**Telangana SCERT Solution Class IX (9) Math Chapter 4 Lines and Angles Exercise 4.4**

**(1)**

**(i)**

From the adjacent figure,

∠ABC + ∠BCA + ∠BAC = 180^{o} [∵ sum of interior angles of a triangle is 180^{o}]

Or, 60^{o} + ∠BCA + 50^{o} =180^{o}

Or, ∠BCA = 180^{o} -110^{o}

Or, ∠BCA = 70^{o}

Now, x + ∠BCA = 180^{o} [∵ supplementary angle]

x = 180^{o} – 70^{o}

= 110^{o}

Alternate method

Exterior x^{o} = ∠BAC + ∠ABC [∵ CD is extension of side BC of the △ABF]

= 50 + 60

= 110^{o}

**(ii)**

PH is the extension of the side EF of △EFG

Exterior Z^{o} = ∠EGF + ∠FEG

=70^{o} + 60^{o}

= 130^{o}

**(iii)**

RS is the extension of the side QR of PQR

Extension y^{o} = QPR + PQR

= 35^{o} + 45^{o}

= 80^{o}

**(3) In the given figure AB || CD; BC || DE then find the values of x and y.**

**Solution:**

Given,

AB ∥ CD

BC ∥ DE

3x = 105 [∵alternate angle]

Or, x = 105^{0}/3

= 35^{o}

y^{o} + 105^{o} + 24^{o} = 180^{o} [∵co interior angle]

Or, y = 180^{o} – (105 + 24^{o})

= 180^{o} – 129^{o}

= 51^{o}

**(4) In the adjacent figure BE ⊥ DA and CD⊥DA then prove that m∠1 ≅ m∠3.**

**Solution:**

Given,

BE ⊥ DA

CD ⊥ DA

BE ∥ CD

∵This perpendiculars drawn on the same line is parallel to each other.

∠1 = ∠3 [∵alternate angle]

**(5) Find the values of x, y for which the lines AD and BC become parallel**

**Solution:**

Given,

AD II BC

5y^{o} = 2x^{o} [∵alternate angles]

Or, y^{o} = 2x^{o}/5 ———– (i)

x – y = 30^{o} [corresponding angle]

Or, x – 2x/5 = 30^{o} [putting the value of y from equation i]

Or, (5x-2x )/5 = 30^{o}

Or, 3x = 150

Or, x = 150^{o}/3 = 50^{o}

∴ x – y = 30^{o}

Or, 50^{o} – y = 30^{o}

Or, y = 50^{o} – 30^{o}

= 20^{o}

**(6) Find the values of x and y in the figure.**

From the adjacent we see that

Exterior x^{o} = 30^{o} + y^{o}

y^{o} + 140^{o} = 180^{o} [∵supplementary angles]

y^{o} = 180^{o} – 140^{o}

= 40^{o}

x^{o} = 30^{o} + y^{o}

= 30^{o} + 40^{o}

= 70^{o}

**(7) In the given figure segments shown by arrow heads are parallel. Find the values of x and y.**

**Solution:** Segment with arrow heads are parallel

X^{o} = 30^{o} [alternate angles]

y^{o} = 45^{o} + x^{o} [exterior angle of a triangle is y]

= 45^{o} + 30^{o}

= 75^{o}

**(8) In the given figure sides QP and RQ of ΔPQR are produced to points S and T respectively.**

**If ∠RPS = 135° and ∠PQT = 110°, find ∠PRQ.**

**Solution:**

In the adjacent figure the sided QR and RQ are produced to point S and T respecting

∠PQT = 110^{o}

∠RPS = 135^{o}

∠QPR = 180^{o} – ∠PRS

= 180^{o} – 135^{o}

= 45^{o}

∠PRQ + ∠QPR = ∠PQT [Exterior angle of △PQR]

Or, ∠PRQ + 45^{o} = 110^{o}

Or, ∠PRQ = 110^{o} – 45^{o}

= 65^{o}

**(9) In the given figure, ∠X = 62°, ∠ZYX = 54°. In ΔXYZ If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively find ∠OZY and ∠YOZ.**

**Solution**

**(10) In the given figure if AB || DE, ∠ BAC = 35° and ∠CDE = 53°, find ∠DCE.**

Given,

AB ∥ DE

∠BAC = 35^{o}, ∠CDE = 53^{o}

∴∠CED = ∠BAC = 35^{o} [∵alternate angles]

∴∠DCE + ∠CDE + ∠CED = 180^{o} [∵sum of interior angles of a triangle is 180^{o}]

Or, ∠DCE + 53^{o} + 35^{o} = 180

Or, ∠OCE = 180^{o} – 88^{o}

Or, ∠DCE = 92^{o}

**(11) In the given figure if line segments PQ and RS intersect at point T, such that ∠TRP = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.**

**Solution:**

Exterior of △PTR

∠PTS = ∠PRT + ∠RPT

= 40^{o} + 95^{o}

= 135^{o}

Exterior of △QTS

∠PTS = ∠TSQ + ∠SQT

Or, 135^{o} = 75^{o} + ∠SQT

Or, ∠SQT = 135^{o} – 75^{o}

Or, ∠SQT = 60^{o}

**(12) In the adjacent figure, ABC is a triangle in which ∠B = 50° and ∠C = 70°. Sides AB and AC are produced. If ‘z’ is the measure of the angle between the bisectors of the exterior angles so formed, then find ‘z’.**

**Solution:** ∠BAC = 180^{o} – (50^{o} + 70^{o}) [∵sum of interior angles of a triangle is = 180^{o}]

Therefore, ∠BAC = 180^{o} – 120^{o}

= 60^{o}

Exterior of △ABC

x + y = ∠BAC + ∠ACB

Or, 2x = 60^{o} + 70^{o}

Or, x = 130^{o}/2

= 65^{o}

Exterior of △ABC

y + y = ∠ABC + ∠BAC

Or, 2y = 50^{o} + 60^{o}

Or, y = = 55^{o}

In △BOC

x^{o} + y^{o} + z^{o} = 180^{o} [∵sum of interior angles of a is 180^{o}]

Or, z = 180^{o} – 65^{o} – 55^{o}

Or, z = 180^{o} – 120^{o}

Or, z = 60^{o}

**(13) In the given figure if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠TRQ = 65°, then find the values of x**

**and y**

**Solution:**

Given,

PQ ⊥ PS, PQ ∥ SR

∠SQR = 28^{o}, ∠TRQ = 65^{o}

Exterior of ∠QSR

∠QRT = ∠QSR + ∠SQR

Or, 65^{o} = ∠QSR + 28^{o}

Or, ∠QSR = 65^{o} – 28^{o}

Or, ∠QSR = 37^{o}

X = QSR = 37^{o} [alternate angle]

In △PQS

∠QPS + ∠PSQ + ∠PQS = 180^{o} [sum of interior angles of △ is 180]

Or, 90^{o} + y^{o} + x^{o} = 180^{o} [∵PQ ⊥ PS]

Or, y = 180^{o} – (90^{o} + 37^{o})

Or, y = 180^{o} – 127^{o}

Or, y = 53^{o}

**(14) In the given figure ΔABC side AC has been produced to D. ∠ BCD = 125 degree and ∠A : ∠B = 2 : 3, find the measure of ∠A and ∠B**

**Solution:**

Given,

∠A : ∠B = 2 : 3

Let,

∠A = 2x; ∠B = 3x

Exterior of △ABC

∠BCD = ∠A + ∠B

Or, 125^{o} = 2x + 3x

Or, 5x = 125^{o}

Or, x =

Or, x = 25^{o}

∠A = 2x = 2 x 25^{o} = 50^{o}

∠B = 3x = 3 x 25^{o} = 75^{o}

**(15) In the adjacent figure, it is given that, BC || DE, ∠BAC = 35 degree and ∠BCE = 102 degree . Find the measure of (i) ∠BCA (ii) ∠ADE and (iii) ∠CED.**

**Solution:**

Given,

BC ∥ DE

∠BAC = 35^{o}, ∠BCE = 102^{o}

In △BCA

Exterior ∠BCE = 35^{o} + ∠ABC

Or, 102^{o} = 35^{o} + ∠ABC

Or, ∠ABC = 102^{o} – 35^{o}

Or, ∠ABC = 67^{o}

(i) ∴∠BCA + ∠ABC + ∠BAC = 180^{o} [∵sum of interior angles of a △ is 180^{o}]

Or, ∠BCA + 67^{o} + 35^{o} = 180^{o}

Or, ∠BCA = 180^{o} – 102^{o}

Or, ∠BCA = 78^{o}

(ii) ∠ADE = ∠ABC = 67^{o} [∵corresponding angle]

(iii) ∠CED + ∠BCE = 180^{o} [co interior angle]

Or, ∠CED + 102 = 180^{o}

Or, ∠CED = 180^{o} – 120^{o}

Or, ∠CED = 78^{o}

**(16) In the adjacent figure, it is given that AB =AC, ∠BAC = 36 degree , ∠BDA = 45 degree and ∠ AEC = 40 degree. Find (i) ∠ABC (ii) ∠ACB (iii) ∠DAB (iv) ∠EAC**

**Solution:**

Given,

AB = AC

∠BAC = 36^{o}, ∠BDA = 45^{o}

∠AEC = 40^{o}

In △ABC,

∠ACB and ∠ABC are equal since AB = AC

Let, ∠ABC = ∠ACB = x

In △ABC,

x + x + 36^{o} = 180^{o} [∵sum of interior angles of a is 180]

Or, 2x = 180^{o} – 36^{o}

Or, x = 144/2

Or, x = 72^{o}

(i) ∠ABC = 72^{o}

(ii) ∠ACB = 72^{o}

∠ABD + ∠ABC = 180^{o} [supplementary angles]

Or, ∠ABD = 180^{o} – 72^{o}

= 108^{o}

Similarly, ∠ACE + ∠ACB = 180^{o}

Or, ∠ACE + 72^{o} = 180^{o}

Or, ∠ACE = 108^{o}

(iii) ∠DAB + ∠ADB + ∠ABD = 180^{o} [∵sum of interior angles of a △is 180^{o}]

Or, ∠DAB + 108^{o} + 45^{o} = 180^{o}

Or, <DAB = 180^{o} – 153^{o}

Or, ∠DAB = 27^{o}

(iv) Similarly, ∠CAE + ∠ACE + ∠AEC = 180^{o}

Or, ∠CAE + 180^{o} + 40^{o} = 180^{o}

Or, ∠CAE = 180^{o} – 148^{o}

Or, ∠CAE = 32^{o}

**(17.) Using information given in the figure, calculate the value of x and y**

**Solution:**

Side BC of △ABC is clearly produced to CD

Side CE of △CDE is clearly produced to EA

In △ABC

Exterior ACD = x = ∠CAB + ∠CBA

= 34^{o} + 62^{o}

= 96^{o}

In △CDE

Exterior ∠AED = y = ∠EDC + ∠ECD

= 24^{o} + ∠ACD

= 24^{o} + x

= 120^{o}

Therefore, x = 96^{0} and y = 120^{0} (Ans)