Telangana SCERT Solution Class IX (9) Math Chapter 4 Lines and Angles Exercise 4.4
(1)
(i)
From the adjacent figure,
∠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,
AD II BC
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)