The angle between the North and West $$=90^o$$
The angle between the South and East $$=90^o$$
Therefore, the angle between North & West and South & East are supplementary angles as they add up to $$180^o$$.
Question 10
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The angle which makes a linear pair with an angle of $$61^\circ$$ is of

A
$$29^\circ$$

B
$$61^\circ$$

C
$$122^\circ$$

D
$$119^\circ$$

Solution

Let the angle making linear pair with $$61^\circ$$ be $$y$$. Then,

$$y+61^\circ =180^\circ$$ [Given as linear pair]
$$\Rightarrow y=180^\circ -61^\circ$$
$$\Rightarrow y=119^\circ$$

Hence, option $$D$$ is correct.