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CBSE (CCE Type) MCQs
Class 9 NCERT Mathematics Solutions - Class IX CBSE Board Maths Questions | Lines and Angles - Exercise 6.3 Answers
CBSE Class IX NCERT Mathematics Solutions
Chapter 6 - Lines and Angles
IXth NCERT Mathematics Textbook Exercise 6.3 Solved
(Page 107, 108)
Q1: In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135º and ∠PQT = 110º, find ∠PRQ.
Ans: In ∆ PQR side QP and RQ are produced to point S and T respectively such that / SPR = 135O, / PQT = 110O. We have to find / PRQ.
As / SPR + / RPQ = 180O (linear pair of angles).
Or, 135O + / RPQ = 180O
Or, / RPQ = 180O – 135O = 45O
Now, / TPQ = / QPR + / PRQ
Or, 110O = 45O + / PRQ
Or, / PRQ = 65O
Q2: In the given figure, ∠X = 62º, ∠XYZ = 54º. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.
As the sum of all interior angles of a triangle is 180º, therefore, for ΔXYZ,
∠X + ∠Y + ∠Z = 180º
Or, 62º + 54º + ∠Z = 180º
Or, ∠Z = 180º − 116º = 64O
Now, ∠OZY = 64O/2 = 32º (as OZ is the angle bisector of ∠Z)
Similarly, ∠OYZ = 54O/2 = 27O
∠OYZ + ∠YOZ + ∠OZY = 180º
Or, 27º + ∠YOZ + 32º = 180º
Or, ∠YOZ = 180º − 59º = 121º
Q3: In the given figure, if AB || DE, ∠BAC = 35º and ∠CDE = 53º, find ∠DCE.
/ DEC = / BAC = 35O ………. (i) [Alternate interior angles]
/ CDE = 53O ………. (ii) [given]
In ∆CDE using angle sum property we have,
/ CDE + / DEC + / DCE = 180O
Or, 53O + 35O + / DCE = 180O
Or, / DCE = 92O
Q4: In the given figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40º, ∠RPT = 95º and ∠TSQ = 75º, find ∠SQT.
Ans: Applying angle sum property in ∆PRT we have,
/ PTR + / PRT + / RPT = 180O
Or, / PTR + 40O + 95O = 180O
Or, / PTR = 45O
Or, / QTS = / PTR = 45O [vertically opposite angles]
Applying angle sum property in ∆TSQ we have,
/ QTS + / TSQ + / SQT = 180O
Or, / SQT + 45O + 75O = 180O
Or, / SQT = 60O
Q5: In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28º and ∠QRT = 65º, then find the values of x and y.
/ QRT = / RQS + / QSR [as the exterior angle is equal to the sum of the two interior opposite angles].
Or, 65O = 28O + / QSR
Or, / QSR = 37O
Given that PQ ⊥ PS i.e. / QPS = 90O
Or, PQ || SR.
So, / QPS + / PSR = 180o [sum of the consecutive interior angles on the same side of the traversal is 180O].
Therefore, 90O + / PSR = 180O
Or, / PSR = 90O
Or, / PSR + / QSR = 90O
Or, y + / QSR = 90O
Or, y + 37O = 90O
Or, y = 53O
Now consider ∆PQS,
/ PQS + / QSP + / QPS = 180O
Or, x + 53O + 90O = 180O
Or, x = 37O
Q6: In the given 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.
/ QTR = ½ / QPR
As RT and QT are the bisectors of / PRS and / PQR respectively,
Therefore, / PRT = / TRS and / PQT = / TQR.
=> / TRS = / T + / TQR
=> 2/ TRS = 2/ T + 2/ TQR
=> / PRS = 2/ T + / PQR
=> / P + / PQR = 2/ T + / PQR
Or, / P = 2/ T
Or, / T = / P/2
[Exterior angle = Sum of opposite interior angle].
Posted by Dr. Abhijit Joardar
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