Lines and Angles: 9 Math

mathematics Class Nine

NCERT Exercise 6.3: 9th math

Lines and Angles Class ninth math NCERT Exercise 6.3 Question (1) In figure, sides OP and RQ of △PQR are produced to point S and T respectively. If ∠SPR = 135^o and ∠PQT = 110^o, find ∠PRQ.

class 9th math lines and angles ncert exercise 6.3 question 1

Solution

Given, PQR is a triangle

In which ∠PQT = 110^o

And, ∠PRQ = 135^o

Then ∠PRQ = ?

∠PQT and ∠PQR are linear pair of angle and together form a straight angle.

Thus, ∠PQT + ∠PQR = 180^o

⇒ 110^o + ∠PQR = 180^o

⇒ ∠PQR = 180^o – 110^o

⇒ ∠PQR = 70^o - - - - - (i)

Similarly, ∠SPR and ∠QPR are linear pair of angle and together form a straight angle.

Thus, ∠SPR + ∠QPR = 180^o

⇒ 135^o + ∠QPR = 180^o

⇒ ∠QPR = 180^o – 135^o

⇒ ∠QPR = 45^o - - - - - - (ii)

Now, from the angle sum property of a triangle theorem, we know that The sum of the angles of a triangle is 180^o

Thus, in △PQR,

∠PQR + ∠QPR + ∠PRQ = 180^o

⇒ 70^o + 45^o + ∠PRQ = 180^o

⇒ 115^o + ∠PRQ = 180^o

⇒ ∠PRQ = 180^o – 115^o

⇒ ∠PRQ = 65^o Answer

Lines and Angles Class ninth math NCERT Exercise 6.3 Question (2) In figure, ∠X = 62^o, ∠XYZ = 54^o. If YO and ZO are the bisector of ∠XYZ and ∠XZY respectively of △XYZ, find ∠OZY and ∠YOZ.

class 9th math lines and angles ncert exercise 6.3 question 2

Solution

Given, ∠X = 62^o

And, ∠XYZ = 54^o

And YO and ZO are the bisector of ∠XYZ and ∠XZY respectively

Thus, ∠OZY and ∠YOZ = ?

From the angle sum property of a triangle theorem, we know that The sum of the angles of a triangle is 180^o

In △XYZ,

∠X + ∠Y + ∠Z = 180^o

⇒ 62^o + 54^o + ∠Z = 180^o

⇒ 116^o + ∠Z = 180^o

⇒ ∠Z = 180^o – 116^o

⇒ ∠Z = ∠XZY = 64^o - - - - - - (i)

As given in the question OY is the bisector of ∠XYZ

Thus, ∠OYZ = 1/2 ∠XYZ

⇒ ∠OYZ = 1/2 × 54^o

⇒ ∠OYZ = 27^o - - - - - - (i)

And according to question, OZ is the bisector of ∠XZY

Thus, ∠OZY = 1/2 ∠XZY

⇒ ∠OZY = 1/2 × 64^o

[From equation (i) ∠XZY = 64^o]

⇒ ∠OZY = 32^o - - - - - (iii)

Now, in △ OYZ,

∠OYZ + ∠OZY + ∠YOZ = 180^o

⇒ 27^o + 32^o + ∠YOZ = 180^o

⇒ 59^o + ∠YOZ = 180^o

⇒ ∠YOZ = 180^o – 59^o

⇒ ∠YOZ = 121^o

Thus, ∠OZY = 32^o and ∠YOZ = 121^o Answer

Lines and Angles Class ninth math NCERT Exercise 6.3 Question (3) In figure, if AB||DE, ∠BAC = 35^o and ∠CDE = 53^o, find ∠DCE.

class 9th math lines and angles ncert exercise 6.3 question 3

Solution

Given, AB||DE

And, ∠BAC = 35^o

And ∠CDE = 53^o

Then, ∠DCE = ?

Now, since AB||DE

And ∠CAB and ∠DEC are pair of alternate interior angles, and hence are equal.

i.e. ∠CAB = ∠DEC = 35^o

i.e. ∠DEC = 35^o

Now, in △

From the angle sum property of a triangle theorem, we know that The sum of the angles of a triangle is 180^o

∠CDE + ∠DEC + ∠DCE = 180^o

⇒ 53^o + 35^o + ∠DCE = 180^o

⇒ 88^o + ∠DCE = 180^o

⇒ ∠DCE = 180^{o – 88^o}

⇒ ∠DCE = 92^o Answer

Lines and Angles Class ninth math NCERT Exercise 6.3 Question (4) In figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40^o, ∠RPT = 95^o and ∠TSQ = 75^o, find ∠SQT.

class 9th math lines and angles ncert exercise 6.3 question 4

Solution

Given, ∠PRT = 40^o

And, ∠RPT = 95^o

And ∠TSQ = 75^o

Then, ∠SQT =?

From the angle sum property of a triangle theorem, we know that The sum of the angles of a triangle is 180^o

In triangle PRT,

∠RPT + ∠PRT + ∠RTP = 180^o

⇒ 95^o + 40^o + ∠RTP = 180^o

⇒ 135^o + ∠RTP = 180^o

⇒ ∠RTP = 180^o – 135^o

⇒ ∠RTP = 45^o - - - - - - (i)

In triangle PRT and triangle TQS,

∠RTP and ∠STP are vertically opposite angles and hence are equal

Thus, ∠RTP = ∠STP = 45^o

⇒ ∠STP = 45^o - - - - - - (ii)

Now, in triangle TQS,

∠SQT + ∠STQ + ∠TSQ = 180^o

[Because sum of all angles of a triangle is equal to 180^o]

⇒ ∠SQT + 45^o + 75^o = 180^o

⇒ ∠SQT + 120^o = 180^o

⇒ ∠SQT = 180^o – 120^o

⇒ ∠SQT = 60^o Answer

Lines and Angles Class ninth math NCERT Exercise 6.3 Question (5) In figure, if PQ `_|_` PS, PQ || SR, ∠SQR = 28^o and ∠QRT = 65^o, then find the values of x and y.

class 9th math lines and angles ncert exercise 6.3 question 5

Solution

Given, PQ `_|_` PS and PQ || SR

And, ∠SQR = 28^o

And ∠QRT = 65^o

Then values of x and y = ?

Since, PQ || SR and QR is transversal passing through these parallel lines

Thus, ∠QRT and ∠RQP are alternate interior angles, and hence are equal.

i.e. ∠RQP = ∠QRT

⇒ ∠RQP = 65^o - - - - - - (i)

Now, among angle RQP and angle SQR and angle SQP

∠SQP = ∠RQP – ∠SQR

⇒ SQP = x = 65^o – 28^o

[From equation (i) ∠RQP = 65^o and as given in question ∠SQR = 28^o]

⇒ SQP = x = 37^o - - - - (ii)

Now, in triangle SQP,

∠QPS = 90^o [Because PQ is perpendicular to PS]

From the angle sum property of a triangle theorem, we know that The sum of the angles of a triangle is 180^o

Thus, ∠y + ∠x + ∠QPS = 180^o

⇒ ∠y + 37^o + 90^o = 180^o

[From equation (ii) x = 37^o]

⇒ ∠y + 127^o = 180^o

⇒ ∠y = 180^o – 127^o

⇒ ∠y = 53^o

Thus, x = 37^o and y = 53^o Answer

Lines and Angles Class ninth math NCERT Exercise 6.3 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 = 1/2 ∠QPR.

class 9th math lines and angles ncert exercise 6.3 question 6

Solution

Given, QT is the bisector of ∠PQR

And, RT is the bisector of angle PRS

Thus, to prove ∠QTR = 1/2 ∠QPR

We know from the theorem of angle sum property of triangles which says that If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

Thus now, in triangle PQR,

Exterior angle PRS = sum of interior angles QPR and PQR

⇒ ∠PRS = ∠QPR + ∠PQR

After dividing both sides by 1/2, we get

class 9th math lines and angles ncert exercise 6.3 solution of question 6