Contents
 Question 1:
 Question 2:
 Question 3:
 Question 4:
 Question 5:
 Question 6:
 Question 7:
 Question 8:
 Question 9:
 Question 10:
 Question 11:
 Question 12:
 Question 13:
 Question 14:
 Question 15:
 Question 1:
 Question 2:
 Question 3:
 Question 4:
 Question 5:
 Question 6:
 Question 7:
 Question 8:
 Question 9:
 Question 10:
 Question 11:
 Question 12:
 Question 13:
 Question 14:
 Question 15:
 Question 1:
 Question 2:
 Question 3:
 Question 4:
 Question 5:
 Question 6:
 Question 7:
 Question 8:
 Question 9:
 Question 10:
 Question 11:
 Question 12:
 Question 13:
 Question 14:
 Question 15:
 Question 16:
 Question 17:
 Question 1:
 Question 2:
 Question 3:
 Question 4:
 Question 5:
 Question 6:
 Question 7:
 Question 8:
 Question 9:
 Question 10:
 Question 11:
 Question 12:
 Question 13:
 Question 14:
 Question 15:
 Question 16:
 Question 17:
 Question 18:
 Question 19:
 Question 20:
 Question 21:
 Question 22:
 Question 23:
 Question 24:
 Question 25:
 Question 26:
Angles, Linesand Triangles
Question 1:
Exercise 4A
 Angle: Two rays having a common end point form an angle.
 Interior of an angle: The interior of âˆ AOB is the set of all points in its plane, which lie on the same side of OA as B and also on same side of OB as A.
 Obtuse angle: An angle whose measure is more than 90Â° but less than 180Â°, is called an obtuse angle.
 Reflex angle: An angle whose measure is more than 180Â° but less than 360Â° is called a reflex angle.
 Complementary angles: Two angles are said to be complementary, if the sum of their measures is 90o.
 Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180Â°.
Question 2:
âˆ A = 36Â° 27â€² 46â€³ and âˆ B = 28Â° 43â€² 39â€³
âˆ´ Their sum = (36Â° 27â€² 46â€³) + (28Â° 43â€² 39â€³)
Therefore, the sum âˆ A + âˆ B = 65Â° 11â€² 25â€³
Question 3:
Let âˆ A = 36Â° and âˆ B = 24Â° 28â€² 30â€³ Their difference = 36Â° â€“ 24Â° 28â€² 30â€³
Thus the difference between two angles is âˆ A â€“ âˆ B = 11Â° 31â€² 30â€³
Question 4:

 Complement of 58o = 90o â€“ 58o = 32o
 Complement of 16o = 90 â€“ 16o = 74o

of a right angle =
Ã— 90o = 60o Complement of 60o = 90o â€“ 60o = 30o  1o = 60â€²
â‡’ 90o = 89o 60â€²
Complement of 46o 30â€² = 90o â€“ 46o 30â€² = 43o 30â€² (v) 90o = 89o 59â€² 60â€³
Complement of 52o 43â€² 20â€³ = 90o â€“ 52o 43â€² 20â€³
= 37o 16â€² 40â€³
(vi) 90o = 89o 59â€² 60â€³
âˆ´ Complement of (68o 35â€² 45â€³)
= 90o â€“ (68o 35â€² 45â€³)
= 89o 59â€² 60â€³ â€“ (68o 35â€² 45â€³)
= 21o 24â€² 15â€³
Question 5:
 Supplement of 63o = 180o â€“ 63o = 117o
 Supplement of 138o = 180o â€“ 138o = 42o

of a right angle =
Ã— 90o = 54o
âˆ´ Supplement of 54o = 180o â€“ 54o = 126o
 1o = 60â€²
â‡’ 180o = 179o 60â€²
Supplement of 75o 36â€² = 180o â€“ 75o 36â€² = 104o 24â€² (v) 1o = 60â€², 1â€² = 60â€³
â‡’ 180o = 179o 59â€² 60â€³
Supplement of 124o 20â€² 40â€³ = 180o â€“ 124o 20â€² 40â€³
= 55o 39â€² 20â€³
(vi) 1o = 60â€², 1â€² = 60â€³
â‡’ 180o = 179o 59â€² 60â€³
âˆ´ Supplement of 108o 48â€² 32â€³ = 180o â€“ 108o 48â€² 32â€³
= 71o 11â€² 28â€³.
Question 6:


Let the required angle be xo Then, its complement = 90o â€“ xo

âˆ´ The measure of an angle which is equal to its complement is 45o.


Let the required angle be xo Then, its supplement = 180o â€“ xo

âˆ´ The measure of an angle which is equal to its supplement is 90o.
Question 7:
Let the required angle be xo Then its complement is 90o â€“ xo
âˆ´ The measure of an angle which is 36o more than its complement is 63o.
Question 8:
Let the required angle be xo Then its supplement is 180o â€“ xo
âˆ´ The measure of an angle which is 25o less than its supplement is
Question 9:
Let the required angle be xo Then, its complement = 90o â€“ xo
âˆ´ The required angle is 72o.
Question 10:
Let the required angle be xo Then, its supplement is 180o â€“ xo
âˆ´ The required angle is 150o.
Question 11:
Let the required angle be xo
Then, its complement is 90o â€“ xo and its supplement is 180o â€“ xo That is we have,
âˆ´ The required angle is 60o.
Question 12:
Let the required angle be xo
Then, its complement is 90o â€“ xo and its supplement is 180o â€“ xo
âˆ´ The required angle is 45o.
Question 13:
Let the two required angles be xo and 180o â€“ xo. Then,
â‡’ 2x = 3(180 â€“ x)
â‡’ 2x = 540 â€“ 3x
â‡’ 3x + 2x = 540
â‡’ 5x = 540
â‡’ x = 108
Thus, the required angles are 108o and 180o â€“ xo = 180 o â€“ 108o = 72o.
Question 14:
Let the two required angles be xo and 90o â€“ xo. Then
â‡’ 5x = 4(90 â€“ x)
â‡’ 5x = 360 â€“ 4x
â‡’ 5x + 4x = 360
â‡’ 9x = 360
â‡’ x =
= 40
Thus, the required angles are 40o and 90o â€“ xo = 90 o â€“ 40o = 50o.
Question 15:
Let the required angle be xo.
Then, its complementary and supplementary angles are (90o â€“ x) and (180o â€“ x) respectively.
Then, 7(90o â€“ x) = 3 (180o â€“ x) â€“ 10o
â‡’ 630o â€“ 7x = 540o â€“ 3x â€“ 10o
â‡’ 7x â€“ 3x = 630o â€“ 530o
â‡’ 4x = 100o
â‡’ x = 25o
Thus, the required angle is 25o.
Question 1:
Exercise 4B
Since âˆ BOC and âˆ COA form a linear pair of angles, we have
âˆ BOC + âˆ COA = 180o
â‡’ xo + 62o = 180o
â‡’ x = 180 â€“ 62
âˆ´ x = 118o
Question 2:
Since, âˆ BOD and âˆ DOA form a linear pair.
âˆ BOD + âˆ DOA = 180o
âˆ´ âˆ BOD + âˆ DOC + âˆ COA = 180o
â‡’ (x + 20)o + 55o + (3x â€“ 5)o = 180o
â‡’ x + 20 + 55 + 3x â€“ 5 = 180
â‡’ 4x + 70 = 180
â‡’ 4x = 180 â€“ 70 = 110
â‡’ x =
= 27.5
âˆ´ âˆ AOC = (3 Ã— 27.5 â€“ 5)o = 82.55 = 77.5o And, âˆ BOD = (x + 20)o = 27.5o + 20o = 47.5o.
Question 3:
Since âˆ BOD and âˆ DOA from a linear pair of angles.
â‡’ âˆ BOD + âˆ DOA = 180o
â‡’ âˆ BOD + âˆ DOC + âˆ COA = 180o
â‡’ xo + (2x â€“ 19)o + (3x + 7)o = 180o
â‡’ 6x â€“ 12 = 180
â‡’ 6x = 180 + 12 = 192
â‡’ x =
= 32
â‡’ x = 32
â‡’ âˆ AOC = (3x + 7)o = (3 32 + 7)o = 103o
â‡’ âˆ COD = (2x â€“ 19)o = (2 32 â€“ 19)o = 45o
and âˆ BOD = xo = 32o
Question 4:
x: y: z = 5: 4: 6
The sum of their ratios = 5 + 4 + 6 = 15 But x + y + z = 180o
[Since, XOY is a straight line]
So, if the total sum of the measures is 15, then the measure of x is 5. If the sum of angles is 180o, then, measure of x =
Ã— 180 = 60
And, if the total sum of the measures is 15, then the measure of y is 4. If the sum of the angles is 180o, then, measure of y =
Ã— 180 = 48 And âˆ z = 180o â€“ âˆ x â€“ âˆ y
= 180o â€“ 60o â€“ 48o
= 180o â€“ 108o = 72o
âˆ´ x = 60, y = 48 and z = 72.
Question 5:
AOB will be a straight line, if two adjacent angles form a linear pair.
âˆ´ âˆ BOC + âˆ AOC = 180o
â‡’ (4x â€“ 36)o + (3x + 20)o = 180o
â‡’ 4x â€“ 36 + 3x + 20 = 180
â‡’ 7x â€“ 16 = 180o
â‡’ 7x = 180 + 16 = 196
â‡’ x =
= 28
âˆ´ The value of x = 28.
Question 6:
Since âˆ AOC and âˆ AOD form a linear pair.
âˆ´ âˆ AOC + âˆ AOD = 180o
â‡’ 50o + âˆ AOD = 180o
â‡’ âˆ AOD = 180o â€“ 50o = 130o
âˆ AOD and âˆ BOC are vertically opposite angles.
âˆ AOD = âˆ BOC
â‡’ âˆ BOC = 130o
âˆ BOD and âˆ AOC are vertically opposite angles.
âˆ´ âˆ BOD = âˆ AOC
â‡’ âˆ BOD = 50o
Question 7:
Since âˆ COE and âˆ DOF are vertically opposite angles, we have,
âˆ COE = âˆ DOF
â‡’ âˆ z = 50o
Also âˆ BOD and âˆ COA are vertically opposite angles. So, âˆ BOD = âˆ COA
â‡’ âˆ t = 90o
As âˆ COA and âˆ AOD form a linear pair,
âˆ COA + âˆ AOD = 180o
â‡’ âˆ COA + âˆ AOF + âˆ FOD = 180o [âˆ t = 90o]
â‡’ t + x + 50o = 180o
â‡’ 90o + xo + 50o = 180o
â‡’ x + 140 = 180
â‡’ x = 180 â€“ 140 = 40
Since âˆ EOB and âˆ AOF are vertically opposite angles So, âˆ EOB = âˆ AOF
â‡’ y = x = 40
Thus, x = 40 = y = 40, z = 50 and t = 90
Question 8:
Since âˆ COE and âˆ EOD form a linear pair of angles.
â‡’ âˆ COE + âˆ EOD = 180o
â‡’ âˆ COE + âˆ EOA + âˆ AOD = 180o
â‡’ 5x + âˆ EOA + 2x = 180
â‡’ 5x + âˆ BOF + 2x = 180
[âˆ´ âˆ EOA and BOF are vertically opposite angles so, âˆ EOA = âˆ BOF]
â‡’ 5x + 3x + 2x = 180
â‡’ 10x = 180
â‡’ x = 18
Now âˆ AOD = 2xo = 2 Ã— 18o = 36o
âˆ COE = 5xo = 5 Ã— 18o = 90o
and, âˆ EOA = âˆ BOF = 3xo = 3 Ã— 18o = 54o
Question 9:
Let the two adjacent angles be 5x and 4x. Now, since these angles form a linear pair. So, 5x + 4x = 180o
â‡’ 9x = 180o
â‡’ x =
= 20
âˆ´ The required angles are 5x = 5x = 5 20o = 100o and 4x = 4 Ã— 20o = 80o
Question 10:
Let two straight lines AB and CD intersect at O and let âˆ AOC = 90o.
Now, âˆ AOC = âˆ BOD [Vertically opposite angles]
â‡’ âˆ BOD = 90o
Also, as âˆ AOC and âˆ AOD form a linear pair.
â‡’ 90o + âˆ AOD = 180o
â‡’ âˆ AOD = 180o â€“ 90o = 90o
Since, âˆ BOC = âˆ AOD [Verticallty opposite angles]
â‡’ âˆ BOC = 90o
Thus, each of the remaining angles is 90o.
Question 11:
Since, âˆ AOD and âˆ BOC are vertically opposite angles.
âˆ´ âˆ AOD = âˆ BOC
Now, âˆ AOD + âˆ BOC = 280o [Given]
â‡’ âˆ AOD + âˆ AOD = 280o
â‡’ 2âˆ AOD = 280o
â‡’ âˆ AOD =
= 140o
â‡’ âˆ BOC = âˆ AOD = 140o
As, âˆ AOC and âˆ AOD form a linear pair. So, âˆ AOC + âˆ AOD = 180o
â‡’ âˆ AOC + 140o = 180o
â‡’ âˆ AOC = 180o â€“ 140o = 40o
Since, âˆ AOC and âˆ BOD are vertically opposite angles.
âˆ´ âˆ AOC = âˆ BOD
â‡’ âˆ BOD = 40o
âˆ´ âˆ BOC = 140o, âˆ AOC = 40o , âˆ AOD = 140o and âˆ BOD = 40o.
Question 12:
Since âˆ COB and âˆ BOD form a linear pair So, âˆ COB + âˆ BOD = 180o
â‡’ âˆ BOD = 180o â€“ âˆ COB â€¦. (1)
Also, as âˆ COA and âˆ AOD form a linear pair. So, âˆ COA + âˆ AOD = 180o
â‡’ âˆ AOD = 180o â€“ âˆ COA
â‡’ âˆ AOD = 180o â€“ âˆ COB â€¦. (2)
[Since, OC is the bisector of âˆ AOB, âˆ BOC = âˆ AOC] From (1) and (2), we get,
âˆ AOD = âˆ BOD (Proved)
Question 13:
Let QS be a perpendicular to AB. Now, âˆ PQS = âˆ SQR
Because angle of incident = angle of reflection
â‡’ âˆ PQS = âˆ SQR =
= 56o
Since QS is perpendicular to AB, âˆ PQA and âˆ PQS are complementary angles. Thus, âˆ PQA + âˆ PQS = 90o
â‡’ âˆ PQA + 56o = 90o
â‡’ âˆ PQA = 90o â€“ 56o = 34o
Question 14:
Given : AB and CD are two lines which are intersecting at O. OE is a ray bisecting the
âˆ BOD. OF is a ray opposite to ray OE.
To Prove: âˆ AOF = âˆ COF
Proof : Since
and
are two opposite rays,
is a straight line passing through O.
âˆ´ âˆ AOF = âˆ BOE and âˆ COF = âˆ DOE
[Vertically opposite angles] But âˆ BOE = âˆ DOE (Given)
âˆ´ âˆ AOF = âˆ COF
Hence, proved.
Question 15:
Given:
is the bisector of âˆ BCD and
is the bisector of âˆ ACD. To Prove: âˆ ECF = 90o
Proof: Since âˆ ACD and âˆ BCD forms a linear pair.
âˆ ACD + âˆ BCD = 180o
âˆ ACE + âˆ ECD + âˆ DCF + âˆ FCB = 180o
âˆ ECD + âˆ ECD + âˆ DCF + âˆ DCF = 180o
because âˆ ACE = âˆ ECD and âˆ DCF = âˆ FCB
2(âˆ ECD) + 2 (âˆ CDF) = 180o
2(âˆ ECD + âˆ DCF) = 180o
âˆ ECD + âˆ DCF =
= 90o
âˆ ECF = 90o (Proved)
Question 1:
Exercise 4C
Since AB and CD are given to be parallel lines and t is a transversal. So, âˆ 5 = âˆ 1 = 70o [Corresponding angles are equal]
âˆ 3 = âˆ 1 = 70o [Vertically opp. Angles]
âˆ 3 + âˆ 6 = 180o [Cointerior angles on same side]
âˆ´ âˆ 6 = 180o â€“ âˆ 3
= 180o â€“ 70o = 110o
âˆ 6 = âˆ 8 [Vertically opp. Angles]
â‡’ âˆ 8 = 110o
â‡’ âˆ 4 + âˆ 5 = 180o [Cointerior angles on same side]
âˆ 4 = 180o â€“ 70o = 110o
âˆ 2 = âˆ 4 = 110o [ Vertically opposite angles]
âˆ 5 = âˆ 7 [Vertically opposite angles] So, âˆ 7 = 70o
âˆ´ âˆ 2 = 110o, âˆ 3 = 70o , âˆ 4 = 110o, âˆ 5 = 70o, âˆ 6 = 110o, âˆ 7 = 70o and âˆ 8 = 110o.
Question 2:
Since âˆ 2 : âˆ 1 = 5 : 4.
Let âˆ 2 and âˆ 1 be 5x and 4x respectively.
Now, âˆ 2 + âˆ 1 = 180o , because âˆ 2 and âˆ 1 form a linear pair. So, 5x + 4x = 180o
â‡’ 9x = 180o
â‡’ x = 20o
âˆ´ âˆ 1 = 4x = 4 Ã— 20o = 80o And âˆ 2 = 5x = 5 Ã— 20o = 100o
âˆ 3 = âˆ 1 = 80o [Vertically opposite angles]
And âˆ 4 = âˆ 2 = 100o [Vertically opposite angles]
âˆ 1 = âˆ 5 and âˆ 2 = âˆ 6 [Corresponding angles] So, âˆ 5 = 80o and âˆ 6 = 100o
âˆ 8 = âˆ 6 = 100o [Vertically opposite angles] And âˆ 7 = âˆ 5 = 80o [Vertically opposite angles]
Thus, âˆ 1 = 80o, âˆ 2 = 100o, âˆ 3 = âˆ 80o, âˆ 4 = 100o, âˆ 5 = 80o, âˆ 6 = 100o, âˆ 7 = 80o and
âˆ 8 = 100o.
Question 3:
Given: AB  CD and AD  BC To Prove: âˆ ADC = âˆ ABC
Proof: Since AB  CD and AD is a transversal. So sum of consecutive interior angles is
180o.
â‡’ âˆ BAD + âˆ ADC = 180o â€¦.(i)
Also, AD  BC and AB is transversal. So, âˆ BAD + âˆ ABC = 180o â€¦.(ii) From (i) and (ii) we get:
âˆ BAD + âˆ ADC = âˆ BAD + âˆ ABC
â‡’ âˆ ADC = âˆ ABC (Proved)
Question 4:

Through E draw EG  CD. Now since EGCD and ED is a transversal.
So, âˆ GED = âˆ EDC = 65o [Alternate interior angles] Since EG  CD and AB  CD,
EGAB and EB is transversal.
So, âˆ BEG = âˆ ABE = 35o [Alternate interior angles] So, âˆ DEB = xo
â‡’ âˆ BEG + âˆ GED = 35o + 65o = 100o.
Hence, x = 100.

Through O draw OFCD.
Now since OF  CD and OD is transversal.
âˆ CDO + âˆ FOD = 180o
[sum of consecutive interior angles is 180o]
â‡’ 25o + âˆ FOD = 180o
â‡’ âˆ FOD = 180o â€“ 25o = 155o
As OF  CD and AB  CD [Given] Thus, OF  AB and OB is a transversal.
So, âˆ ABO + âˆ FOB = 180o [sum of consecutive interior angles is 180o]
â‡’ 55o + âˆ FOB = 180o
â‡’ âˆ FOB = 180o â€“ 55o = 125o
Now, xo = âˆ FOB + âˆ FOD = 125o + 155o = 280o.
Hence, x = 280.

Through E, draw EF  CD.
Now since EF  CD and EC is transversal.
âˆ FEC + âˆ ECD = 180o
[sum of consecutive interior angles is 180o]
â‡’ âˆ FEC + 124o = 180o
â‡’ âˆ FEC = 180o â€“ 124o = 56o
Since EF  CD and AB CD
So, EF  AB and AE is a trasveral.
So, âˆ BAE + âˆ FEA = 180o
[sum of consecutive interior angles is 180o]
âˆ´ 116o + âˆ FEA = 180o
â‡’ âˆ FEA = 180o â€“ 116o = 64o
Thus, xo = âˆ FEA + âˆ FEC
= 64o + 56o = 120o.
Hence, x = 120.
Question 5:
Since AB  CD and BC is a transversal.
So, âˆ ABC = âˆ BCD [atternate interior angles]
â‡’ 70o = xo + âˆ ECD â€¦.(i)
Now, CD  EF and CE is transversal.
So, âˆ ECD + âˆ CEF = 180o
âˆ´ âˆ ECD + 130o = 180o
[sum of consecutive interior angles is 180o]
â‡’ âˆ ECD = 180o â€“ 130o = 50o
Putting âˆ ECD = 50o in (i) we get, 70o = xo + 50o
â‡’ x = 70 â€“ 50 = 20
Question 6:
Through C draw FG  AE
Now, since CG  BE and CE is a transversal.
So, âˆ GCE = âˆ CEA = 20o
âˆ´ âˆ DCG = 130o â€“ âˆ GCE
= 130o â€“ 20o = 110o
[Alternate angles]
Also, we have AB  CD and FG is a transversal.
So, âˆ BFC = âˆ DCG = 110o
As, FG  AE, AF is a transversal.
[Corresponding angles]
âˆ BFG = âˆ FAE [Corresponding angles]
âˆ´ xo = âˆ FAE = 110o. Hence, x = 110
Question 7:
Given: AB  CD
To Prove: âˆ BAE â€“ âˆ DCE = âˆ AEC
Construction : Through E draw EF  AB Proof : Since EF  AB, AE is a transversal. So, âˆ BAE + âˆ AEF = 180O â€¦.(i)
[sum of consecutive interior angles is 180o] As EF  AB and AB  CD [Given]
So, EF  CD and EC is a transversal.
So, âˆ FEC + âˆ DCE = 180o â€¦.(ii)
[sum of consecutive interior angles is 180o] From (i) and (ii) we get,
âˆ BAE + âˆ AEF = âˆ FEC + âˆ DCE
â‡’ âˆ BAE â€“ âˆ DCE = âˆ FEC â€“ âˆ AEF = âˆ AEC [Proved]
Question 8:
Since AB  CD and BC is a transversal.
So, âˆ BCD = âˆ ABC = xo
[Alternate angles]
As BC  ED and CD is a transversal.
âˆ BCD + âˆ EDC = 180o
â‡’ âˆ BCD + 75o =180o
â‡’ âˆ BCD = 180o â€“ 75o = 105o
âˆ ABC = 105o
âˆ´ xo = âˆ ABC = 105o Hence, x = 105.
[since âˆ BCD = âˆ ABC]
Question 9:
Through F, draw KH  AB  CD
Now, KF  CD and FG is a transversal.
â‡’ âˆ KFG = âˆ FGD = ro â€¦. (i) [alternate angles]
Again AE  KF, and EF is a transversal.
So, âˆ AEF + âˆ KFE = 180o
âˆ KFE = 180o â€“ po â€¦. (ii) Adding (i) and (ii) we get,
âˆ KFG + âˆ KFE = 180 â€“ p + r
â‡’ âˆ EFG = 180 â€“ p + r
â‡’ q = 180 â€“ p + r i.e., p + q â€“ r = 180
Question 10:
Since AB  PQ and EF is a transversal.
So, âˆ CEB = âˆ EFQ [Corresponding angles]
â‡’ âˆ EFQ = 75o
â‡’ âˆ EFG + âˆ GFQ = 75o
â‡’ 25o + yo = 75o
â‡’ y = 75 â€“ 25 = 50
Also, âˆ BEF + âˆ EFQ = 180o
âˆ BEF = 180o â€“ âˆ EFQ
= 180o â€“ 75o
âˆ BEF = 105o
[sum of consecutive interior angles is 180o]
âˆ´ âˆ FEG + âˆ GEB = âˆ BEF = 105o
â‡’ âˆ FEG = 105o â€“ âˆ GEB = 105o â€“ 20o = 85o
In âˆ†EFG we have,
xo + 25o + âˆ FEG = 180o
Hence, x = 70.
Question 11:
Since AB  CD and AC is a transversal.
So, âˆ BAC + âˆ ACD = 180o
â‡’ âˆ ACD = 180o â€“ âˆ BAC
= 180o â€“ 75o = 105o
[sum of consecutive interior angles is 180o]
â‡’ âˆ ECF = âˆ ACD [Vertically opposite angles]
âˆ ECF = 105o
Now in âˆ†CEF,
âˆ ECF + âˆ CEF + âˆ EFC =180o
â‡’ 105o + xo + 30o = 180o
â‡’ x = 180 â€“ 30 â€“ 105 = 45
Hence, x = 45.
Question 12:
Since AB  CD and PQ a transversal.
So, âˆ PEF = âˆ EGH [Corresponding angles]
â‡’ âˆ EGH = 85o
âˆ EGH and âˆ QGH form a linear pair. So, âˆ EGH + âˆ QGH = 180o
â‡’ âˆ QGH = 180o â€“ 85o = 95o
Similarly, âˆ GHQ + 115o = 180o
â‡’ âˆ GHQ = 180o â€“ 115o = 65o
In âˆ†GHQ, we have, xo + 65o + 95o = 180o
â‡’ x = 180 â€“ 65 â€“ 95 = 180 â€“ 160
âˆ´ x = 20
Question 13:
Since AB  CD and BC is a transversal. So, âˆ ABC = âˆ BCD
â‡’ x = 35
Also, AB  CD and AD is a transversal. So, âˆ BAD = âˆ ADC
â‡’ z = 75
In âˆ†ABO, we have,
âˆ AOB + âˆ BAO + âˆ BOA = 180o
â‡’ xo + 75o + yo = 180o
â‡’ 35 + 75 + y = 180
â‡’ y = 180 â€“ 110 = 70
âˆ´ x = 35, y = 70 and z = 75.
Question 14:
Since AB  CD and PQ is a transversal. So, y = 75 [Alternate angle]
Since PQ is a transversal and AB  CD, so x + APQ = 180o
[Sum of consecutive interior angles]
â‡’ xo = 180o â€“ APQ
â‡’ x = 180 â€“ 75 = 105
Also, AB  CD and PR is a transversal.
So, âˆ APR = âˆ PRD [Alternate angle]
â‡’ âˆ APQ + âˆ QPR = âˆ PRD [Since âˆ APR = âˆ APQ + âˆ QPR]
â‡’ 75o + zo = 125o
â‡’ z = 125 â€“ 75 = 50
âˆ´ x = 105, y = 75 and z = 50.
Question 15:
âˆ PRQ = xo = 60o
[vertically opposite angles]
Since EF  GH, and RQ is a transversal. So, âˆ x = âˆ y [Alternate angles]
â‡’ y = 60
AB  CD and PR is a transversal.
So, âˆ PRD = âˆ APR [Alternate angles]
â‡’ âˆ PRQ + âˆ QRD = âˆ APR [since âˆ PRD = âˆ PRQ + âˆ QRD]
â‡’ x + âˆ QRD = 110o
â‡’ âˆ QRD = 110o â€“ 60o = 50o
In âˆ†QRS, we have,
âˆ QRD + to + yo = 180o
â‡’ 50 + t + 60 = 180
â‡’ t = 180 â€“ 110 = 70
Since, AB  CD and GH is a transversal So, zo = to = 70o [Alternate angles]
âˆ´ x = 60 , y = 60, z = 70 and t = 70
Question 16:
 Lines l and m will be parallel if 3x â€“ 20 = 2x + 10
[Since, if corresponding angles are equal, lines are parallel]
â‡’ 3x â€“ 2x = 10 + 20
â‡’ x = 30
 Lines will be parallel if (3x + 5)o + 4xo = 180o
[if sum of pairs of consecutive interior angles is 180o, the lines are parallel] So, (3x + 5) + 4x = 180
â‡’ 3x + 5 + 4x = 180
â‡’ 7x = 180 â€“ 5 = 175
â‡’ x =
= 25
Question 17:
Given: Two lines m and n are perpendicular to a given line l.
To Prove: m  n Proof : Since m âŠ¥ l So, âˆ 1 = 90o Again, since n âŠ¥ l
âˆ 2 = 90o
âˆ´ âˆ 1 = âˆ 2 = 90o
But âˆ 1 and âˆ 2 are the corresponding angles made by the transversal l with lines m and n and they are proved to be equal.
Thus, m  n.
Question 1:
Exercise 4D
Since, sum of the angles of a triangle is 180o
âˆ A + âˆ B + âˆ C = 180o
â‡’ âˆ A + 76o + 48o = 180o
â‡’ âˆ A = 180o â€“ 124o = 56o
âˆ´ âˆ A = 56o
Question 2:
Let the measures of the angles of a triangle are (2x)o, (3x)o and (4x)o. Then, 2x + 3x + 4x = 180 [sum of the angles of a triangle is 180o ]
â‡’ 9x = 180
â‡’ x =
= 20
âˆ´ The measures of the required angles are: 2x = (2 Ã— 20)o = 40o
3x = (3 Ã— 20)o = 60o
4x = (4 Ã— 20)o = 80o
Question 3:
Let 3âˆ A = 4âˆ B = 6âˆ C = x (say) Then, 3âˆ A = x
â‡’ âˆ A =
4âˆ B = x
â‡’ âˆ B =
and 6âˆ C = x
â‡’ âˆ C =
As âˆ A + âˆ B + âˆ C = 180o
Question 4:
âˆ A + âˆ B = 108o [Given]
But as âˆ A, âˆ B and âˆ C are the angles of a triangle,
âˆ A + âˆ B + âˆ C = 180o
â‡’ 108o + âˆ C = 180o
â‡’ C = 180o â€“ 108o = 72o
Also, âˆ B + âˆ C = 130o [Given]
â‡’ âˆ B + 72o = 130o
â‡’ âˆ B = 130o â€“ 72o = 58o Now as, âˆ A + âˆ B = 108o
â‡’ âˆ A + 58o = 108o
â‡’ âˆ A = 108o â€“ 58o = 50o
âˆ´ âˆ A = 50o, âˆ B = 58o and âˆ C = 72o.
Question 5:
Since. âˆ A , âˆ B and âˆ C are the angles of a triangle . So, âˆ A + âˆ B + âˆ C = 180o
Now, âˆ A + âˆ B = 125o [Given]
âˆ´ 125o + âˆ C = 180o
â‡’ âˆ C = 180o â€“ 125o = 55o
Also, âˆ A + âˆ C = 113o [Given]
â‡’ âˆ A + 55o = 113o
â‡’ âˆ A = 113o â€“ 55o = 58o Now as âˆ A + âˆ B = 125o
â‡’ 58o + âˆ B = 125o
â‡’ âˆ B = 125o â€“ 58o = 67o
âˆ´ âˆ A = 58o, âˆ B = 67o and âˆ C = 55o.
Question 6:
Since, âˆ P, âˆ Q and âˆ R are the angles of a triangle. So, âˆ P + âˆ Q + âˆ R = 180o â€¦.(i)
Now, âˆ P â€“ âˆ Q = 42o [Given]
â‡’ âˆ P = 42o + âˆ Q â€¦.(ii)
and âˆ Q â€“ âˆ R = 21o [Given]
â‡’ âˆ R = âˆ Q â€“ 21o â€¦.(iii)
Substituting the value of âˆ P and âˆ R from (ii) and (iii) in (i), we get,
â‡’ 42o + âˆ Q + âˆ Q + âˆ Q â€“ 21o = 180o
â‡’ 3âˆ Q + 21o = 180o
â‡’ 3âˆ Q = 180o â€“ 21o = 159o
âˆ Q =
= 53o
âˆ´ âˆ P = 42o + âˆ Q
= 42o + 53o = 95o
âˆ R = âˆ Q â€“ 21o
= 53o â€“ 21o = 32o
âˆ´ âˆ P = 95o, âˆ Q = 53o and âˆ R = 32o.
Question 7:
Given that the sum of the angles A and B of a ABC is 116o, i.e., âˆ A + âˆ B = 116o. Since, âˆ A + âˆ B + âˆ C = 180o
So, 116o + âˆ C = 180o
â‡’ âˆ C = 180o â€“ 116o = 64o
Also, it is given that:
âˆ A â€“ âˆ B = 24o
â‡’ âˆ A = 24o + âˆ B
Putting, âˆ A = 24o + âˆ B in âˆ A + âˆ B = 116o, we get,
â‡’ 24o + âˆ B + âˆ B = 116o
â‡’ 2âˆ B + 24o = 116o
â‡’ 2âˆ B = 116o â€“ 24o = 92o
âˆ B =
= 46o
Therefore, âˆ A = 24o + 46o = 70o
âˆ´ âˆ A = 70o, âˆ B = 46o and âˆ C = 64o.
Question 8:
Let the two equal angles, A and B, of the triangle be xo each. We know,
âˆ A + âˆ B + âˆ C = 180o
â‡’ xo + xo + âˆ C = 180o
â‡’ 2xo + âˆ C = 180o â€¦.(i) Also, it is given that,
âˆ C = xo + 18o â€¦.(ii)
Substituting âˆ C from (ii) in (i), we get,
â‡’ 2xo + xo + 18o = 180o
â‡’ 3xo = 180o â€“ 18o = 162o x =
= 54o
Thus, the required angles of the triangle are 54o, 54o and xo + 18o = 54o + 18o = 72o.
Question 9:
Let âˆ C be the smallest angle of ABC. Then, âˆ A = 2âˆ C and B = 3âˆ C
Also, âˆ A + âˆ B + âˆ C = 180o
â‡’ 2âˆ C + 3âˆ C + âˆ C = 180o
â‡’ 6âˆ C = 180o
â‡’ âˆ C = 30o
So, âˆ A = 2âˆ C = 2 (30o) = 60o
âˆ B = 3âˆ C = 3 (30o) = 90o
âˆ´ The required angles of the triangle are 60o, 90o, 30o.
Question 10:
Let ABC be a right angled triangle and âˆ C = 90o Since, âˆ A + âˆ B + âˆ C = 180o
â‡’ âˆ A + âˆ B = 180o â€“ âˆ C = 180o â€“ 90o = 90o
Suppose âˆ A = 53o
Then, 53o + âˆ B = 90o
â‡’ âˆ B = 90o â€“ 53o = 37o
âˆ´ The required angles are 53o, 37o and 90o.
Question 11:
Let ABC be a triangle. Given, âˆ A + âˆ B = âˆ C
We know, âˆ A + âˆ B + âˆ C = 180o
â‡’ âˆ C + âˆ C = 180o
â‡’ 2âˆ C = 180o
â‡’ âˆ C =
= 90o
So, we find that ABC is a right triangle, right angled at C.
Question 12:
Given : âˆ†ABC in which âˆ A = 90o, AL âŠ¥ BC To Prove: âˆ BAL = âˆ ACB
Proof :
In right triangle âˆ†ABC,
â‡’ âˆ ABC + âˆ BAC + âˆ ACB = 180o
â‡’ âˆ ABC + 90o + âˆ ACB = 180o
â‡’ âˆ ABC + âˆ ACB = 180o â€“ 90o
âˆ´ âˆ ABC + âˆ ACB = 90o
â‡’ âˆ ACB = 90o â€“ âˆ ABC â€¦.(1)
Similarly since âˆ†ABL is a right triangle, we find that,
âˆ BAL = 90o â€“ âˆ ABC â€¦(2)
Thus from (1) and (2), we have
âˆ´ âˆ BAL = âˆ ACB (Proved)
Question 13:
Let ABC be a triangle. So, âˆ A < âˆ B + âˆ C
Adding A to both sides of the inequality,
â‡’ 2âˆ A < âˆ A + âˆ B + âˆ C
â‡’ 2âˆ A < 180o
â‡’ âˆ A <
= 90o
[Since âˆ A + âˆ B + âˆ C = 180o]
Similarly, âˆ B < âˆ A + âˆ C
â‡’ âˆ B < 90o
and âˆ C < âˆ A + âˆ B
â‡’ âˆ C < 90o
âˆ†ABC is an acute angled triangle.
Question 14:
Let ABC be a triangle and âˆ B > âˆ A + âˆ C Since, âˆ A + âˆ B + âˆ C = 180o
â‡’ âˆ A + âˆ C = 180o â€“ âˆ B Therefore, we get
âˆ B > 180o â€“ âˆ B
Adding âˆ B on both sides of the inequality, we get,
â‡’ âˆ B + âˆ B > 180o â€“ âˆ B + âˆ B
â‡’ 2âˆ B > 180o
â‡’ âˆ B >
= 90o
i.e., âˆ B > 90o which means âˆ B is an obtuse angle.
âˆ†ABC is an obtuse angled triangle.
Question 15:
Since âˆ ACB and âˆ ACD form a linear pair. So, âˆ ACB + âˆ ACD = 180o
â‡’ âˆ ACB + 128o = 180o
â‡’ âˆ ACB = 180o â€“ 128 = 52o
Also, âˆ ABC + âˆ ACB + âˆ BAC = 180o
â‡’ 43o + 52o + âˆ BAC = 180o
â‡’ 95o + âˆ BAC = 180o
â‡’ âˆ BAC = 180o â€“ 95o = 85o
âˆ´ âˆ ACB = 52o and âˆ BAC = 85o.
Question 16:
As âˆ DBA and âˆ ABC form a linear pair. So, âˆ DBA + âˆ ABC = 180o
â‡’ 106o + âˆ ABC = 180o
â‡’ âˆ ABC = 180o â€“ 106o = 74o
Also, âˆ ACB and âˆ ACE form a linear pair. So, âˆ ACB + âˆ ACE = 180o
â‡’ âˆ ACB + 118o = 180o
â‡’ âˆ ACB = 180o â€“ 118o = 62o
In âˆ ABC, we have,
âˆ ABC + âˆ ACB + âˆ BAC = 180o 74o + 62o + âˆ BAC = 180o
â‡’ 136o + âˆ BAC = 180o
â‡’ âˆ BAC = 180o â€“ 136o = 44o
âˆ´ In triangle ABC, âˆ A = 44o, âˆ B = 74o and âˆ C = 62o
Question 17:
(i) âˆ EAB + âˆ BAC = 180o [Linear pair angles]
110o + âˆ BAC = 180o
â‡’ âˆ BAC = 180o â€“ 110o = 70o
Again, âˆ BCA + âˆ ACD = 180o [Linear pair angles]
â‡’ âˆ BCA + 120o = 180o
â‡’ âˆ BCA = 180o â€“ 120o = 60o
Now, in âˆ†ABC,
âˆ ABC + âˆ BAC + âˆ ACB = 180o
xo + 70o + 60o = 180o
â‡’ x + 130o = 180o
â‡’ x = 180o â€“ 130o = 50o
âˆ´ x = 50 (ii)
In âˆ†ABC,
âˆ A + âˆ B + âˆ C = 180o
â‡’ 30o + 40o + âˆ C = 180o
â‡’ 70o + âˆ C = 180o
â‡’ âˆ C = 180o â€“ 70o = 110o
Now âˆ BCA + âˆ ACD = 180o [Linear pair]
â‡’ 110o + âˆ ACD = 180o
â‡’ âˆ ACD = 180o â€“ 110o = 70o In âˆ†ECD,
â‡’ âˆ ECD + âˆ CDE + âˆ CED = 180o
â‡’ 70o + 50o + âˆ CED = 180o
â‡’ 120o + âˆ CED = 180o
âˆ CED = 180o â€“ 120o = 60o
Since âˆ AED and âˆ CED from a linear pair So, âˆ AED + âˆ CED = 180o
â‡’ xo + 60o = 180o
â‡’ xo = 180o â€“ 60o = 120o
âˆ´ x = 120 (iii)
âˆ EAF = âˆ BAC [Vertically opposite angles]
â‡’ âˆ BAC = 60o
In âˆ†ABC, exterior âˆ ACD is equal to the sum of two opposite interior angles. So, âˆ ACD = âˆ BAC + âˆ ABC
â‡’ 115o = 60o + xo
â‡’ xo = 115o â€“ 60o = 55o
âˆ´ x = 55 (iv)
Since AB  CD and AD is a transversal. So, âˆ BAD = âˆ ADC
â‡’ âˆ ADC = 60o
In âˆ ECD, we have,
âˆ E + âˆ C + âˆ D = 180o
â‡’ xo + 45o + 60o = 180o
â‡’ xo + 105o = 180o
â‡’ xo = 180o â€“ 105o = 75o
âˆ´ x = 75 (v)
In âˆ†AEF,
Exterior âˆ BED = âˆ EAF + âˆ EFA
â‡’ 100o = 40o + âˆ EFA
â‡’ âˆ EFA = 100o â€“ 40o = 60o
Also, âˆ CFD = âˆ EFA [Vertically Opposite angles]
â‡’ âˆ CFD = 60o Now in âˆ†FCD,
Exterior âˆ BCF = âˆ CFD + âˆ CDF
â‡’ 90o = 60o + xo
â‡’ xo = 90o â€“ 60o = 30o
âˆ´ x = 30 (vi)
In âˆ†ABE, we have,
âˆ A + âˆ B + âˆ E = 180o
â‡’ 75o + 65o + âˆ E = 180o
â‡’ 140o + âˆ E = 180o
â‡’ âˆ E = 180o â€“ 140o = 40o
Now, âˆ CED = âˆ AEB [Vertically opposite angles]
â‡’ âˆ CED = 40o
Now, in âˆ†CED, we have,
âˆ C + âˆ E + âˆ D = 180o
â‡’ 110o + 40o + xo = 180o
â‡’ 150o + xo = 180o
â‡’ xo = 180o â€“ 150o = 30o
âˆ´ x = 30
Question 18:
Produce CD to cut AB at E.
Now, in âˆ†BDE, we have,
Exterior âˆ CDB = âˆ CEB + âˆ DBE
â‡’ xo = âˆ CEB + 45o In âˆ†AEC, we have,
â€¦..(i)
Exterior âˆ CEB = âˆ CAB + âˆ ACE
= 55o + 30o = 85o
Putting âˆ CEB = 85o in (i), we get, xo = 85o + 45o = 130o
âˆ´ x = 130
Question 19:
The angle âˆ BAC is divided by AD in the ratio 1 : 3. Let âˆ BAD and âˆ DAC be y and 3y, respectively. As BAE is a straight line,
âˆ BAC + âˆ CAE = 180o
[linear pair]
â‡’ âˆ BAD + âˆ DAC + âˆ CAE = 180o
â‡’ y + 3y + 108o = 180o
â‡’ 4y = 180o â€“ 108o = 72o
â‡’ y =
= 18o Now, in âˆ†ABC,
âˆ ABC + âˆ BCA + âˆ BAC = 180o
y + x + 4y = 180o
[Since, âˆ ABC = âˆ BAD (given AD = DB) and âˆ BAC = y + 3y = 4y]
â‡’ 5y + x = 180
â‡’ 5 Ã— 18 + x = 180
â‡’ 90 + x = 180
âˆ´ x = 180 â€“ 90 = 90
Question 20:
Given : A âˆ†ABC in which BC, CA and AB are produced to D, E and F respectively. To prove : Exterior âˆ DCA + Exterior âˆ BAE + Exterior âˆ FBD = 360o
Proof : Exterior âˆ DCA = âˆ A + âˆ B â€¦.(i) Exterior âˆ FAE = âˆ B + âˆ C â€¦.(ii) Exterior âˆ FBD = âˆ A + âˆ C â€¦.(iii) Adding (i), (ii) and (iii), we get,
Ext. âˆ DCA + Ext. âˆ FAE + Ext. âˆ FBD
= âˆ A + âˆ B + âˆ B + âˆ C + âˆ A + âˆ C
= 2âˆ A + 2âˆ B + 2âˆ C
= 2 (âˆ A + âˆ B + âˆ C)
= 2 Ã— 180o
[Since, in triangle the sum of all three angle is 180o]
= 360o
Hence, proved.
Question 21:
In âˆ†ACE, we have,
âˆ A + âˆ C + âˆ E = 180o â€¦.(i) In âˆ†BDF, we have,
âˆ B + âˆ D + âˆ F = 180o â€¦.(ii)
Adding both sides of (i) and (ii), we get,
âˆ A + âˆ C +âˆ E + âˆ B + âˆ D + âˆ F = 180o + 180o
â‡’ âˆ A + âˆ B + âˆ C + âˆ D + âˆ E + âˆ F = 360o.
Question 22:
Given : In âˆ†ABC, bisectors of âˆ B and âˆ C meet at O and âˆ A = 70o In âˆ†BOC, we have,
âˆ BOC + âˆ OBC + âˆ OCB = 180o
= 180o â€“ 55o = 125o
âˆ´ âˆ BOC = 125o.
Question 23:
We have a âˆ†ABC whose sides AB and AC have been procued to D and E. A = 40o and bisectors of âˆ CBD and âˆ BCE meet at O.
In âˆ†ABC, we have, Exterior âˆ CBD = C + 40o
And exterior âˆ BCE = B + 40o
Now, in âˆ†BCO, we have,
= 50o + 20o
= 70o
Thus, âˆ BOC = 70o
Question 24:
In the given âˆ†ABC, we have,
âˆ A : âˆ B : âˆ C = 3 : 2 : 1
Let âˆ A = 3x, âˆ B = 2x, âˆ C = x. Then,
âˆ A + âˆ B + âˆ C = 180o
â‡’ 3x + 2x + x = 180o
â‡’ 6x = 180o
â‡’ x = 30o
âˆ A = 3x = 3 30o = 90o
âˆ B = 2x = 2 30o = 60o
and, âˆ C = x = 30o
Now, in âˆ†ABC, we have,
Ext âˆ ACE = âˆ A + âˆ B = 90o + 60o = 150o
âˆ ACD + âˆ ECD = 150o
â‡’ âˆ ECD = 150o â€“ âˆ ACD
â‡’ âˆ ECD = 150o â€“ 90o
â‡’ âˆ ECD= 60o
[since , AD âŠ¥ CD, âˆ ACD = 90o]
Question 25:
In âˆ†ABC, AN is the bisector of âˆ A and AM âŠ¥ BC. Now in âˆ†ABC we have;
âˆ A = 180o â€“ âˆ B â€“ âˆ C
â‡’ âˆ A = 180o â€“ 65o â€“ 30o
= 180o â€“ 95o
= 85o
Now, in âˆ†ANC we have;
Thus, âˆ MAN =
Question 26:
(i) False (ii) True (iii) False (iv) False (v) True (vi) True.