Lines and Angles Test

Result

Lines and Angles Test - 28

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Question 1
1 / -0

In the given figure, $$\angle QPB$$ is,

A
$$60^{\circ}$$

B
$$45^{\circ}$$

C
$$30^{\circ}$$

D
$$15^{\circ}$$

Solution

In the given figure,
$$\angle APR + \angle RPQ + \angle QPB = 180^{\circ}$$
$$2x + 3x+x = 180^{\circ}$$
$$6x = 180^{\circ}$$
$$x = 30^{\circ}$$
Question 2
1 / -0

In the figure, if OP||RS, $$\angle OPQ=110^0 and \angle QRS =130^0$$, then $$\angle PQR$$ is equal to

A
$$40^0$$

B
$$50^0$$

C
$$60^0$$

D
$$70^0$$

Solution

As OP||RS,
$$\angle$$ PQR $$ =180 -(180-110)-(180-130) $$=$$60$$
Question 3
1 / -0

If two supplementary angles are in the ratio 2 : 7, then the angles are :

A
$$35^{\circ}, 145^{\circ}$$

B
$$70^{\circ}, 110^{\circ}$$

C
$$40^{\circ}, 140^{\circ}$$

D
$$50^{\circ}, 130^{\circ}$$

Solution

$$\textbf{ Hint: Sum of supplementary angles is}$$ $$180^\circ$$

$$\textbf{Step1: Evaluate using addition of ratio}$$
$$\text{Given}$$
$$\text{ratio}$$ $$=2:7$$
$$\text{ Let angles as }$$ $$2x$$ $$\text{and}$$ $$7x$$
$$\text{As we know that sum of supplementary angle is }180$$
$$\Rightarrow 2x+7x=180$$
$$\Rightarrow 9x=180$$
$$\Rightarrow x=20$$
$$2x\Rightarrow 2(20)=40$$
$$7x\Rightarrow 7(20)=140$$
$$\text{ Hence, the angles are}$$ $$40^\circ$$ $$140^\circ.$$
$$\textbf{Option C is correct.}$$
Question 4
1 / -0

In the figure, if $$\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = k\times $$ right angle, then $$k$$ is :

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

$$\Rightarrow$$ In $$\triangle ABC$$

$$\Rightarrow$$ $$\angle A+\angle B+\angle C=180^\circ$$ [Sum of interior angles of triangle is $$180^\circ$$] --- ( 1 )

$$\Rightarrow$$ In $$\triangle DEF$$

$$\Rightarrow$$ $$\angle D+\angle E+\angle F=180^\circ$$ [Sum of interior angles of triangle is $$180^\circ$$] ---( 2 )

$$\Rightarrow$$ $$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=180^\circ+180^\circ$$ [Adding ( 1 ) and ( 2 )]

$$\Rightarrow$$ $$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=360^\circ$$

$$\Rightarrow$$ $$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=k\times (right\, angle)$$.

$$\therefore$$ $$k=\dfrac{360^o}{90^o}$$

$$\therefore$$ $$k=4$$
Question 5
1 / -0

If one of the angles of a triangle is $$130^0$$, then the angle between the bisectors of the other two angles can be

A
$$50^0$$

B
$$65^0$$

C
$$145^0$$

D
$$155^0$$

Solution

Let $$x$$ and $$y$$ be the bisected angles.
So, in the original triangle,
sum of angles $$=180^o$$
Therefore, $$2x+2y+130=180$$
$$\Rightarrow 2(x+y)=50$$
$$\Rightarrow x+y=25$$
In the smallest triangle, consisting of original side opposite $$130^o$$.
Therefore, $$25^o+$$ Angle between bisectors $$=180^o$$
$$\Rightarrow $$ Angle between bisectors of other two angles $$=155^o$$.
Question 6
1 / -0

Two supplementary angles are in the ratio 4 : 5. The angles are

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

B
$$80^{\circ}, 100^{\circ}$$

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

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

Solution

Let the angles be $$4x$$ and $$5x$$
Angles are supplementary
$$\therefore 4x+5x={ 180 }^{ \circ }\\ \Rightarrow 9x={ 180 }^{ \circ }\\ \Rightarrow x={ 20 }^{ \circ }$$
So the angles are
$$4\times { 20 }^{ \circ }={ 80 }^{ \circ }\\ 5\times { 20 }^{ \circ }={ 100 }^{ \circ }$$
Question 7
1 / -0

The angle which exceeds its complement by $$20^{\circ}$$ is:

A
$$45^{\circ}$$

B
$$55^{\circ}$$

C
$$70^{\circ}$$

D
$$110^{\circ}$$

Solution

Let the required angle be $$x$$.
We knows, complementary angles $$=$$ Sum of two angles is $$90^o$$.
$$\therefore x = (90^o - x ) + 20^o $$
$$\therefore$$ $$x = 90^o - x + 20^o$$
$$\therefore$$ $$ 2x = 110^o$$
$$\therefore x = 55^o$$.
Hence, option $$B$$ is correct.
Question 8
1 / -0

From the adjoining figure, calculate the values of $$a$$.

A
$$46^o$$

B
$$38^o$$

C
$$56^o$$

D
none of the above

Solution

We know, by angle sum property, the sum of angles of triangle $$= 180^o$$
Then, $$a + 110^o + 32^o = 180^o $$
$$\implies$$ $$a + 142^o = 180^o $$
$$a =180^o-142^o=38^{\circ}$$.

Thus measure of angle $$a$$ is $$38^o$$.

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

In the given figure, AB // CD, EH // BC, $$\angle BAC = 60^{\circ}$$ and $$\ \angle DGH = 40^{\circ}$$. Find the measures of $$\angle AFG$$.

A
$$96^o$$

B
$$86^o$$

C
$$100^o$$

D
none of the above

Solution

Given that $$\angle BAC=60^o$$ and $$\angle DGH=40^o$$.
$$\implies$$ $$\angle EAF=60^o$$.

Also, given, $$AE \parallel DG$$.

Now, $$\angle AEF = \angle DGH = 40^{\circ}$$ (Corresponding angles).

Then, $$\angle AFG = \angle AEF + \angle EAF$$ (Exterior angle property)

$$\implies$$ $$\angle AFG = 40^{\circ} + 60^{\circ}$$

$$\implies$$ $$\angle AFG = 100^{\circ}$$.

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

In the given figure, AB // CD, EH // BC, $$\angle BAC = 60^{\circ}$$ and $$\ \angle DGH = 40^{\circ}$$. Find the measures of $$\angle AEF$$ .

A
$$36^o$$

B
$$40^o$$

C
$$46^o$$

D
none of the above

Solution

Given, $$\angle DGH = 40^{\circ}$$.

Since $$AB \parallel CD$$,
$$\implies$$ $$AE \parallel DG$$.

By corresponding angles axiom,
$$\angle AEF = \angle DGH = 40^{\circ}$$.

Hence, option $$B$$ is correct.