Class 9 NCERT Solutions – Chapter 6 Lines And Angles – Exercise 6.3
Question1: In Figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Solution:
Given: ∠TQP = 110°, ∠SPR = 135°
TQR is a Straight line as we can see in the figure
As we have studied in this chapter, TQP and PQR will form a linear pair
⇒ ∠TQP + ∠PQR = 180° ———-(i)
Putting the value of ∠TQP = 110° in Equation (i) we get,
⇒ 110° + ∠PQR = 180°
⇒ ∠PQR = 70°
Consider the ΔPQR,
Here, the side QP is extended to S and so, SPR forms the exterior angle.
Thus, ∠SPR (∠SPR = 135°) is equal to the sum of interior opposite angles. (Triangle property)
Or, ∠PQR + ∠PRQ = 135° ———(ii)
Now, putting the value of PQR = 70° in equation (ii) we get,
∠PRQ = 135° – 70°
Hence, ∠PRQ = 65°
Question 2: In Figure, ∠X = 62°, ∠ XYZ = 54°. If YO and ZO are the bisectors of XYZ and XZY respectively of Δ XYZ, find OZY and YOZ.
Solution:
Given: ∠X = 62°, ∠XYZ = 54°
As we have studied in this chapter,
We know that the sum of the interior angles of the triangle is 180°.
So, ∠X +∠XYZ +∠XZY = 180°
Putting the values as given in the question we get,
62°+54° + ∠XZY = 180°
Or, ∠XZY = 64°
Now, we know that ZO is the bisector so,
∠OZY = ½ XZY
∴ ∠OZY = 32°
Similarly, YO is a bisector and so,
∠OYZ = ½ XYZ
Or, ∠OYZ = 27° (As XYZ = 54°)
Now, as the sum of the interior angles of the triangle,
∠OZY +∠OYZ +O = 180°
Putting their respective values, we get,
∠O = 180°-32°-27°
Hence, ∠O = 121°
Question 3: In Figure, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Solution:
Given: AB || DE, ∠BAC = 35° and ∠CDE = 53°
Since, we know that AE is a transversal of AB and DE
Here, BAC and AED are alternate interior angles.
Hence, ∠BAC = ∠AED
∠BAC = 35° (Given)
∠AED = 35°
Now consider the triangle CDE. We know that the sum of the interior angles of a triangle is 180°.
∴ ∠DCE + ∠CED + ∠CDE = 180°
Putting the values, we get
∠DCE + 35° + 53° = 180°
Hence, ∠DCE = 92°
Question 4: In Figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Solution:
Given: ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°
In △PRT.
∠PRT +∠RPT + ∠PTR = 180° (The Sum of all the angles of Triangle is 180°)
⇒ ∠PTR = 45°
Now ∠PTR will be equal to STQ as they are vertically opposite angles.
⇒ ∠PTR = ∠STQ = 45°
Again, in triangle STQ,
⇒ ∠TSQ +∠PTR + ∠SQT = 180° (The Sum of all the angles of Triangle is 180°)
Solving this we get,
⇒ ∠SQT = 60°
Question 5: In Figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Solution:
Given: PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°
x + SQR = QRT (As they are alternate angles since QR is transversal)
Now, Putting the value of ∠SQR = 28° and ∠QRT = 65°
⇒ x + 28° = 65°
∴ x = 37°
It is also known that alternate interior angles are same and
⇒ QSR = x = 37°
Also,
⇒ QRS + QRT = 180° (As they form a Linear pair)
Putting the value of ∠QRT = 65° we get,
⇒ QRS + 65° = 180°
⇒ QRS = 115°
As we know that the sum of the angles in a quadrilateral is 360°.
⇒ P + Q + R + S = 360°
Putting their respective values, we get,
⇒ S = 360° – 90° – 65° – 115° = 900
In Δ SPQ
⇒ ∠SPQ + x + y = 1800
⇒ 900 + 370 + y = 1800
⇒ y = 1800 – 1270 = 530
Hence, y = 53°
Question 6: In Figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = ½ ∠QPR.
Solution:
Given: T is the bisector of ∠PQR and ∠PRS,
To Prove: ∠QTR = ½ ∠QPR
Proof:
Consider the ΔPQR.
∠PRS is an exterior angle.
∠QPR and ∠PQR are interior angles.
⇒ ∠PRS = ∠QPR + ∠PQR (According to triangle property)
⇒ ∠PRS – ∠PQR = ∠QPR ————(i)
Now, consider the ΔQRT,
∠TRS = ∠TQR + ∠QTR (Since exterior angle are equal)
⇒ ∠QTR = ∠TRS – ∠TQR
We know that QT and RT bisect ∠PQR and ∠PRS respectively.
So, ∠PRS = 2 ∠TRS and ∠PQR = 2∠TQR
Now,
⇒ ∠QTR = ½ ∠PRS – ½ ∠PQR
⇒ ∠QTR = ½ ∠(PRS – PQR)
From equation (i) we know that ∠PRS – ∠PQR = ∠QPR,
⇒ ∠QTR = ½ ∠QPR
Hence, Proved.
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