# Telangana SCERT Class 9 Math Solution Chapter 4 Lines and Angles Exercise 4.4

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

(1)

(i) ∠ABC + ∠BCA + ∠BAC = 180o [∵ sum of interior angles of a triangle is 180o]

Or, 60o + ∠BCA + 50o =180o

Or, ∠BCA = 180o -110o

Or, ∠BCA = 70o

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

x = 180o – 70o

= 110o

Alternate method

Exterior xo = ∠BAC + ∠ABC [∵ CD is extension of side BC of the △ABF]

= 50 + 60

= 110o

(ii) PH is the extension of the side EF of △EFG

Exterior Zo = ∠EGF + ∠FEG

=70o + 60o

= 130o

(iii) RS is the extension of the side QR of PQR

Extension yo = QPR + PQR

= 35o + 45o

= 80o (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 = 1050/3

= 35o

yo + 105o + 24o = 180o [∵co interior angle]

Or, y = 180o – (105 + 24o)

= 180o – 129o

= 51o

(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,

5yo = 2xo [∵alternate angles]

Or, yo = 2xo/5  ———– (i)

x – y = 30o [corresponding angle]

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

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

Or, 3x = 150

Or, x =  150o/3 = 50o

∴ x – y = 30o

Or, 50o – y = 30o

Or, y = 50o – 30o

= 20o

(6) Find the values of x and y in the figure. From the adjacent we see that

Exterior xo = 30o + yo

yo + 140o = 180o [∵supplementary angles]

yo = 180o – 140o

= 40o

xo = 30o + yo

= 30o + 40o

= 70o

(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

Xo = 30o [alternate angles]

yo = 45o + xo [exterior angle of a triangle is y]

= 45o + 30o

= 75o

(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 = 110o

∠RPS = 135o

∠QPR = 180o – ∠PRS

= 180o – 135o

= 45o

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

Or, ∠PRQ + 45o = 110o

Or, ∠PRQ = 110o – 45o

= 65o

(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 = 35o, ∠CDE = 53o

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

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

Or, ∠DCE + 53o + 35o = 180

Or, ∠OCE = 180o – 88o

Or, ∠DCE = 92o

(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

= 40o + 95o

= 135o

Exterior of △QTS

∠PTS = ∠TSQ + ∠SQT

Or, 135o = 75o + ∠SQT

Or, ∠SQT = 135o – 75o

Or, ∠SQT = 60o

(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 = 180o – (50o + 70o)     [∵sum of interior angles of a triangle is = 180o]

Therefore, ∠BAC = 180o – 120o

= 60o

Exterior of △ABC

x + y = ∠BAC + ∠ACB

Or, 2x = 60o + 70o

Or, x = 130o/2

= 65o

Exterior of △ABC

y + y = ∠ABC + ∠BAC

Or, 2y = 50o + 60o

Or, y =  = 55o

In △BOC

xo + yo + zo = 180o              [∵sum of interior angles of a is 180o]

Or, z = 180o – 65o – 55o

Or, z = 180o – 120o

Or, z = 60o

(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 = 28o, ∠TRQ = 65o

Exterior of ∠QSR

∠QRT = ∠QSR + ∠SQR

Or, 65o = ∠QSR + 28o

Or, ∠QSR = 65o – 28o

Or, ∠QSR = 37o

X = QSR = 37o [alternate angle]

In △PQS

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

Or, 90o + yo + xo = 180o    [∵PQ ⊥ PS]

Or, y = 180o – (90o + 37o)

Or, y = 180o – 127o

Or, y = 53o

(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, 125o = 2x + 3x

Or, 5x = 125o

Or, x =

Or, x = 25o

∠A = 2x = 2 x 25o = 50o

∠B = 3x = 3 x 25o = 75o

(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 = 35o, ∠BCE = 102o

In △BCA

Exterior ∠BCE = 35o + ∠ABC

Or, 102o = 35o + ∠ABC

Or, ∠ABC = 102o – 35o

Or, ∠ABC = 67o

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

Or, ∠BCA + 67o + 35o = 180o

Or, ∠BCA = 180o – 102o

Or, ∠BCA = 78o

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

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

Or, ∠CED + 102 = 180o

Or, ∠CED = 180o – 120o

Or, ∠CED = 78o

(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 = 36o, ∠BDA = 45o

∠AEC = 40o

In △ABC,

∠ACB and ∠ABC are equal since AB = AC

Let, ∠ABC = ∠ACB = x

In △ABC,

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

Or, 2x = 180o – 36o

Or, x = 144/2

Or, x = 72o

(i) ∠ABC = 72o

(ii) ∠ACB = 72o

∠ABD + ∠ABC = 180o [supplementary angles]

Or, ∠ABD = 180o – 72o

= 108o

Similarly, ∠ACE + ∠ACB = 180o

Or, ∠ACE + 72o = 180o

Or, ∠ACE = 108o

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

Or, ∠DAB + 108o + 45o = 180o

Or, <DAB = 180o – 153o

Or, ∠DAB = 27o

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

Or, ∠CAE + 180o + 40o = 180o

Or, ∠CAE = 180o – 148o

Or, ∠CAE = 32o

(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

= 34o + 62o

= 96o

In △CDE

Exterior ∠AED = y = ∠EDC + ∠ECD

= 24o + ∠ACD

= 24o + x

= 120o

Therefore, x = 960 and y = 1200 (Ans)

Updated: September 21, 2021 — 4:32 pm