**1.(v)** Find the remainder when is divided by

(v)

**6. **In Fig. 6.44, the side QR of PQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR = QPR.

We have,
PQR is produced to a point S and bisectors of PQR and PRS meet at point T,
By exterior angle sum property,
PRS = P + PQR
Now,
................(i)
Since QT and QR are the bisectors of PQR and PRS respectively.
Now, in QRT,
..............(ii)
From eq (i) and eq (ii), we get
Hence proved

**5. **In Fig. 6.43, if PQ PS, PQ SR, SQR = 28° and QRT = 65°, then find the values of *x* and *y*.

We have,
PQ PS, PQ || SR, SQR = 28° and QRT = 65°
Now, In QRS, the side SR produced to T and PQ || RS
therefore, QRT = =
So,
Also, QRT = RSQ + SQR (By exterior angle property of a triangle)
Therefore, RSQ = QRT - SQR
Now, in PQS,
P + PQS + PSQ =

**4. **In Fig. 6.42, if lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75°, find SQT.

We have,
lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75°
In PRT, by using angle sum property
PRT + PTR + TPR =
So, PTR =
Since lines, PQ and RS intersect at point T
therefore, PTR = QTS (Vertically opposite angles)
QTS =
Now, in QTS,
By using angle sum property
TSQ + STQ + SQT =
So, SQT =

**3. **In Fig. 6.41, if AB DE, BAC = 35° and CDE = 53°, find DCE.

We have,
AB || DE, BAC = 35° and CDE = 53°
AE is a transversal so, BAC = AED =
Now, In CDE,
CDE + DEC + ECD = (By angle sum property)
Therefore, ECD =

**2. **In Fig. 6.40, X = 62°, XYZ = 54°. If YO and ZO are the bisectors of XYZ and XZY respectively of XYZ, find OZY and YOZ.

We have
X = , XYZ =
YO and ZO bisects the XYZ and XZY
Now, In XYZ, by using angle sum property
XYZ + YZX + ZXY =
So, YZX =
YZX =
and, OYZ = also, OZY =
Now, in OYZ
Y + O + Z = [Y = and Z = ]
So, YOZ =

**1. **In Fig. 6.39, sides QP and RQ of PQR are produced to points S and T respectively. If SPR = 135° and PQT = 110°, find PRQ.

Given,
PQR is a triangle, SPR =, PQT =
Now, TQP + PQR = (Linear pair)
So, PQR =
Since the side of QP of the triangle, PQR is produced to S
So, PQR + PRQ = (Exterior angle property of triangle)

**6. **In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB CD.

Draw a ray BL PQ and CM RS
Since PQ || RS (Given)
So, BL || CM and BC is a transversal
LBC = MCB (Alternate interior angles).............(i)
It is known that, angle of incidence = angle of reflection
So, ABL = LBC and MCB = MCD
..................(ii)
Adding eq (i) and eq (ii), we get
ABC = DCB
Both the interior angles are equal
Hence AB || CD

**5. **In Fig. 6.32, if AB CD, APQ = 50° and PRD = 127°, find *x* and *y*.

Given, AB || CD, APQ = and PRD =
PQ is a transversal. So,
APQ = PQR= (alternate interior angles)
Again, PR is a transversal. So,
y + = (Alternate interior angles)

**4. **In Fig. 6.31, if PQ ST, PQR = 110° and RST = 130°, find QRS.

[**Hint** : Draw a line parallel to ST through point R.]

Draw a line EF parallel to the ST through R.
Since PQ || ST and ST || EF
EF || PQ
PQR = QRF = (Alternate interior angles)
QRF = QRS + SRF .............(i)
Again, RST + SRF = (Interior angles of two parallels ST and RF)
(RST = , given)
Thus, QRS =

**3. **In Fig. 6.30, if AB CD, EF CD and GED = 126°, find AGE, GEF and FGE.

Given AB || CD, EFCD and GED =
In the above figure,
GE is transversal. So, that AGE = GED = [Alternate interior angles]
Also, GEF = GED - FED
=
Since AB is a straight line
Therefore, AGE + FGE =
So, FGE =

**2. **In Fig. 6.29, if AB CD, CD EF and *y : z = 3 : 7*, find *x*.

Given AB || CD and CD || EF and
therefore, AB || EF and (alternate interior angles)..............(i)
Again, CD || AB
.............(ii)
Put the value of in equation (ii), we get
Then
By equation (i), we get the value of

**1. **In Fig. 6.28, find the values of *x* and *y* and then show that AB CD.

Given that,
In the figure, CD and PQ intersect at F
Therefore, (vertically opposite angles)
PQ is a straight line. So,
Hence AB || CD (since and are alternate interior angles)

**6.**** **It is given that XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ZYP, find XYQ and reflex QYP.

Given that,
XYZ = and XY produced to point P and Ray YQ bisects ZYP
Now, XYP is a straight line
So, XYZ + ZYQ + QYP =
Thus reflex of QYP =
Since XYQ = XYZ + ZYQ [
=

**5. **In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

Given that,
POQ is a line, OR PQ and ROQ is a right angle.
Now, POS + ROS + ROQ = [since POQ is a straight line]
.............(i)
and, ROS + ROQ = QOS
..............(ii)
Add the eq (i ) and eq (ii), we get
hence proved

**4. **In Fig. 6.16, if* x + y = w + z*, then prove that AOB is a line.

Given that,
..............(i)
It is known that, the sum of all the angles at a point =
..............(ii)
From eq (i) and eq (ii), we get
Hence proved AOB is a line.

**3. **In Fig. 6.15, PQR = PRQ, then prove that PQS = PRT.

Given that,
ABC is a triangle such that PQR = PRQ and ST is a straight line.
Now, PQR + PQS = {Linear pair}............(i)
Similarly, PRQ + PRT = ..................(ii)
equating the eq (i) and eq (ii), we get
{but PQR = PRQ }
Therefore, PQS = PRT
Hence proved.

**2. **In Fig. 6.14, lines XY and MN intersect at O. If and a : b = 2 : 3, find c.

Given that,
Line XY and MN intersect at O and POY = also ..............(i)
Since XY is a straight line
Therefore, ...........(ii)
Thus, from eq (i) and eq (ii), we get
So,
Since MOY = c [vertically opposite angles]
a + POY = c

**1. **In Fig. 6.13, lines AB and CD intersect at O. If AOC + BOE = 70° and BOD = 40°, find BOE and reflex COE.

Given that,
AB is a straight line. Lines AB and CD intersect at O. and BOD =
Since AB is a straight line
AOC + COE + EOB =
[since ]
So, reflex COE =
It is given that AB and CD intersect at O
Therefore, AOC = BOD [vertically opposite angle]
[ GIven BOD = ]
Also,
So, BOE =

15.12) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?

(i) (a) All the points vibrate with the same frequency of 60 Hz.
(b) They all have the same phase as it depends upon time.
(c) At different points, the amplitude is different and is equal to A(x) given by
(ii)

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