Lines and Angles Test

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Lines and Angles Test - 18

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Question 1
1 / -0

Two angles are called adjacent if

A
they have a common vertex

B
they have a ray in common
C

their other arms lie on the opposite sides of the common arm
D
all the above

Solution

We know that, adjacent angles are angles that have a common vertex and a common side, but do not overlap. Also, their other arms lie on the opposite side of the common arm.

So, all the given options are correct.

Hence, option D is the answer.
Question 2
1 / -0

At $$4:15$$, the angle formed between the two hands of a clock is:

A
acute

B
obtuse

C
right angle

D
none of these

Solution

At $$4:15$$, the hour will be between $$4$$ and $$5$$.
And minute hand will be on $$3$$.
Thus, the angle formed between these two hands will be less than $$90^{o}$$.
Hence, they form a acute angle.
Question 3
1 / -0

$$\angle ABC$$ in the following figure is a / an:

A
acute angle

B
obtuse angle

C
reflex angle

D
straight angle

Solution

Angles larger than a straight angle but less than complete angle (i.e. between $$180^o$$ and $$360^o$$) are called reflex angles.

Here, the given $$\angle ABC$$ measures greater than $$180^o$$ but less than $$360^o$$.
Hence, the given angle is a reflex angle.
Therefore, option $$C$$ is correct.
Question 4
1 / -0

The complement of $$(90^o-a)$$ is :

A
$$-a$$

B
$$90^o+a$$

C
$$90^o-a$$

D
$$a$$

Solution

Let the complement be $$y$$.
We know, two angles are complementary if their sum is $${ 90 }^{ \circ }$$.
$$\therefore { 90 }^{ \circ }-a+y={ 90 }^{ \circ }\\ \Rightarrow y=a$$.
Hence, option $$D$$ is correct.
Question 5
1 / -0

In Fig, POQ is a line. The value of x is:

A
$$20^0$$

B
$$25^0$$

C
$$30^0$$

D
$$35^0$$

Solution

Sum of the angles forming a linear pair is $$180^\circ$$
Here angles measuring $$40^\circ, 4x$$ and $$3x$$ form a linear pair.
So, $$40^\circ+4x+3x=180^\circ$$
$$7x+40^\circ=180^\circ$$
$$x=20^\circ$$
Question 6
1 / -0

If the sum of two adjacent angles is $$100^{\circ}$$ and one of them is $$35^{\circ}$$, then the other is :

A
$$70^{\circ}$$

B
$$65^{\circ}$$

C
$$135^{\circ}$$

D
$$145^{\circ}$$

Solution

Let the other angle be $$x$$
Now their sum $$=100^{\circ}$$
$$\Rightarrow x+{ 35 }^{ \circ }={ 100 }^{ \circ }\\ \Rightarrow x={ 100 }^{ \circ }-{ 35 }^{ \circ }={ 65 }^{ \circ }$$
So the other angle is $$65^{\circ}$$
Question 7
1 / -0

Find the complement of the angle :
$$\dfrac{1}{4}$$ of a right angle.

A
$$67 \cdot 5^o$$

B
$$57 \cdot 5^o$$

C
$$37 \cdot 5^o$$

D
none of the above

Solution

We know, two angles are complementary when they add up to $$90^o.$$
Given, measure of one complementary angle is $$=\dfrac{1}{4}\times 90^o=\dfrac{45}{2}^o.$$
$$\Rightarrow$$ Measure of other complementary angle $$=90^o-$$ $$\dfrac{45}{2}^o$$ $$=\dfrac{180-45}{2}^o$$ $$=\dfrac{135}{2}^o$$ $$=67.5^o.$$
$$\therefore$$ Measure of a complementary angle of $$\dfrac{1}{4}$$ of a right angle $$=67.5^o$$.
Question 8
1 / -0

Find the angle which is $$20^o$$ more than its supplement.

A
$$90$$

B
$$95$$

C
$$100$$

D
$$105$$

Solution

Let the required angle be $$x$$, then its supplement $$=(180-x)\quad $$
Given that $$ x=(180-x)+20$$
$$ \Rightarrow 2x=200\\ \Rightarrow x={ 100 }^{ o }
$$
Question 9
1 / -0

Find the angle which is $$30^o$$ more than its complement.

A
$$50$$

B
$$55$$

C
$$60$$

D
$$65$$

Solution

Let the required angle be $$ x$$.
Then, its complement $$=(90^o-x)$$.
Given, the angle is $$30^o$$ more than its complement.
Then, $$ x=(90^o-x)+30^o$$
$$\Rightarrow x+x=90^o+30^o$$
$$ \Rightarrow 2x=120^o\\ \Rightarrow x=\dfrac { 120 }{ 2 } ^o\\ \Rightarrow x={ 60 }^{ o }$$.

Therefore, option $$C$$ is correct.
Question 10
1 / -0

If the sides of a triangle are produced, then the sum of the exterior angle i.e. $$\angle a+\angle b+\angle c$$ is equal to:

A
$${ 180 }^{ 0 }$$

B
$${ 360 }^{ 0 }$$

C
$$90^{ 0 }$$

D
$$270^{ 0 }$$

Solution

In $$\triangle ABC$$,
$$\angle a = \angle ABC + \angle ACB$$ (Exterior angle property)
$$\angle b = \angle ACB + \angle BAC$$ (Exterior angle property)
$$\angle C = \angle ABC + \angle BAC$$ (Exterior angle property)

Adding all the equations:
$$\angle a + \angle b + \angle c = 2 (\angle ABC + \angle ACB + \angle BAC)$$
$$\angle a + \angle b + \angle c = 2 (180^o)$$
$$\angle a + \angle b + \angle c = 360^{\circ}$$.

Therefore, option $$B$$ is correct.