NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3
NCERT Solutions for Class
9 Maths Chapter 6 Lines
and Angles Ex 6.3
are the part of NCERT
Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 6 Lines
and Angles Ex 6.3.
- NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1
- NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2
- NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3
Ex 6.3 Class 9 Maths Question
1.
In the given
figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Solution:
∠RPS + ∠RPQ = 180° [Linear pair of angles]
135°
+ ∠RPQ = 180° [∵ ∠RPS = 135° (given)]
⇒ ∠RPQ = 180° –
135° = 45°
Now, ∠RPQ + ∠PRQ = ∠PQT [Exterior angle = sum of interior opposite
angles]
⇒ 45° + ∠PRQ = 110°
⇒ ∠PRQ = 110° –
45° = 65°
Ex 6.3 Class 9 Maths Question
2.
In the
given figure, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the
bisectors of ∠XYZ
and ∠XZY respectively of
ΔXYZ, find ∠OZY
and ∠YOZ.
Solution:
But ∠XYZ = 54° and ∠ZXY = 62°
∴ 54° + ∠YZX + 62° = 180°
⇒ ∠YZX = 180° – 54° – 62° = 64°
YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
∴ ∠OYZ = ½ ∠XYZ = ½ (54°) = 27°
and ∠OZY = ½ ∠YZX = ½ (64°) = 32°
Now, in ∆OYZ, we have
∠YOZ + ∠OYZ + ∠OZY = 180° [Angle sum property of a triangle]
⇒ ∠YOZ + 27° + 32° = 180°
⇒ ∠YOZ = 180° – 27° – 32° = 121°
Thus, ∠OZY = 32° and ∠YOZ = 121°
Ex 6.3 Class 9 Maths Question
3.
In figure,
if AB || DE, ∠BAC = 35°and ∠CDE = 53°, find ∠DCE.
Solution:
We have, AB
|| DE
⇒ ∠AED = ∠BAE [Alternate interior angles]
Now, ∠BAE = ∠BAC
⇒ ∠BAE = 35° [ ∵ ∠BAC =
35°Given)]
⇒ ∠AED = 35°
In ΔDCE, ∠DCE + ∠CED + ∠EDC = 180°
[∵ Sum of all the
angles of a triangle is equal to 180°]
⇒ ∠DCE + 35° +
53° = 180° (∵∠AED = ∠CED = 35°)
⇒ ∠DCE = 180 °
– (35° + 53°)
⇒ ∠DCE = 92°
Ex 6.3 Class 9 Maths Question
4.
In the
given figure, if lines PQ and RS intersect at point T such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, then find ∠SQT.
Solution:
Given,
∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°
In ΔPRT, ∠PRT + ∠RPT + ∠PTR = 180° …(i) [Using angle sum property of a triangle]
On putting ∠PRT = 40° and ∠RPT = 95° in eq. (i), we get
40° + 95° + ∠PTR = 180°
⇒ 135° + ∠PTR = 180°
⇒ ∠PTR = 180° – 135°
⇒ ∠PTR = 45°
Now, ∠PTR = ∠QTS [Vertically opposite angles]
∴ ∠QTS = 45°
In ΔTQS, ∠QTS + ∠TSQ + ∠SQT = 180° … (ii)
[Using
angle sum property of a triangle]
On
putting ∠QTS = 45°
and ∠TSQ = 75° in eq. (ii),
we get
45° + 75°
+ ∠SQT = 180°
⇒ 120° + ∠SQT = 180°
⇒ ∠SQT = 180° –
120°
∴ ∠SQT = 60°
Ex 6.3 Class 9 Maths Question
5.
In the
given figure, if PQ ⊥ PS, PQ ||
SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x
and y.
Solution:
For ΔQSR,
∠QRT is an exterior angle.
∠QRT = ∠SQR + ∠QSR [∵ Exterior angle = sum of interior opposite angles]
⇒ 65° = 28° + ∠QSR [∵ ∠QRT = 65° and ∠SQR = 28° (given)]
∠QSR = 65° – 28°
⇒ ∠QSR = 37°
Given PQ ||
SR and SQ is the transversal which intersects PQ and ST at Q and S,
respectively.
∴ ∠QSR = ∠PQS [Alternate interior angles]
⇒ x = 37°
Now, in
ΔPQS, ∠SPQ + ∠PQS + ∠PSQ = 180°
[Since, sum of all the angles of a triangle is 180°]
⇒ 90° + 37° + y =
180° [∵ PQ ⊥ PS ⇒ ∠SPQ = 90°]
⇒ 127° + y = 180°
⇒ y = 180° – 127° =
53°
Hence, x =
37° and y = 53°.
Ex 6.3 Class 9 Maths Question 6.
In figure,
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 = ½ ∠QPR.
In ∆PQR,
side QR is produced to S, so by exterior angle property,
∠PRS = ∠QPR + ∠PQR
⇒ ½ ∠PRS = ½ ∠QPR + ½ ∠PQR
⇒ ∠TRS = ½ ∠QPR + ∠TQR …(1)
[∵ QT and RT are bisectors of ∠PQR and ∠PRS respectively.]
Now, in
∆QRT, we have
∠TRS = ∠TQR + ∠QTR …(2)
[Exterior
angle property of a triangle]
From (1) and (2), we have ½ ∠QPR + ∠TQR = ∠TQR + ∠QTR
⇒ ½ ∠QPR = ∠QTR
Thus, ∠QTR = ½ ∠QPR. Hence proved.
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