myhelper: RD Sharma 2020 solution class 9 chapter 10 Lines and Angles FBQS

RD Sharma 2020 solution class 9 chapter 10 Lines and Angles FBQS

FBQS

Page-10.57

Question 1:

An angle is $\frac{2}{3}$ times its supplementary angle, the measure of the angle is __________.

Answer 1:

Let the measure of angle be x.

∴ Supplement of the angle = 180º − x

It is given that,

Measure of the angle = $\frac{2}{3}$ × Supplement of the angle
$∴ x = \frac{2}{3} (180 ° - x) \Rightarrow 3 x = 360 ° - 2 x \Rightarrow 5 x = 360 ° \Rightarrow x = \frac{360 °}{5} = 72 °$

Thus, the measure of the angle is 72º.

An angle is $\frac{2}{3}$ times its supplementary angle, the measure of the angle is ___72º___.

Question 2:

If two complementary angles are in the ratio 7 : 11, the supplement of the larger angle is _________.

Answer 2:

Let the two complementary angle be 7x and 11x.

Now,

7x + 11x = 90º (Two angles are complementary if their sum is 90º)

⇒ 18x = 90º

$\Rightarrow x = \frac{90 °}{18} = 5 °$

$∴ 7 x = 7 \times 5 ° = 35 °$ and $11 x = 11 \times 5 ° = 55 °$

So, the measure of the larger angle is 55º.

∴ Supplement of the larger angle = 180º − 55º = 125º

If two complementary angles are in the ratio 7 : 11, the supplement of the larger angle is ___125º___.

Question 3:

If x° is the measure of an angle which is equal to its complement and y° is the measure of an angle which is equal to its supplement, then $\frac{x °}{y °}$ is equal to ___________.

Answer 3:

Two angles are complementary if their sum is 90º.
$∴ x ° = 90 ° - x ° \Rightarrow 2 x ° = 90 ° \Rightarrow x ° = 45 °$

Two angles are supplementary if their sum is 180º.
$∴ y ° = 180 ° - y ° \Rightarrow 2 y ° = 180 ° \Rightarrow y ° = 90 °$

$∴ \frac{x °}{y °} = \frac{45 °}{90 °} = \frac{1}{2}$

If x° is the measure of an angle which is equal to its complement and y° is the measure of an angle which is equal to its supplement, then $\frac{x °}{y °}$ is equal to $\frac{1}{2}$ .

Question 4:

In the given figure, if AB||CD || EF, PQ||RS, ∠ROD = 25° and ∠CQP = 60°, then ∠QRS = ____________.

Answer 4:

In the given figure, AB || CD and RQ is the transversal.

∴ ∠ARQ = ∠RQD (Pair of alternate interior angles)

∠RQD = 25º                (Given)

⇒ ∠ARQ = 25º .....(1)

Also, PQ || RS and QR is the transversal.

∴ ∠BRS = ∠PQC (Pair of alternate exterior angles)

∠PQC = 60º                (Given)

⇒ ∠BRS = 60º

Also,

∠BRS + ∠ARS = 180º         (Linear pair)

⇒ 60º + ∠ARS = 180º

⇒ ∠ARS = 180º − 60º = 120º    .....(2)

Now,

∠QRS = ∠ARQ + ∠ARS = 25º + 120º = 145º               [Using (1) and (2)]

Thus, the measure of ∠QRS is 145º.

In the given figure, if AB|| CD || EF, PQ || RS, ∠ROD = 25° and ∠CQP = 60°, then ∠QRS = ___145º___.

Page-10.58

Question 5:

In the given figure, POQ is a line. The value of x is ___________.

Answer 5:

It is given that, POQ is a straight line.

∴ ∠POR + ∠ROQ = 180º (Linear pair)

⇒ 40º + (4x + 3x) = 180º (∠ROQ = ∠ROS + ∠SOQ)

⇒ 7x = 180º − 40º = 140º

⇒ x = $\frac{140 °}{7}$ = 20º

Thus, the value of x is 20º.

In the given figure, POQ is a line. The value of x is ___20º___.

Question 6:

If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3, then the greater of the two angles is ________.

Answer 6:

Let the two interior angle be 2x and 3x.

If a transversal intersects two parallel lines, then each pair of consecutive interior angles are supplementary.

$∴ 2 x + 3 x = 180 ° \Rightarrow 5 x = 180 ° \Rightarrow x = \frac{180 °}{5} = 36 °$

So, 2x = 2 × 36º = 72º and 3x = 3 × 36º = 108º

Thus, the greater of the two angles is 108º.

If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3, then the greater of the two angles is ___108º___.

Question 7:

The bisectors of two adjacent supplementary angels are always at ________.

Answer 7:

OP is the bisector of ∠BOC.

∴ ∠BOP = ∠COP = $\frac{∠ BOC}{2}$ .....(1)

OQ is the bisector of ∠AOC

∴ ∠AOQ = ∠COQ = $\frac{∠ AOC}{2}$ .....(2)

It is given that,

∠AOC and ∠BOC are two adjacent supplementary angles.

∴ ∠BOC + ∠AOC = 180º

⇒ 2∠POC + 2∠COQ = 180º

⇒ ∠POC + ∠COQ = $\frac{180 °}{2}$ = 90º

⇒ ∠POQ = 90º

Thus, the bisectors of two adjacent supplementary angles are always at right angle i.e. 90º.

The bisectors of two adjacent supplementary angels are always at __right angle__.

Question 8:

The bisectors of a pair of vertically opposite angels are in the __________.

Answer 8:

Let AB and CD be two lines intersecting at O. Suppose OP and OQ are the bisectors of the vertically opposite angles ∠AOD and ∠BOC, respectively.

∴ ∠AOP = ∠DOP     .....(1)

∠BOQ = ∠COQ      .....(2)

Also, ∠BOD = ∠AOC        .....(3)            (Vertically opposite angles)

Now,

∠AOP + ∠DOP + ∠BOD + ∠BOQ + ∠COQ + ∠AOC = 360º                 (Sum of angles formed at a point is 360º)

⇒ ∠DOP + ∠DOP + ∠BOD + ∠BOQ + ∠BOQ + ∠BOD = 360º            [Using (1), (2) and (3)]

⇒ 2∠DOP + 2∠BOD + 2∠BOQ = 360º

⇒ 2(∠DOP + ∠BOD + ∠BOQ) = 360º

⇒ ∠POQ = $\frac{360 °}{2}$ = 180º

Thus, OP and OQ i.e. the bisectors of the pair of vertically opposite angles are in the same line.

The bisectors of a pair of vertically opposite angels are in the __same line__.

Question 9:

If a transversal intersects two parallel lines then each pair of consecutive interior angle is _________.

Answer 9:

AB and CD are two parallel lines intersected by a transversal PQ at R and S, respectively. Here, the pair of consecutive interior angles are ∠BRS and ∠DSR & ∠ARS and ∠CSR.

Now,

∠PRB + ∠BRS = 180º .....(1) (Linear pair)

∠PRB = ∠DSR .....(2) (Corresponding angles)

From (1) and (2), we have

∠DSR + ∠BRS = 180º

Similarly, ∠ARS + ∠CSR = 180º

Thus, if a transversal intersects two parallel lines, then each pair of consecutive interior angles are supplementary.

If a transversal intersects two parallel lines then each pair of consecutive interior angle is __supplementary__.

Question 10:

The angle whose supplement is three times the complement is __________.

Answer 10:

Let the measure of angle be x.

Supplement of the angle = 180º − x

Complement of the angle = 90º − x

It is given that,

Supplement of the angle = 3 × Complement of the angle

⇒ 180º − x = 3(90º − x)

⇒ 180º − x = 270º − 3x

⇒ 3x − x = 270º − 180º

⇒ 2x = 90º

⇒ x = $\frac{90 °}{2}$ = 45º

The angle whose supplement is three times the complement is ____45º____.

Question 11:

The supplement of an angle and the complement of another have a sum equal to half of a complete angle. If the greater angle is 10° more than the smaller, then the measures of the angles are ___________.

Answer 11:

Let the two angles be x and y (x > y).

Supplement of angle x = 180º − x

Complement of angle y = 90º − y

Now,

180º − x + 90º − y = $\frac{1}{2}$ × 360º

⇒ 270º − x − y = 180º

⇒ x + y = 90º .....(1)

Also,

x = y + 10º

⇒ x − y = 10º .....(2)

Adding (1) and (2), we get

x + y + x − y = 90º + 10º

⇒ 2x = 100º

⇒ x = $\frac{100 °}{2}$ = 50º

Substituting the value of x in (1), we get

50º + y = 90º

⇒ y = 90º − 50º = 40º

Thus, the measure of the angles are 40º and 50º.

The supplement of an angle and the complement of another have a sum equal to half of a complete angle. If the greater angle is 10° more than the smaller, then the measures of the angles are ___40º and 50º___.

Question 12:

If two parallel lines are intersected by a transversal then the bisectors of the interior angles form a _________.

Answer 12:

AB and CD are two parallel lines intersected by a transversal PQ at R and S, respectively.

RX is the bisector of ∠BRS and SX is the bisector of ∠DSR.

∴ ∠BRS = 2∠XRS            .....(1)

∠DSR = 2∠XSR            .....(2)

RY is the bisector of ∠ARS and SY is the bisector of ∠CSR.

∴ ∠ARS = 2∠YRS        .....(3) (RY is the bisector of ∠ARS)

∠CSR = 2∠YSR            .....(4) (SY is the bisector of ∠CSR)

We know that if a transversal intersects two parallel lines, then each pair of consecutive interior angles are supplementary.

∴ ∠BRS + ∠DSR = 180º

⇒ 2∠XRS + 2∠XSR = 180º         [Using (1) and (2)]

⇒ 2(∠XRS + ∠XSR) = 180º

⇒ ∠XRS + ∠XSR = 90º           .....(5)

In ∆RSX,

∠XRS + ∠XSR + ∠RXS = 180º              (Angle sum property of triangle)

⇒ 90º + ∠RXS = 180º                             [Using (5)]

⇒ ∠RXS = 180º − 90º = 90º         .....(6)

Similarly, ∠RYS = 90º                  .....(7)

Now,

∠BRS + ∠ARS = 180º                  (Linear pair)

⇒ 2∠XRS + 2∠YRS = 180º         [Using (1) and (3)]

⇒ 2(∠XRS + ∠YRS) = 180º

⇒ ∠XRY = 90º                          .....(8)

Similarly, ∠XSY = 90º              .....(9)

In quadrilateral XRYS,

∠RXS = ∠XRY = ∠RYS = ∠XSY = 90º     [From (6), (7), (8) and (9)]

∴ Quadrilateral XRYS is a rectangle.    (If each angle of a quadrilateral is 90º, then it is a rectangle)

If two parallel lines are intersected by a transversal then the bisectors of the interior angles form a __rectangle__.

Question 13:

AB is a straight line and P is a point on it. If a line PQ is drawn not coinciding with PA and PB, then ∠APQ and ∠BPQ are ___________.

Answer 13:

Here, ∠APQ and ∠BPQ are adjacent angles. Also, AP and BP are two opposite rays.

We know that two adjacent angles are said to form a linear pair, if their non-common arms are two opposite rays.

∴ ∠APQ + ∠BPQ = 180º (Linear pair of angles)

AB is a straight line and P is a point on it. If a line PQ is drawn not coinciding with PA and PB, then ∠APQ and ∠BPQ are __supplementary__.

Question 14:

If the arms of one angle are respectively parallel to the arms of another angle, then the two angles are either _______ or _______.

Answer 14:

Let ∠ABC and ∠PQR be two angles such that AB || PQ and BC || QR.

Suppose PQ and BC intersect at O.


Figure 1	Figure 2

Consider figure 1.

Here,

∠ABC = ∠POC    .....(1)            (Corresponding angle)

∠POC = ∠PQR     .....(2)           (Corresponding angle)

From (1) and (2), we have

∠ABC = ∠PQR

Consider figure 2.

∠ABC + ∠QOB = 180º      .....(3)       (If two lines are parallel, then the pair of adjacent interior angles are supplementary)

Also, ∠QOB = ∠PQR    .....(4)    (Alternate angles)

From (3) and (4), we have

∠ABC + ∠PQR = 180º

If the arms of one angle are respectively parallel to the arms of another angle, then the two angles are either __equal__ or __supplementary__.

Question 15:

In given figure, if OP || RS, ∠OPQ = 110° and ∠QRS = 130°, then ∠PQR = _________.

Answer 15:

Extend OP to intersect QR at T.

Now, OU || RS and RT is the transversal.

∴ ∠PTR = ∠SRT        (Pair of alternate angles)

⇒ ∠PTR = 130º    (∠SRT = 130º)

Now,

∠PTR + ∠PTQ = 180º          (Linear pair)

⇒ 130º + ∠PTQ = 180º

⇒ ∠PTQ = 180º − 130º = 50º

In ∆PQT,

∠OPQ = ∠PTQ + ∠PQT              (The exterior angle of a triangle is equal to the sum of its opposite interior angles)

⇒ 110º = 50º + ∠PQT

⇒ ∠PQT = 110º − 50º = 60º

Or ∠PQR = 60º

Thus, the measure of ∠PQR is 60º.

In given figure, if OP || RS, ∠OPQ = 110° and ∠QRS = 130°, then ∠PQR = ___60º___.