RS Aggarwal 2017 Solutions for Class 9 Math Chapter 4 Angles, Lines And Triangles are provided here with simple step-by-step explanations. These solutions for Angles, Lines And Triangles are extremely popular among class 9 students for Math Angles, Lines And Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2017 Book of class 9 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2017 Solutions. All RS Aggarwal 2017 Solutions for class 9 Math are prepared by experts and are 100% accurate.

Page No 113:

Question 1:

Define the following terms:
(i) Angle
(ii) Interior of an angle
(iii) Obtuse angle
(iv) Reflex angle
(v) Complementary angles
(vi) Supplementary angles

Answer:

(i) Two rays OA and OB, with a common end-point O, form an angle AOB that is represented as AOB.


(ii) The interior of an angle is the set of all points in its plane, which lie on the same side of OA as B and also on the same side of OB as A.


(iii) An angle greater than 90° but less than 180° is called an obtuse angle.



(iv) An angle greater than 180° but less than 360° is called a reflex angle.



(v) Two angles are said to be complementary if the sum of their measures is 90°.

(vi) Two angles are said to be supplementary if the sum of their measures is 180°.

Page No 113:

Question 2:

If ∠A = 36°27'46" and ∠B = 28°43'39", find ∠A + ∠B.

Answer:

A+B=36°27'46''+28°43'39''
We know that 1°=60 min or 60'1'=60 sec or 60''
 

Deg Min Sec
  36 27 46
+28 43 39
  65 11 25
 
Hence, the measure of the required angle = 65° 11' 25''

Page No 113:

Question 3:

Find the difference between two angles measuring 36° and 24°28' 30".

Answer:

We know that 36°=35° 59' 60''

Deg Min Sec
35 59 60
24 28 30
11 31 30

Hence, the measure of the required angle = 11° 31' 30''.

Page No 113:

Question 4:

Find the complement of each of the following angles.
(i) 58°
(ii) 16°
(iii) 23 of a right angle
(iv) 46°30'
(v) 52°43'20"
(vi) 68°35'45"

Answer:

(i) Complement of 58°=90-58°
                                                 =32°

(ii) Complement of 16°=90-16°
                                                  =74°

(iii) 23 of a right angle90×23°=60°
Complement of 23 of a right angle=90-60°
                                                            =30°

(iv) We know that 90°=89° 60'
Complement of 46° 30'=89° 60'-46° 30'°
                                                =43° 30'

(v) We know that 90°=89° 59' 60''
Complement of 52° 43' 20''=89° 59' 60''-52° 43' 20''

Deg Min Sec
89 59 60
52 43 20
37 16 40

Hence, the measure of the angle=37° 16' 40''.

(vi) We know that 90°=89° 59' 60''
Complement of 68° 35' 45''=89° 59' 60''-68° 35' 45''
Deg Min Sec
89 59 60
68 35 45
21 24 15
 
Hence, the measure of the required angle =21° 24' 15''.

Page No 113:

Question 5:

Find the supplement of each of the following angles.
(i) 63°
(ii) 138°
(iii) 35 of a right angle
(iv) 75°36'
(v) 124°20'40"
(vi) 108°48'32"

Answer:

(i) Supplement of 63°=180-63°
                                               =117°

(ii) Supplement of 138°=180-138°
                                                  =42°

(iii) 35 of a right angle35×90=54°
Supplement of 35 of a right angle=180-54°=126°

(iv) We know that 180°=179° 60'
Supplement of 75° 36'=179° 60'-75° 36'
                                                =104°24'

(v) We know that 180°=179° 59' 60''
Supplement of 124° 20' 40''=179° 59' 60''-124° 20' 40''

Deg Min Sec
179 59 60
124 20 40
55 39 20

Hence, the measure of the required angle=55° 39' 20'.

(vi) We know that 180=179° 59' 60'' 
Supplement of 108° 48' 32''=179° 59' 60''-108° 48' 32''
Deg Min Sec
179 59 60
108 48 32
71 11 28

Hence, the measure of the required angle=71° 11' 28''.

Page No 113:

Question 6:

Find the measure of an angle which is
(i) equal to its complement
(ii) equal to its supplement

Answer:

(i) Let the measure of the required angle be x°.
Then, in case of complementary angles:
x+x=90°2x=90°x=45°
Hence, measure of the angle that is equal to its complement is 45°.

(ii) Let the measure of the required angle be x°​.
Then, in case of supplementary angles:
x+x=180°2x=180°x=90°
Hence, measure of the angle that is equal to its supplement is 90°.

Page No 113:

Question 7:

Find the measure of an angle which is 36° more than its complement.

Answer:

Let the measure of the required angle be x°.
Then, measure of its complement =90-x°.
Therefore,
x-90°-x=36°2x=126°x=63°
Hence, the measure of the required angle is 63°.

Page No 113:

Question 8:

Find the measure of an angle that is 25° less than its supplement.

Answer:

Let the measure of the required angle be x.
Then, measure of its supplement =180°-x.
Therefore,
180°-x-x=25°2x=155°x=77.5°
Hence, the measure of the required angle is 77.5°.

Page No 113:

Question 9:

Find the angle which is four times its complement.

Answer:

Let the measure of the required angle be x.
Then, measure of its complement =90°-x.
Therefore,
x=90°-x4x=360°-4x5x=360°x=72°
Hence, the measure of the required angle is 72°.

Page No 113:

Question 10:

Find the angle which is five times its supplement.

Answer:

Let the measure of the required angle be x.
Then, measure of its supplement =180°-x.
Therefore,
x=180°-x5x=900°-5x6x=900°x=150°
Hence, the measure of the required angle is 150°.

Page No 113:

Question 11:

Find the angle whose supplement is four times its complement.

Answer:

Let the measure of the required angle be x°.
Then, measure of its complement =90-x°.
And, measure of its supplement=180-x°.
Therefore,
180-x=490-x180-x=360-4x3x=180x=60
Hence, the measure of the required angle is 60°.

Page No 113:

Question 12:

Find the angle whose complement is one-third of its supplement.

Answer:

Let the measure of the required angle be x°.
Then, the measure of its complement =90-x°.
And the measure of its supplement=180-x°.
Therefore,
90-x=13180-x390-x=180-x270-3x=180-x2x=90x=45
Hence, the measure of the required angle is 45°.

Page No 113:

Question 13:

Two supplementary angles are in the ratio 3 : 2. Find the angles.

Answer:

Let the two angles be 3x and 2x, respectively.
Then,
3x+2x=1805x=180x=36°
Therefore, the two angles are 
       3x=3×36°=108° and 2x=2×36°=72°.

Page No 113:

Question 14:

Two complementary angles are in the ratio 4 : 5. Find the angles.

Answer:

Let the two angles be 4x and 5x, respectively.
Then,
4x+5x=909x=90x=10°
Hence, the two angles are 4x=4×10°=40° and 5x=5×10°=50°.

Page No 113:

Question 15:

Find the measure of an angle, if seven times its complement is 10° less than three times its supplement.

Answer:

Let the measure of the required angle be x°.
Then, the measure of its complement =90-x°.
And the measure of its supplement=180-x°.
Therefore,
790-x=3180-x-10630-7x=540-3x-104x=100x=25
Hence, the measure of the required angle is 25°.



Page No 118:

Question 1:

In the adjoining figure, AOB is a straight line. Find the value of x.

Answer:

We know that the sum of angles in a linear pair is 180°.
Therefore,
AOC+BOC=180°62°+x°=180°x°=180°-62°x=118°
Hence, the value of x is 118°.

Page No 118:

Question 2:

In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ∠AOC and ∠BOD.

Answer:

As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is 180°.
Therefore,
 AOC+COD+BOD=180°3x-6°+55°+x+20°=1804x=111°x=27.5°
Hence,
AOC=3x-6
         =3×27.5-6=76.5°
and BOD=x+20
                =27.5+20=47.5°

Page No 118:

Question 3:

In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ∠AOC, ∠COD and ∠BOD.

Answer:

AOB is a straight line. Therefore,
AOC+COD+BOD=180°3x+7°+2x-19°+x°=180°6x=192°x=32°
Therefore,
AOC=3×32°+7=103°COD=2×32°-19=45° andBOD=32°

Page No 118:

Question 4:

In the adjoining figure, x:y:z = 5:4:6. If XOY is a straight line, find the values of x, y and z.

Answer:

Let x=5a, y=4a and z=6a
XOY is a straight line. Therefore,

XOP+POQ+YOQ=180°5a+4a+6a=180°15a=180°a=12°
Therefore,

      x5×12°=60°      y4×12°=48°and z6×12°=72°

Page No 118:

Question 5:

In the adjoining figure, what value of x will make AOB a straight line?

Answer:

AOB will be a straight line if
3x+20+4x-36=180°7x=196°x=28°
Hence, x = 28 will make AOB a straight line.



Page No 119:

Question 6:

Two lines AB and CD intersect at O. If ∠AOC = 50°, find ∠AOD, ∠BOD and ∠BOC.

Answer:

We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, AOC=BOD=50°
Let AOD=BOC=x°
Also, we know that the sum of all angles around a point is 360°.
Therefore, 
AOC+AOD+BOD+BOC=360°50+x+50+x=360°2x=260°x=130°
Hence, AOD=BOC=130°
Therefore, AOD=130°, BOD=50° and BOC=130°.

Page No 119:

Question 7:

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.

Answer:

We know that if two lines intersect, then the vertically opposite angles are equal.
BOD=AOC=90°
Hence, t=90°
Also, 
DOF=COE=50°
Hence, z=50°
Since, AOB is a straight line, we have:
AOC+COE+BOE=180°90+50+y=180°140+y=180°y=40°
Also,
BOE=AOF=40°
Hence, x=40°
 x=40°, y=40°, z=50° and t=90°

Page No 119:

Question 8:

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Also, find ∠AOD, ∠COE and ∠AOE.

Answer:

We know that if two lines intersect, then the vertically-opposite angles are equal.
DOF=COE=5x°AOD=BOC=2x° andAOE=BOF=3x°

Since, AOB is a straight line, we have:
AOE+COE+BOC=180°3x+5x+2x=180°10x=180°x=18°

Therefore,
AOD=2×18°=36°COE=5×18°=90°AOE=3×18°=54°

Page No 119:

Question 9:

Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each of these angles.

Answer:

Let the two adjacent angles be 5x and 4x, respectively.
Then,
5x+4x=180°9x=180°x=20°
Hence, the two angles are 5×20°=100° and 4×20°=80°.

Page No 119:

Question 10:

If two straight lines intersect in such a way that one of the angles formed measures 90°, show that each of the remaining angles measures 90°.

Answer:

We know that if two lines intersect, then the vertically-opposite angles are equal.

AOC=90°. Then, AOC=BOD=90°.
And let BOC=AOD=x
Also, we know that the sum of all angles around a point is 360°
AOC+BOD+AOD+BOC=360°90°+90°+x+x=360°2x=180°x=90°
Hence, BOC=AOD=90°
AOC=BOD=BOC=AOD=90°
Hence, the measure of each of the remaining angles is 90o.

Page No 119:

Question 11:

Two lines AB and CD intersect at a point O, such that ∠BOC + ∠AOD = 280°, as shown in the figure. Find all the four angles.

Answer:

We know that if two lines intersect, then the vertically-opposite angles are equal.
Let BOC=AOD=x°
Then,
x+x=2802x=280x=140°BOC=AOD=140°
Also, let AOC=BOD=y°
We know that the sum of all angles around a point is 360°.
AOC+BOC+BOD+AOD=360°y+140+y+140=360°2y=80°y=40°
Hence, AOC=BOD=40°
BOC=AOD=140° and AOC=BOD=40°

Page No 119:

Question 12:

In the given figure, ray OC is the bisector of ∠AOB and OD is the ray opposite OC. Show that ∠AOD = ∠BOD.

Answer:

Since DOC is a straight line, we have:
BOD+COB=180°               ...(1)AOC+AOD=180°               ...(2)From (1) and (2), we get:BOD+COB=AOC+AOD

Also, COB=AOC     [Since OC is a bisector of  AOB]
AOD=BOD

Page No 119:

Question 13:

In the given figure, AB is a mirror; PQ is the incident ray and QR, the reflected ray. If ∠PQR = 112°, find ∠PQA.

Answer:

We know that the angle of incidence = angle of reflection.
Hence, let PQA=BQR=x°
Since, AQB is a straight line, we have:
PQA+PQR+BQR=180°x+112+x=180°2x=68°x=34°PQA=34°

Page No 119:

Question 14:

If two straight lines intersect each other, then prove that the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

Answer:

Let AB and CD be the two lines intersecting at a point O and let ray OE bisect AOC. Now, draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let COE=1, AOE=2, BOF=3 and DOF=4.
We know that vertically-opposite angles are equal.
1=4 and 2=3
But, 1=2    [Since OE bisects AOC ]
4=3
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

Page No 119:

Question 15:

Prove that the bisectors of two adjacent supplementary angles include a right angle.

Answer:

Let AOB denote a straight line and let AOC and BOC be the supplementary angles.
Then, we have:
AOC=x° and BOC=180-x°
Let OE bisect AOC and OF bisect BOC.
Then, we have:
AOE=COE=12x° andBOF=FOC=12180-x°
Therefore,
COE+FOC=12x+12180°-x
                       =12x+180°-x=12180°=90°



Page No 129:

Question 1:

In the adjoining figure, AB || CD. The lines are cut by a transversal t at E and F, respectively. If ∠1 = 70°, find the measure of each of the remaining marked angles.

Answer:

We have, 1=70°. Then,
1=5   Corresponding angles5=70°1=3   Vertically-opposite angles3=70°

5=7   Vertically-opposite angles7=70°1+2=180°   Since AFB is a straight line70°+2=180°

2=110°2=4   Vertically-opposite angles4=110°2=6   Corresponding angles

6 =110°6=8   Vertically-opposite angles8=110°1=70°, 2=110°, 3=70°, 4=110°, 5=70°,6 =110°, 7=70° and 8=110°

Page No 129:

Question 2:

In the adjoining figure, AB || CD. The lines are cut by a transversal t at E and F, respectively. If ∠2 : ∠1 = 5 : 4, find the measure of each one of the marked angles.

Answer:

Let 2=5x and 1=4x. Then,
1+2=180°   Since AEB is a straight line5x+4x=180°9x=180°x=20°1=4×20°=80° and 2=5×20°=100°

1=5   Corresponding angles5=80°1=3   Vertically-opposite angles3=80°

5=7   Vertically-opposite angles7=80°

2=4   Vertically-opposite angles4=100°2=6   Corresponding angles

6 =100°6=8   Vertically-opposite angles8=100°1=80°, 2=100°, 3=80°, 4=100°, 5=80°,6 =100°, 7=80° and 8=100°

Page No 129:

Question 3:

In the adjoining figure, ABCD is a quadrilateral in which AB || DC and AD || BC. Prove that ∠ADC = ∠ABC.

Answer:

Let ADBC and CD is the transversal. Then,
ADC+DCB=180°   ...i   Consecutive Interior Angles
Also,
ABCD and BC is the transversal. Then,
DCB+ABC=180°   ...ii   Consecutive Interior Angles
From (i) and (ii), we get:
ADC+DCB=DCB + ABC
ADC=ABC

Page No 129:

Question 4:

In each of the figures given below, AB || CD. Find the value of x in each case.

Answer:

(i)

Draw EFABCD.
Now, ABEF and BE is the transversal.
Then,
ABE=BEF   Alternate Interior AnglesBEF=35°
Again, EFCD and DE is the transversal.
Then,
DEF=FEDFED=65°x°=BEF+FED     =35+65°     =100°or, x=100

(ii)

Draw EOABCD.
Then, EOB+EOD=x°
Now, EOAB and BO is the transversal.
EOB+ABO=180°   Consecutive Interior AnglesEOB+55°=180°EOB=125°
Again, EOCD and DO is the transversal.
EOD+CDO=180°   Consecutive Interior AnglesEOD+25°=180°EOD=155°
Therefore,
x°=EOB+EOD =125+155° =280°or, x=280

(iii)

Draw EFABCD.
Then, AEF+CEF=x°
Now, EFAB and AE is the transversal.
 AEF+BAE=180°   Consecutive Interior Angles AEF+116=180AEF=64°
Again, EFCD and CE is the transversal.
CEF+ECD=180°   Consecutive Interior AnglesCEF+124=180CEF=56°

Therefore,
x°=AEF+CEF  =64+56°  =120°or, x=120



Page No 130:

Question 5:

In the given figure, AB || CD || EF. Find the value of x.

Answer:

EFCD and CE is the transversal.
Then,
ECD+CEF=180°   Consecutive Interior AnglesECD+130°=180°ECD=50°
Again, ABCD and BC is the transversal.
Then,
ABC=BCD   Alternate Interior Angles70°=x+50°  BCD=BCE+ECDx=20°

Page No 130:

Question 6:

In the given figures, AB || CD. Find the value of x.

Answer:


Draw EFABCD.
EFCD and CE is the transversal.
Then,
ECD+CEF=180°   Angles on the same side of a transversal line are supplementary130°+CEF=180°CEF=50°
Again, EFAB and AE is the transversal.
Then,
BAE+AEF=180°  Angles on the same side of a transversal line are supplementaryx°+20°+50°=180°   AEF=AEC+CEFx°+70°=180°x°=110°x=110

Page No 130:

Question 7:

In the given figure, AB || CD. Prove that ∠BAE − ∠DCE = ∠AEC.

Answer:


Draw EFABCD through E.
Now, EFAB and AE is the transversal.
Then, BAE+AEF=180°   Angles on the same side of a transversal line are supplementary
Again, EFCD and CE is the transversal.
Then,
DCE+CEF=180°   Angles on the same side of a transversal line are supplementaryDCE+AEC+AEF=180°DCE+AEC+180°-BAE=180°BAE-DCE=AEC

Page No 130:

Question 8:

In the given figure, AB || CD and BC || ED. Find the value of x.

Answer:

BCED and CD is the transversal.
Then,

BCD+CDE=180°   Angles on the same side of a transversal line are supplementaryBCD+75=180BCD=105°

ABCD and BC is the transversal.

ABC=BCD  (alternate angles) x°=105°x=105

Page No 130:

Question 9:

In the given figure, AB || CD. Prove that p + qr = 180.

Answer:


Draw PFQABCD.
Now, PFQAB and EF is the transversal.
Then,
AEF+EFP=180°.....(1)                                                     Angles on the same side of a transversal line are supplementary
Also, PFQCD.

PFG=FGD=r°Alternate  Anglesand EFP=EFG-PFG=q°-r°putting the value of EFP in eqn. (i)we get,p°+q°-r°=180°p+q-r=180



Page No 131:

Question 10:

In the given figure, AB || PQ. Find the values of x and y.

Answer:



Given, ABPQ.
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,
CEB+BEG+GEF=180°   Since CD is a straight line75°+20°+GEF=180°GEF=85°
We know that the sum of angles of a triangle is 180°.
GEF+EGF+EFG=18085°+x+25°=180°110°+x=180°x=70°
And
FEG+BEG=DFQ   Corresponding Angles85°+20°=DFQDFQ=105°EFG+GFQ+DFQ=180°   Since CD is a straight line25°+y+105°=180°y=50°x=70° and y=50°

Page No 131:

Question 11:

In the given figure, AB || CD. Find the value of x.

Answer:

ABCD and AC is the transversal.
Then,
BAC+ACD=180°   Consecutive Interior Angles75+ACD=180ACD=105°
And,
ACD=ECF   Vertically-Opposite AnglesECF=105°
We know that the sum of the angles of a triangle is 180°.
ECF+CFE+CEF=180°105°+30°+x=180°135°+x=180°x=45°

Page No 131:

Question 12:

In the given figure, AB || CD. Find the value of x.

Answer:

ABCD and PQ is the transversal.
Then,
PEF=EGH   Corresponding AnglesEGH=85°
And,
EGH+QGH=180°   Since PQ is a straight line85°+QGH=180°QGH=95°
Also,
CHQ+GHQ=180°   Since CD is a straight line115°+GHQ=180°GHQ=65°
We know that the sum of angles of a triangle is 180°.
QGH+GHQ+GQH=180°95°+65°+x=180°x=20°x=20°

Page No 131:

Question 13:

In the given figure, AB || CD. Find the values of x, y and z.

Answer:

ADC=DAB   Alternate Interior Anglesz=75°ABC=BCD   Alternate Interior Anglesx=35°
We know that the sum of the angles of a triangle is 180°.
35°+y+75°=180°y=70°x=35°, y=70° and z=75°.

Page No 131:

Question 14:

In the given figure, AB || CD. Find the values of x, y and z.

Answer:



ABCD and let EF and EG be the transversals.
Now, ABCD  and EF is the transversal.
Then,
AEF=EFG   Alternate Anglesy°=75°y=75
Also,
EFC+EFD=180°   Since CFD is a straight linex+y=180x+75=180x=105
And,
EGF+EGD=180°   Since CFGD is a straight lineEGF+125=180EGF=55°
We know that the sum of angles of a triangle is 180°
EFG+GEF+EGF=180°y+z+55=18075+z+55=180z=50x=105, y=75 and z=50



Page No 132:

Question 15:

In the given figure, AB || CD and EF || GH. Find the values of x, y, z and t.

Answer:

In the given figure,
x=60°   Vertically-Opposite AnglesPRQ=SQR   Alternate Anglesy=60°APR=PQS   Corresponding Angles110°=PQR+60°   PQS=PQR+RQSPQR=50°
PQR+RQS+BQS=180°   Since AB is a straight line50°+60°+z=180°110°+z=180°z=70°
DSH=z   Corresponding AnglesDSH=70°DSH=t   Vertically-Opposite Anglest=70° x=60°, y=60°, z=70° and t=70°.

Page No 132:

Question 16:

For what value of x will the lines l and m be parallel to each other?

Answer:

For the lines l and m to be parallel
(i)
3x-20=2x+10   Corresponding Anglesx=30

(ii)
3x+5+4x=180   Consecutive Interior Angles7x=175x=25

Page No 132:

Question 17:

If two straight lines are perpendicular to the same line, prove that they are parallel to each other.

Answer:

Let m and n be the two parallel lines perpendicular to the line l.
Since, l is perpendicular to m,
1=90°
Also, since ​l is perpendicular to m,
2=90°
Therefore, 1=2
This implies that the corresponding angles are equal.
Therefore, the lines m and n are parallel to each other.



Page No 144:

Question 1:

In ∆ABC, if ∠B = 76° and ∠C = 48°, find ∠A.

Answer:

InABC,A+B+C=180°   Sum of the angles of a triangleA+76°+48°=180°A+124°=180°A=56°

Page No 144:

Question 2:

The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles.

Answer:

Let the angles of the given triangle measure 2x°, 3x° and 4x°, respectively.
Then,
2x+3x+4x=180°   Sum of the angles of a triangle9x=180°x=20°
Hence, the measures of the angles are 2×20°=40°, 3×20°=60° and 4×20°=80°.

Page No 144:

Question 3:

In ∆ABC, if 3∠A = 4 ∠B = 6 ∠C, calculate ∠A, ∠B and ∠C.

Answer:

Let 3A=4B=6C=x°.
Then,
A=x3°, B=x4° andC=x6° x3+x4+x6=180°   Sum of the angles of a triangle4x+3x+2x=2160°9x=2160°x=240°
Therefore,
A=2403°=80°, B=2404°=60° andC=2406°=40°

Page No 144:

Question 4:

In ∆ABC, if ∠A + ∠B = 108° and ∠B + ∠C = 130°, find ∠A, ∠B and ∠C.

Answer:

Let A+B=108° and B+C=130°.
A+B+B+C=108+130°A+B+C+B=238°   A+B+C=180°180°+B=238°B=58°

 C=130°-B
          =130-58°=72°

 A=108°-B
         =108-58°=50°

Page No 144:

Question 5:

In ∆ABC, ∠A + ∠B = 125° and ∠A + ∠C = 113°. Find ∠A, ∠B and ∠C.

Answer:

Let A+B=125° and A+C=113°.
Then,
A+B+A+C=125+113°A+B+C+A=238°180°+A=238°A=58°

 B=125°-A
           =125-58°=67°

C=113°- A
         =113-58°=55°

Page No 144:

Question 6:

In ∆PQR, if ∠P − ∠Q = 42° and ∠Q − ∠R = 21°, find ∠P, ∠Q and ∠R.

Answer:

 Given: P-Q=42° and Q-R=21°
Then,
P=42°+Q and R=Q-21°42°+Q+Q+Q-21°=180°   Sum of the angles of a triangle3Q=159°Q=53°

P=42°+Q
         =42+53°=95°

R=Q-21°
         =53-21°=32°

Page No 144:

Question 7:

The sum of two angles of a triangle is 116° and their difference is 24°. Find the measure of each angle of the triangle.

Answer:

Let A+B=116° and A-B=24°
Then,
A+B+A-B=116+24°2A=140°A=70°

B=116°-A
          =116-70°=46°

Also, in ∆ ABC:
A+B+C=180°   Sum of the angles of a triangle70°+46°+C=180°C=64°

Page No 144:

Question 8:

Two angles of a triangle are equal and the third angle is greater than each one of them by 18°. Find the angles.

Answer:

Let A=B and C=A+18°.
Then,
A+B+C=180°   Sum of the angles of a triangleA+A+A+18°=180°3A=162°A=54°
Since,
A=BB=54°C=A+18°

        =54+18°=72°

Page No 144:

Question 9:

Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.

Answer:

Let the smallest angle of the triangle be C and let A=2C and B=3C.
Then,
A+B+C=180°   Sum of the angles of a triangle2C+3C+C=180°6=180°C=30°

A=2C
         =230°=60°
Also,
B=3C     =330°     =90°

Page No 144:

Question 10:

In a right-angled triangle, one of the acute angles measures 53°. Find the measure of each angle of the triangle.

Answer:

Let ABC be a triangle right-angled at B.
Then, B=90°  and let A=53°.
A+B+C=180°   Sum of the angles of a triangle53°+90°+C=180°C=37°
Hence, A=53°, B=90° and C=37°.

Page No 144:

Question 11:

If one angle of a triangle is equal to the sum of the other two, show that the triangle is right-angled.

Answer:

Let ABC be a triangle.
Then,A=B+C
A+B+C=180°   Sum of the angles of a triangleB+C+B+C=180°2B+C=180°B+C=90°A=90°   A=B+C
This implies that the triangle is right-angled at A.

Page No 144:

Question 12:

ABC is right-angled at A. If ALBC, prove that ∠BAL = ∠ACB.

Answer:

We know that the sum of two acute angles of a right angled triangle is 90°
From the right ABL, we have:
BAL+ABL=90°BAL=90°-ABLBAL=90°-ABC       ...(1)

Also, from the right ABC, we have:
ABC+ACB=90°ACB=90°-ABC    ...(2)From (1) and (2), we get:ACB=BAL   BAL=90°-ABCBAL=ACB

Page No 144:

Question 13:

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute-angled.

Answer:

Let ABC be the triangle.
Let A<B+C
Then,
2A<A+B+C   Adding A to both sides2A<180°   A+B+C =180°A<90°

Also, let B<A+C
Then,
2B<A+B+C   Adding B to both sides2B<180°   A+B+C =180°B<90°

And let C<A+B
Then,
2C<A+B+C   Adding C to both sides2C<180°   A+B+C =180°C<90°

Hence, each angle of the triangle is less than 90°.
Therefore, the triangle is acute-angled.

Page No 144:

Question 14:

If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse-angled.

Answer:

Let ABC be a triangle and let C>A+B.
Then, we have:
2C>A+B+C   Adding C to both sides2C>180°   A+B+C=180°C>90°
Since one of the angles of the triangle is greater than 90°, the triangle is obtuse-angled.



Page No 145:

Question 15:

In the given figure, side BC of ∆ABC is produced to D. If ∠ACD = 128° and ∠ABC = 43°, find ∠BAC and ∠ACB.

Answer:

Side BC of triangle ABC is produced to D.
ACD=A+B        [Exterior angle property]128°=A+43°A=128-43°A=85°BAC=85°
Also, in triangle ABC,
BAC+ABC+ACB=180°   Sum of the angles of a triangle85°+43°+ACB=180°128°+ACB=180°ACB=52°

Page No 145:

Question 16:

In the given figure, the side BC of ∆ ABC has been produced on both sides−on the left to D and on the right to E. If ∠ABD = 106° and ∠ACE = 118°, find the measure of each angle of the triangle.

Answer:

Side BC of triangle ABC is produced to D.

ABC=A+C106°=A+C   ...i

Also, side BC of triangle ABC is produced to E.

ACE=A+B118°=A+B   ...ii

Adding (i) and (ii), we get:A+A+B+C=106+118°
A+B+C+A=224°   A+B+C=180°180°+A=224°A=44°

B=118°-A   Using iiB=118-44°B=74°

And,

C=106°-A   Using iC=106-44°C=62°

Page No 145:

Question 17:

Calculate the value of x in each of the following figures.

Answer:

(i)
Side AC of triangle ABC is produced to E.
EAB=B+C110°=x+C   ...i
Also,
ACD+ACB=180°   linear pair120°+ACB=180°ACB=60°C=60°
Substituting the value of C in i, we get x=50

(ii)
From ABC we have:
A+B+C=180°   Sum of the angles of a triangle30°+40°+C=180°C=110°ACB=110°
Also,
ECB+ECD=180°   linear pair110°+ECD=180°ECD=70°Now, in ECD,AED=ECD+EDC     [exterior angle property]x=70°+50°x=120°

(iii)
ACB+ACD=180°   linear pairACB+115°=180°ACB=65°
Also,
EAF=BAC   Vertically-opposite anglesBAC=60°BAC+ABC+ACB=180°   Sum of the angles of a triangle60°+x+65°=180°x=55°

(iv)
BAE=CDE   Alternate anglesCDE=60°ECD+CDE+CED=180°     Sum of the angles of a triangle45°+60°+x=180°x=75°

(v)
From ABC, we have:
BAC+ABC+ACB=180°   Sum of the angles of a triangle40°+ABC+90°=180°ABC=50°
Also, from EBD, we have:
BED+EBD+BDE=180°   Sum of the angles of a triangle100°+50°+x=180°   ABC=EBDx=30°

(vi)
From ABE, we have:
BAE+ABE+AEB=180°   Sum of the angles of a triangle75°+65°+AEB=180°AEB=40°AEB=CED   Vertically-opposite anglesCED=40°

Also, From CDE, we have
ECD+CDE+CED=180°   Sum of the angles of a triangle110°+x+40°=180°x=30°



Page No 146:

Question 18:

Calculate the value of x in the given figure.

Answer:


Join A and D to produce AD to E.
Then,
CAD+DAB=55° andCDE+EDB=x°
Side AD of triangle ACD is produced to E.
CDE=CAD+ACD   ...i   (Exterior angle property)
Side AD of triangle ABD is produced to E.
EDB=DAB+ABD   ...ii   (Exterior angle property)
Adding i and iiwe get,CDE+EDB=CAD+ACD+DAB+ABD
 
                 x°=CAD+DAB+30°+45°x°=55°+30°+45°x°=130°x=130

Page No 146:

Question 19:

In the given figure, AD divides ∠BAC in the ratio 1 : 3 and AD = DB. Determine the value of x.

Answer:

BAC+CAE=180°   BE is a straight lineBAC+108°=180°BAC=72°

Now, divide 72° in the ratio 1 : 3.
a+3a=72°a=18°a=18° and 3a=54°
Hence, the angles are 18o and 54o
BAD=18° and DAC=54°

Given,
AD=DBDAB=DBA=18°

In ABC, we have:
BAC+ABC+ACB=180°   Sum of the angles of a triangle72°+18°+x°=180°x°=90° x=90

Page No 146:

Question 20:

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles.

Answer:

Side BC of triangle ABC is produced to D.
ACD=B+A   ...i
Side AC of triangle ABC is produced to E.
BAC=B+C   ...i
And side AB of triangle ABC is produced to F.
CBF=C+A   ...iii

Adding (i), (ii) and (iii), we get:ACD+BAE+CBF=2A+B+C
                               =2180°=360°=4×90°=4 right angles
Hence, the sum of the exterior angles so formed is equal to four right angles.

Page No 146:

Question 21:

In the adjoining figure, show that
A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°.

Answer:

In ACE , we have :
A+C+E=180°  ...i   [Sum of the angles of a triangle]
In BDF, we have :
B+D+F=180°  ...ii   [Sum of the angles of a triangle]​

Adding (i) and (ii), we get:A+C+E+B+D+F=180+180°
A+B+C+D+E+F=360°

Page No 146:

Question 22:

In ∆ABC, the angle bisectors of ∠B and ∠C meet at O. If ∠A = 70°, find ∠BOC.

Answer:

Let 
ABO=OBC=1   OB is the angle bisector andACO=OCB=2   OC is the angle bisector

From ABC, we have:
A+B+C=180°   Sum of the angles of a triangle
12A+12B+12C=90°12A+1+2=90°   Here,12B=ABO and12C=ACO35°+1+2=90°1+2=55°   ...i  

Now in OBC, we have:
1+2+BOC=180°   Sum of the angles of a triangle1+2+BOC=180°55°+BOC=180°   Using iBOC=125°



Page No 147:

Question 23:

The sides AB and AC of ∆ABC have been produced to D and E, respectively. The bisectors of ∠CBD and ∠BCE meet at O. If ∠A = 40°, find ∠BOC.

Answer:

Let 
OBC=OBD=1   OB is the angle bisector andOCB=OCE=2   OC is the angle bisector
Then,
B+CBD=180°   linear pair12B+1=90°1=90°-12B   ...i
Again,
C+BCE=180° linear pair  12C+12BCE=90°12C+2=90°2=90°-12C   ...ii
Now, in OBC, we have:
1+2+BOC=180°   Sum of the angles of a triangle90°-12B+90°-12C+BOC=180°BOC=12B+CBOC=12A+B+C-12ABOC=12180°-1240°BOC=70°

Page No 147:

Question 24:

In the given figure, ABC is a triangle in which ∠A : ∠B : ∠C = 3 : 2 : 1 and ACCD. Find the measure of ∠ECD.

Answer:

Let A=3x°, B=2x° and C=x°
From ABC, we have:
A+B+C=180°   Sum of the angles of a triangle3x+2x+x=180°6x=180°x=30°A=330°=60°   B=230°=60° and C=30°
Side BC of triangle ABC is produced to E.
ACE=A+BACD+ECD=90+60°90°+ECD=150°ECD=60°

Page No 147:

Question 25:

In the given figure, AMBC and AN is the bisector of ∠A. Find the measure of ∠MAN.

Answer:

From ABC, we have:
A+B+C=180°   Sum of the angles of a triangleA+65°+30°=180°A=85°
Since AN is the bisector of A, we have:
NAB=12ANAB=1285°NAB=42.5°

From the right AMB, we have:
B+MAB=90°MAB=90°-BMAB=90-65°MAB=25°MAN=NAB-MABMAN=42.5-25°MAN=17.5°

Page No 147:

Question 26:

State 'True' or 'False':
(i) A triangle can have two right angles.
(ii) A triangle cannot have two obtuse angles.
(iii) A triangle cannot have two acute angles.
(iv) A triangle can have each angle less than 60°.
(v) A triangle can have each angle equal to 60°.
(vi) There cannot be a triangle whose angles measure 10°, 80° and 100°.

Answer:

(i) The given statement is false.

(ii) The given statement is true.

(iii) ​The given statement is false.

(iv) The given statement is false.

(v) The given statement is true.

(vi) The given statement is true.



Page No 148:

Question 1:

If two angles are complements of each other, then each angle is
(a) an acute angle
(b) an obtuse angle
(c) a right angle
(d) a reflex angle

Answer:

(a) an acute angle

If two angles are complements of each other, that is, the sum of their measures is 90°, then each angle is an acute angle.

Page No 148:

Question 2:

An angle which measures more than 180° but less than 360° is called
(a) an acute angle
(b) an obtuse angle
(c) a straight angle
(d) a reflex angle

Answer:

(d) a reflex angle

An angle which measures more than 180o but less than 360o is called a reflex angle.



Page No 149:

Question 3:

The complement of 72°40' is
(a) 107°20'
(b) 27°20'
(c) 17°20'
(d) 12°40'

Answer:

(c) 17°20'

We know that 90°=89° 60'
The complement of 72° 40'=89° 60'-72° 40'
                                          =17° 20'

Page No 149:

Question 4:

The supplement of 54°30' is
(a) 35°30'
(b) 125°30'
(c) 45°30'
(d) 65°30'

Answer:

(b) 125°30'

We know that 180°=179° 60'
The supplement of 54° 30'=179° 60'-54° 30'
                                        =125° 30'

Page No 149:

Question 5:

The measure of an angle is five times its complement. The angle measures
(a) 25°
(b) 35°
(c) 65°
(d) 75°

Answer:

(d) 75°

Let the measure of the required angle be x°.
Then, the measure of its complement will be 90-x°.
x=590-xx=450-5x6x=450x=75

Page No 149:

Question 6:

Two complementary angles are such that twice the measure of one is equal to three times the measure of the other. The measure of larger angle is
(a) 72°
(b) 54°
(c) 63°
(d) 36°

Answer:

(b) 54°

Let the measure of the required angle be x°.
Then, the measure of its complement will be90-x°.
2x=390-x2x=270-3x5x=270x=54

Page No 149:

Question 7:

Two straight lines AB and CD cut each other at O. If ∠BOD = 63°, then ∠BOC = ?
(a) 63°
(b) 117°
(c) 17°
(d) 153°

Answer:

(b) 117°

We have:
BOC+BOD=180°   Since COD is a straight lineBOC+63°=180°BOC=117°

Page No 149:

Question 8:

In the given figure, AOB is a straight line. If ∠AOC + ∠BOD = 95°, then ∠COD = ?
(a) 95°
(b) 85°
(c) 90°
(d) 55°

Answer:

(b) 85°

We have :
AOC+BOD+COD=180°   Since AOBis a straight lineAOC+BOD+COD=180°95°+COD=180°COD=85°

Page No 149:

Question 9:

In the given figure, AOB is a straight line. If ∠AOC = 4x° and ∠BOC = 5x°, then ∠AOC = ?
(a) 40°
(b) 60°
(c) 80°
(d) 100°

Answer:

(c) 80°

We have :
AOC+BOC=180°   Since AOB is a straight line4x+5x=180°9x=180°x=20°AOC=4×20°=80°

Page No 149:

Question 10:

In the given figure, AOB is a straight line. If ∠AOC = (3x + 10)° and ∠BOC (4x − 26)°, then ∠BOC = ?
(a) 96°
(b) 86°
(c) 76°
(d) 106°

Answer:

(b) 86°

We have :
AOC+BOC=180°   Since AOB is a straight line 3x+10+4x-26=180°7x=196°x=28°

BOC=4×28-26°Hence, BOC=86°.

Page No 149:

Question 11:

In the given figure, AOB is a straight line. If ∠AOC = 40°, ∠COD = 4x° and ∠BOD = 3x°, then ∠COD = ?
(a) 80°
(b) 100°
(c) 120°
(d) 140°

Answer:

(a) 80°

We have :
AOC+COD+BOD=180°   Since AOBis a straight line 40+4x+3x=1807x=140x=20COD=4×20°=80°



Page No 150:

Question 12:

In the given figure, AOB is a straight line. If ∠AOC = (3x − 10)°, ∠COD = 50° and ∠BOD = (x + 20)°, then ∠AOC = ?
(a) 40°
(b) 60°
(c) 80°
(d) 50°

Answer:

(c) 80°

We have :
AOC+COD+BOD=180°   Since AOB is a straight line 3x-10+50+x+20=1804x=120x=30

AOC=3×30-10°AOC=80°

Page No 150:

Question 13:

Which of the following statements is false?
(a) Through a given point, only one straight line can be drawn.
(b) Through two given points, it is possible to draw one and only one straight line.
(c) Two straight lines can intersect at only one point.
(d) A line segment can be produced to any desired length.

Answer:

(a) Through a given point, only one straight line can be drawn.

Clearly, statement (a) is false because we can draw infinitely many straight lines through a given point.

Page No 150:

Question 14:

An angle is one-fifth of its supplement. The measure of the angle is
(a) 15°
(b) 30°
(c) 75°
(d) 150°

Answer:

(b) 30°

Let the measure of the required angle be x°
Then, the measure of its supplement will be 180-x°
x=15180°-x5x=180°-x6x=180°x=30°

Page No 150:

Question 15:

In the adjoining figure, AOB is a straight line. If x : y : z = 4 : 5 : 6, then y = ?
(a) 60°
(b) 80°
(c) 48°
(d) 72°

Answer:

(a) 60°

Let  
AOB=x°=4a°, COB=y°=5a° and BOD=z°=6a°
Then, we have:
AOB+COB+BOD=180°   Since AOB is a straight line4a+5a+6a=180° 15a=180°a=12°y=5×12°=60°

Page No 150:

Question 16:

In the given figure, straight lines AB and CD intersect at O. If AOC = ɸ, ∠BOC = θ and θ = 3ɸ, then ɸ = ?
(a) 30°
(b) 40°
(c) 45°
(d) 60°

Answer:

(c) 45°

We have :
θ+ϕ=180°   AOD is a straight line3ϕ+ϕ=180°   θ=3ϕ4ϕ=180°ϕ=45°

Page No 150:

Question 17:

In the given figure, straight lines AB and CD intersect at O. If AOC + ∠BOD = 130°, then ∠AOD = ?
(a) 65°
(b) 115°
(c) 110°
(d) 125°

Answer:

(b) 115°

We have :
AOC=BOD   Vertically-Opposite AnglesAOC+BOD=130°AOC+AOC=130°   AOC=BOD2AOC=130°AOC=65°
Now,
AOC+AOD=180°  COD is a straight line 65°+AOD=180°AOC=115°

Page No 150:

Question 18:

In the given figure, AB is a mirror, PQ is the incident ray and QR is the reflected ray. If ∠PQR = 108°, find ∠AQP.
(a) 72°
(b) 18°
(c) 36°
(d) 54°

Answer:

(c) 36°

We know that angle of incidence = angle of reflection.
Then, let AQP=BQR=x°
Now,
AQP+PQR+BQR=180°   AQB is a straight line x+108+x=180°2x=72°x=36°AQP=36°



Page No 151:

Question 19:

In the given figure, AB || CD. If ∠OAB = 124° and ∠OCD = 136°, then ∠AOC = ?
(a) 80°
(b) 90°
(c) 100°
(d) 110°

Answer:

(c) 100°

Draw OLABCD.
Now, OLAB and OA is the transversal.
AOL+OAB=180°   Angles on the same side of a transversal line are supplementaryAOL+124°=180°AOL=56°
Also,
OLCD and OC is the transversal.
DCO+COL=180°   Angles on the same side of a transversal line are supplementary136°+COL=180°COL=44°
Therefore,
AOC=AOL+COLAOC=56+44°AOC=100°

Page No 151:

Question 20:

In the given figure, AB || CD and O is a point joined with B and D, as shown in the figure, such that ∠AOB = 35° and ∠CDO = 40°. What will be the measure of reflex ∠BOD ?
(a) 225°
(b) 265°
(c) 275°
(d) 285°

Answer:

(d) 285°

Draw EOFABCD.
Now, EFAB and OB is the transversal.
ABO+EOB=180°   Angles on the same side of a transversal line are supplementary35°+EOB=180°EOB=145°
Also,
EFCD and OD is the transversal.
CDO+EOD=180°   Angles on the same side of a transversal line are supplementary40°+EOD=180°EOD=140°
Therefore,
ReflexBOD=EOB+EODBOD=145+140°BOD=285°

Page No 151:

Question 21:

In the given figure, AB || CD. If ABO = 130° and ∠OCD = 110, then ∠BOC = ?
(a) 50°
(b) 60°
(c) 70°
(d) 80°

Answer:

(b) 60°

Draw EOFABCD through O.
Now, EOAB and OB is the transversal.
ABO+EOB=180°   Angles on the same side of a transversal line are supplementary130°+EOB=180°EOB=50°

Also, OFCD and OC is the transversal.
OCD+COF=180°   Angles on the same side of a transversal line are supplementary110°+COF=180°COF=70°Now,EOB+BOC+COF=180°   EOF is a straight line50°+BOC+70°=180°BOC=60°

Page No 151:

Question 22:

In the given figure, AB || CD. If BAO = 60° and ∠OCD = 110°, then ∠AOC = ?
(a) 70°
(b) 60°
(c) 50°
(d) 40°

Answer:

(c) 50°

Draw EOFABCD.
Now, EOAB and OA is the transversal.
EOA=OAB=60°  Alternate Interior Angles 
Also,
OFCD and OC is the transversal.
COF+OCD=180°   Angles on the same side of a transversal line are supplementaryCOF+110°=180°COF=70°
Now,
EOA+AOC+COF=180°   EOF is a straight line60°+AOC+70°=180°AOC=50°

Page No 151:

Question 23:

In the given figure, AB || CD. If AOC = 30° and ∠OAB = 100°, then ∠OCD = ?
(a) 130°
(b) 150°
(c) 80°
(d) 100°

Answer:

(a) 130°

Draw OEABCD
Now, OEAB and OA is the transversal.
OAB+AOE=180°   Angles on the same side of a transversal line are supplementaryOAB+AOC+COE=180°100°+30°+COE=180°COE=50°
Also, 
OECD and OC is the transversal.
OCD+COE=180°   Angles on the same side of a transversal line are supplementaryOCD+50°=180°OCD=130°

Page No 151:

Question 24:

In the given figure, AB || CD. If CAB = 80° and ∠EFC = 25°, then ∠CEF = ?
(a) 65°
(b) 55°
(c) 45°
(d) 75°

Answer:

(c) 45°

ABCD and AF is the transversal.
DCF=CAB=80°   Corresponding Angles
Side EC of triangle EFC is produced to D.
CEF+EFC=DCFCEF+25°=80°CEF=55°



Page No 152:

Question 25:

In the given figure, AB || CD. If APQ = 70° and ∠PRD = 120°, then ∠QPR = ?
(a) 50°
(b) 60°
(c) 40°
(d) 35°

Answer:

(a) 50°

ABCD and PQ is the transversal.
PQR=APQ=70°   Alternate Interior Angles
Side QR of traingle PQR is produced to D.
PQR+QPR=PRD70°+QPR=120°QPR=50°

Page No 152:

Question 26:

In the given figure, x = ?
(a) α + β − γ
(b) α − β + γ
(c) α + β + γ
(d) α + γ − β

Answer:

(c) α + β + γ

Join OC to produce it to D.
Let BOC=m°, AOC=n°, BCD=p° and ADC=q°.
Then, m+n=β and p+q=x.
Side OC of triangle BOC is produced to D.
p=α+m   ...i
Side OC of triangle AOC is produced to D.
q=n+γ   ...ii
Adding (i) and (ii), we get:p+q=α+γ+m+nx=α+β+γ

Page No 152:

Question 27:

If 3 A = 4 ∠B = 6 ∠C, then A : B : C = ?
(a) 3 : 4 : 6
(b) 4 : 3 : 2

(c) 2 : 3 : 4
(d) 6 : 4 : 3

Answer:

(b) 4 : 3 : 2

Let 3A=4B= 6C=k°
Then, 
A=k3, B=k4 and C=k6A : B : C=k3 : k4 : k6A : B : C=4 : 3 : 2

Page No 152:

Question 28:

In ABC, if ∠A + ∠B = 125° and ∠A + ∠C = 113°, then ∠A = ?
(a) (62.5)°
(b) (56.5)°
(c) 58°

(d) 63°

Answer:

(c) 58°

Let 
A+B=125°   ...i andA+C=113°   ...ii
Adding (i) and (ii), we get:A+A+B+C=125+113°A+A+B+C=238°   A+B+C=180°A+180°=238°A=58°

Page No 152:

Question 29:

In ABC, if ∠A − ∠B = 42° and ∠B − ∠C = 21°, ∠B = ?
(a) 95°
(b) 53°
(c) 32°
(d) 63°

Answer:

(b) 53°

Let 
A-B=42°   ...i andB-C=21°   ...iiAdding (i) and (ii), we get:A-C=63°B=A-42°   Using iC=A-63°   Using iii

A+B+C=180°   Sum of the angles of a triangleA+A-42°+A-63°=180°3A-105°=180°3A=285°A=95°B=95-42°B=53°

Page No 152:

Question 30:

In ABC, side BC is produced to D. If ABC = 40° and ∠ACD = 120°, then A = ?
(a) 60°
(b) 40°
(c) 80°
(d) 50°

Answer:

(c) 80°

A+B=ACDA+40°=120°A=80°

Page No 152:

Question 31:

Side BC of ABC has been produced to D on left hand side and to E on right hand side such that ∠ABD = 125° and ∠ACE = 130°. Then ∠A = ?
(a) 65°
(b) 75°
(c) 50°
(d) 55°

Answer:

(b) 75°

We have :
ABD+ABC=180°   DE is a straight line125°+ABC=180°ABC=55°
Also,
ACE+ACB=180°130°+ACB=180°ACB=50°BAC+ABC+ACB=180°   Sum of the angles of a triangleBAC+55°+50°=180°BAC=75°A=75°

Page No 152:

Question 32:

In the given figure, BAC = 30°, ∠ABC = 50° and ∠CDE = 40°. Then ∠AED = ?
(a) 120°
(b) 100°
(c) 80°
(d) 110°

Answer:

(a) 120°
From ABC, we have:
BAC+ABC+ACB=180°   Sum of the angles of a triangle30°+50°+ACB=180°ACB=100°
Now,
ACB+ECD=180°   BCD is a straight line100°+ECD=180°ECD=80°
Side CE of triangle CED is produced to A.
AED=ECD+EDC            [Exterior angle property]AED=80+40°AED=120°



Page No 153:

Question 33:

In the given figure, BAC = 40°, ∠ACB = 90° and ∠BED = 100°. Then ∠BDE = ?
(a) 50°
(b) 30°
(c) 40°
(d) 25°

Answer:

(b) 30°

In ABC, we have:
A+B+C=180°   Sum of the angles of a triangle40°+B+90°=180°B=50°
In EBD, we have:
E+B+D=180°   Sum of the angles of a triangle100°+50°+D=180°D=30°BDE=30°

Page No 153:

Question 34:

In the given figure, BO and CO are the bisectors of B and ∠C, respectively. If ∠A = 50°, then ∠BOC = ?
(a) 130°
(b) 100°
(c) 115°
(d) 120°

Answer:

(c) 115°

In ABC, we have:
A+B+C=180°   Sum of the angles of a triangle50°+B+C=180°B+C=130°12B+12C=65°   ...i
In OBC, we have:
OBC+OCB+BOC=180°12B+12C+BOC=180°   Using i65°+BOC=180°BOC=115°

Page No 153:

Question 35:

In the given figure, AB || CD. If EAB = 50° and ∠ECD = 60°, then ∠AEB = ?
(a) 50°
(b) 60°
(c) 70°
(d) 55°

Answer:

(c) 70°

ABCD and BC is the transversal.
ABE=BCD=60°   Alternate Internal Angles
In ABE, we have:
EAB+ABE+AEB=180°   Sum of the angles of a triangle50°+60°+AEB=180°AEB=70°

Page No 153:

Question 36:

In the given figure, ∠OAB = 75°, ∠OBA = 55° and ∠OCD = 100°. Then ∠ODC = ?
(a) 20°
(b) 25°
(c) 30°
(d) 35°

Answer:

(c) 30°
In OAB, we have:
OAB+OBA+AOB=180°   Sum of the angles of a triangle75°+55°+AOB=180°AOB=50°
COD=AOB=50°   Vertically-Opposite Angles
In OCD, we have:
COD+OCD+ODC=180°   Sum of the angles of a triangle50°+100°+ODC=180°ODC=30°

Page No 153:

Question 37:

In ∆ABC,A : ∠B : ∠C = 3 : 2 : 1 and CDAC. Then, ∠ECD = ?
(a) 60°
(b) 45°
(c) 75°
(d) 30°

Answer:

Let A=3x°, B=2x° and C=x°
Then,
3x+2x+x=180°   Sum of the angles of a triangle6x=180°x=30°
Hence, the angles are
A=3×30°=90°, B=2×30°=60° and C=30°
Side BC of triangle ABC is produced to E.
ACE=A+BACD+ECD=90°+60°90°+ECD=150°ECD=60°



Page No 154:

Question 38:

In the given figure, AB || CD. If ABO = 45° and ∠COD = 100°, ∠CDO = ?
(a) 25°
(b) 30°
(c) 35°
(d) 45°

Answer:

(c) 35°

ABCD and BOC is the transversal.
OCD=OBA=45°   Alternate Interior Angles
Now, in OCD, we have:
COD+OCD+CDO=180°   Sum of the angles of a triangle100°+45°+CDO=180°CDO=35°

Page No 154:

Question 39:

In the given figure, AB || DC, BAD = 90°, ∠CBD = 28° and ∠BCE = 65°. Then ∠ABD = ?
(a) 32°
(b) 37°
(c) 43°
(d) 53°

Answer:

Side DC of triangle BCD is produced to E.
BCE=CBD+BDC65°=28°+BDCBDC=37°
Now,
ABDC and BD is the transversal.
ABD=BDC=37°   Alternate Interior AnglesABD=37°

Page No 154:

Question 40:

For what value of x, shall l || m?
(a) x = 50
(b) x = 70
(c) x = 60
(d) x = 45

Answer:

(a) x = 50

for l || m we have,

    2x-30=x+20   Corresponding Angles x=50

Page No 154:

Question 41:

For what value of x, shall l || m?
(a) x = 35
(b) x = 30
(c) x = 25
(d) x = 20

Answer:

(c) x = 25

for l || m we have,

   4x+3x+5=180   Angles on the same side of transversal line are supply mentryor, 7x=175 x=25

Page No 154:

Question 42:

In the given figure, sides CB and BA of ∆ABC have been produced to D and E, respectively, such that ∠ABD = 110° and ∠CAE = 135°. Then ∠ACB = ?
(a) 35°
(b) 45°
(c) 55°
(d) 65°

Answer:

(d) 65°

We have :
ABD+ABC=180°   CBD is a straight line100°+ABC=180°ABC=70°
Side AB of triangle ABC is produced to E.
CAE=ABC+ACB135°=70°+ACBACB=65°

Page No 154:

Question 43:

In ABC, BDAC, ∠CAE = 30° and ∠CBD = 40°. Then, ∠AEB = ?
(a) 70°
(b) 80°
(c) 50°
(d) 60°

Answer:

(b) 80°

In BCD, we have:
CBD+BDC+BCD=180°   Sum of the angles of a triangle40°+90°+BCD=180°BCD=50°ECD=50°
Side CE of triangle AEC is produced to B.
AEB=EAC+ACEAEB=30°+50°AEB=80°



Page No 155:

Question 44:

In the given figure, if AB || CD, CD || EF and y : z = 3 : 7, x = ?
(a) 108°
(b) 126°
(c) 162°
(d) 63°

Answer:

(b) 126°
Let y=3a° and z=7a°
Let the transversal intersect AB at P, CD at O and EF at Q.

Then, we have:
COP=DOF=y   Vertically-Opposite AnglesOQF+DOQ=180°   Consecutive Interior Angles3a+7a=180°10a=180°a=18°y=3×18°=54°
Also,
APO+COP=180°x+54°=180°x=126°

Page No 155:

Question 45:

In the given figure, AB || CD || EF, EAAB and BDE is the transversal, such that DEF = 55°. Then, ∠AEB = ?
(a) 35°
(b) 45°
(c) 25°
(d) 55°

Answer:

(a) 35°

We have :
EAAB
Therefore, BAE=90°
And AB  || EF and BE is transversal
  BAE+AEF=180°  90°+AEF=180° AEF=90°

AEB+BEF=90°AEB+DEF=90°AEB+55°=90°AEB=35°

Page No 155:

Question 46:

In the given figure, AMBC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, then ∠MAN = ?
(a) 20°
(b) 25°
(c) 15°
(d) 30°

Answer:

(b) 25°

In ABC, we have:
A+B+C=180°   Sum of the angles of a triangleA+70°+20°=180°A=90°12A=45°BAN=45°
In ABM, we have:
ABM+AMB+BAM=180°   Sum of the angles of a triangle70°+90°+BAM=180°BAM=20°MAN=BAN-BAMMAN=45°-20°MAN=25°

Page No 155:

Question 47:

An exterior angle of a triangle is 110° and one of its interior opposite angles is 45°. The other interior opposite angle is
(a) 45°
(b) 65°
(c) 25°
(d) 135°

Figure

Answer:

Let the required angle be x°.
We know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.
Then,
110°=45°+x°x°=65°            
Therefore, the other interior opposite angle is 65o

Page No 155:

Question 48:

The sides BC, CA and AB of ∆ABC have been produced to D, E and F, respectively, as shown in the figure, forming exterior angles ∠ACD, ∠BAE and ∠CBF. Then, ∠ACD + ∠BAE + ∠CBF = ?
(a) 240°
(b) 300°
(c) 320°
(d) 360°

Answer:

(d) 360°

We have :
1+BAE=180°   ...i2+CBF=180°   ...ii and3+ACD=180°   ...iii
Adding (i), (ii) and (iii), we get:1+2+3+BAE+CBF+ACD=540°
180°+BAE+CBF+ACD=540°   1+2+3=180°BAE+CBF+ACD=360°

Page No 155:

Question 49:

The angles of a triangle are in the ration 3 : 5 : 7. The triangle is
(a) acute-angled
(b) right-angled
(c) obtuse-angled
(d) isosceles

Answer:

(a) acute-angled

Let the angles measure 3x°, 5x° and 7x°.
Then,
3x+5x+7x=180°15x=180°x=12°
Therefore, the angles are 312°=36°, 512°=60° and 712°=84°.
Hence, the triangle is acute-angled.



Page No 156:

Question 50:

If the vertical angle of a triangle is 130°, then the angle between the bisectors of the base angles of the triangle is
(a) 65°
(b) 100°
(c) 130°
(d) 155°

Answer:

(d) 155°



Let ABC be a triangle and let A=130°.
Then, from ABC, we have:
A+B+C=180°   Sum of the angles of a triangle130°+B+C=180°B+C=50°12B+12C=25°   ...i

Now, in BOC, we have:
12B+12C+BOC=180°   Sum of the angles of a triangle25°+BOC=180°   Using iBOC=155°

Page No 156:

Question 51:

The sides BC, BA and CA of ABC have been produced to D, E and F, respectively, as shown in the given figure. Then, ∠B = ?
(a) 35°
(b) 55°
(c) 65°
(d) 75°

Answer:

(d) 75°

We have:
BAC=EAF=35°   Vertically-Opposite Angles
Side CD of triangle ABC is produced to D.
ACD=BAC+ABC110°=35°+ABCABC=75°B=75°

Page No 156:

Question 52:

In the adjoining figure, y = ?
(a) 36°
(b) 54°
(c) 63°
(d) 72°

Answer:

(b) 54°

We have:
3x+72=180°   AOB is a straight line3x=108x=36
Also,
AOC+COD+BOD=180°   AOB is a straight line36°+90°+y=180°y=54°

Page No 156:

Question 53:

Assertion: If two angles of a triangle measure 50° and 70°, its third angle is 60°.
Reason: The sum of the angles of a triangle is 180°.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Assertion:
Let the third angle be x°.
Then,
50°+70°+x°=180°   Sum of the angles of a trianglex°=60°
Hence, the Assertion is true.
Reason:
Clearly, the given statement is true.
Since, both the Assertion and Reason are true and Reason is a correct explanation of Assertion.



Page No 157:

Question 54:

Assertion: If a ray CD stands on a line AB, such that ∠ACD = ∠BCD, then ∠ACD = 90°.
Reason: If a ray
CD stands on a line AB, then ACD + ∠BCD = 180°.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.



Assertion:
ACD+BCD=180°   linear pair12ACD+12BCD=90°12ACD+12ACD=90°   BCD=ACD212ACD=90°ACD=90°
Hence, the Assertion is true.
Reason:
Clearly, the given statement is true.
Hence, both the Assertion and Reason are true and Reason is a correct explanation of Assertion.

Page No 157:

Question 55:

Assertion: If the side BC of ABC is produced to D, then ∠ACD = ∠A + ∠B.
Reason: The sum of the angles of a triangle is 180°.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.










Assertion:
A+B+C=180°   ...i   Sum of the angles of a triangleACB+ACD=180°   ...ii
From i and ii, we get:A+B+C=ACB+ACDA+B+C=C+ACDA+B=ACD

Hence, the Assertion is true.
Reason:
Clearly, the given statement is true.
Hence, both the Assertion and Reason are true but Reason is not a correct explanation of Assertion.

Page No 157:

Question 56:

Assertion: If two lines AB and CD intersect at O, such that AOC = 40°, then ∠BOC = 140°.

Reason: If two straight lines intersect each other, then vertically-opposite angles are equal.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

Assertion:
AOC+BOC=180°   AOB is a straight line40°+BOC=180°BOC=140°
Hence, the Assertion is true.
Reason:
Clearly, the given statement is true.
Hence, both the Assertion and Reason are true but Reason is not a correct explanation of Assertion.

Page No 157:

Question 57:

Assertion: If AB || CD and t is the transversal as shown, then ∠3 = ∠5.

Reason: If a ray stands on a straight line, then the sum of the adjacent angles so formed is 180°.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(d) Assertion is false and Reason is true.

Assertion:
3+5=180°   Consecutive Interior Angles
If the sum of two angles is 180o, it is not necessary that they are equal.
Hence, the Assertion is false.
Reason: "If a ray stands on a straight line, then the sum of the adjacent angles so formed is 180°."
Clearly, the given statement is true.



Page No 158:

Question 58:

Match the following columns:

Column I Column II
(a) If x° and y° be the measures of two complementary angles, such that 2x = 3y, then x = ....... (p) 45°
(b) If an angle is the supplement of itself, then the measure of the angle is ....... (q) 60°
(c) If an angle is the complement of itself, then the measure of the angle is ....... (r) 54°
(d) If x° and y° be the angles forming a linear pair, such that xy = 60°, then = ........ (s) 90°
(a) ......,
(b) ......,
(c) ......,
(d) ......,

Answer:

(a)
We have :
x+y=90°x+23x=90°   2x=3y3x+2x=270°5x=270°x=54°
Hence, (a)(r).

(b)
We have :
x+x=180°2x=180°x=90°
Hence, (b)(s).

(c)
We have :
x+x=90°2x=90°x=45°
Hence, (c)(p).

(d)
We have:
x+y=180°   ...i   Linear pairx-y=60°   ...ii   GivenAdding i and ii, we get:2y=120°y=60°
Hence, (d)(q).

Page No 158:

Question 59:

Match the following columns:

Column I Column II
(a) In the given figure, ACB is a straight line. Then, ACD = ......
(p) 110°
(b) In the given figure, AOC = (2x − 10)° and ∠BOC = (3x − 10)°. Then, ∠AOD = ...... .
(q) 85°
(c) In the given figure, side BC of ABC has been produced to D. If ∠A = 65° and ∠B = 60°, then ∠ACD = ?
(r) 72°
(d) In the given figure, if AB || DE, CDE = 50° and ∠BAC = 45°, then ∠ACB = ...... .
(s) 125°
(a) ......,
(b) ......,
(c) ......,
(d) ......,

Answer:

(a)
We have:
2x+3x=180°   ACB is a straight line5x=180°x=36°ACD=2×36°=72°
Hence, (a)(r).

(b)
We have:
2x-10+3x-10=180°5x=200°x=40°BOC=3×40°-10=110°
Also,
AOD=BOC=110°   Vertically-Opposite AnglesAOD=110°
Hence, (b)(p)

(c)
We have :
ACD=A+B             Exterior angle propertyACD=65°+60°ACD=125°
Hence, (c)(s).

(d)
ABDE and BD is the transversal.
CDE=ABC=50°   Alternate Interior Angles
From ABC, we have:
BAC+ABC+ACB=180°   Sum of the angles of a triangle45°+50°+ACB=180°ACB=85°
Hence, (d)(q)



Page No 166:

Question 1:

The angles of a triangle are in the ratio 3 : 2 : 7. Find the measure of each of its angles.

Answer:

Let the angles measure 3x°, 2x° and 7x°.
Then,
3x+2x+7x=180°   Sum of the angles of a triangle12x=180°x=15°
Therefore, the measures of its angles are 3×15°=45°, 2×15°=30° and 7×15°=105°.

Page No 166:

Question 2:

In ABC, if ∠A − ∠B = 40° and ∠B − ∠C = 10°, find the measures of ∠A, ∠B and ∠C.

Answer:

A-B=40°   ...iB-C=10°   ...iiFrom i, we have:B=A-40°   ...iiiAdding i and ii, we get: A-C=50°C=A-50°   ...iv
Now,
A+B+C=180°   Sum of the angles of a triangleA+A-40°+A-50°=180°   Using iii and iv3A=270°A=90°B=90°-40°=50° and   C=90°-50°=40°



Page No 167:

Question 3:

The side BC of ABC has been increased on both sides as shown. If ∠ABD = 105° and ∠ACE = 110°, find ∠A.

Answer:

We have:
ACD+ACE=180°   Linear pairACD+110°=180°ACD=70°
Side BC of triangle ABC is produced to D.
ABD=BAC+ACD           [Exterior angleproperty]105°=BAC+70°BAC=35°A=35°         

Page No 167:

Question 4:

Prove that the bisectors of two adjacent supplementary angles include a right angle.

Answer:

Let AOB denote a straight line and let AOC and BOC be the supplementary angles.
Then, AOC=x° and BOC=180-x°.
Let OE bisect AOC and OF bisect BOC.
Then, we have:
AOE=COE=12x°BOF=COF=12180-x°
Now,
EOF=COE+COF
        =12x°+12180-x°=12x+180-x=12×180°=90°
Hence, the bisectors of two supplementary angles include a right angle.

Page No 167:

Question 5:

If one angle of a triangle is equal to the sum of the two other angles, show that the triangle is right-angled.

Answer:

Let ABC be a triangle and let A=B+C.
Then,
A+B+C=180°   Sum of the angles of a triangleA+B+C=180°A+A=180°2A=180°A=90°
Therefore, the triangle is right-angled at A.

Page No 167:

Question 6:

In the given figure, ACB is a straight line and CD is a line segment, such that ACD = (3x − 5)° and ∠BCD = (2x + 10)°. Then, x = ?
(a) 25
(b) 30
(c) 35
(d) 40

Answer:

(c) 35
We have:
ACD+BCD=180°   Linear pair3x-5°+2x+10°=180°5x°=175°x°=35°

Page No 167:

Question 7:

In the given figure, AOB is a straight line. If AOC = 40°, ∠COD = 4x° and ∠BOD = 3x°, then x = ?
(a) 20
(b) 25
(c) 30
(d) 35

Answer:

(a) 20

We have:
AOC+COD+BOD=180°   AOB is a straight line40+4x+3x=180°7x=140°x=20°

Page No 167:

Question 8:

The supplement of an angle is six times its complement. The measure of this angle is
(a) 36°
(b) 54°
(c) 60°
(d) 72°

Answer:

(d) 72°

Let the required angle be x°.
Then, measure of its supplement = 180-x° and
measure of its complement = 90-x°.
180-x=690-x5x=360°x=72°

Page No 167:

Question 9:

In the given figure, AB || CD || EF. If ABC = 85°, ∠BCE = x° and CEF = 130°, then x = ?
(a) 30
(b) 25
(c) 35
(d) 15

Answer:

(c) 35

EFCD and CE is the transversal.
CEF+ECD=180°   Consecutive Interior Angles130°+ECD=180°ECD=50°
Also, ABCD and BC is the transversal.
ABC=BCD   Alternate Interior Angles85°=BCE+ECD85°=x+50°x=35°

Page No 167:

Question 10:

In the given figure, AB || CD, BAD = 30° and ∠ECD = 50°. Find ∠CED.

Answer:

ABCD and AD is the transversal.
CDE=BAD=30°   Alternate  Interior Angles
In CDE, we have:
ECD+CDE+CED=180°   Sum of the angles of a triangle50°+30°+CED=180°CED=100°



Page No 168:

Question 11:

In the given figure, BAD || EF, AEF = 55° and ∠ACB = 25°. Find ∠ABC.

Answer:

BADEF and AE is the transversal.
 AEF+EAD=180°   Consecutive Interior Angles55°+EAD=180°EAD=125°
Also,
BAC=EAD=125°   Vertically-Opposite Angles
From ABC, we have:
BAC+ABC+ACB=180°   Sum of the angles of a triangle125°+ABC+25°=180°ABC=30°

Page No 168:

Question 12:

In the given figure, BE AC, ∠DAC = 30° and ∠DBE = 40°. Find ∠ACB and ∠ADB.

Answer:

From BEC, we have:
CBE+BEC+ECB=180°   Sum of the angles of a triangle40°+90°+ECB=180°ECB=50°Now, ACB=ECB=50°
Side CD of triangle ACD is produced to B.
ADB=DAC+ACDADB=30°+50°   ACD=ECBADB=80°

Page No 168:

Question 13:

In the given figure, AB || CD and EF is a transversal, cutting them at G and H, respectively. If EGB = 35° and QPEF, find the measure of ∠PQH.

Answer:

We have:
EGB=GHD=35°   Corresponding Angles
Also,
GHD=PHQ=35°   Vertically-Opposite Angles
Now, in PQH, we have:
HPQ+PQH+PHQ=180°   Sum of the angles of a triangle90°+PQH+35°=180°PQH=55°

Page No 168:

Question 14:

In the given figure, AB || CD and EF AB. If EG is the transversal, such that ∠GED = 130°, find ∠EGF.

Answer:

We have:
CEG+GED=180°   CED is a straight lineCEG+130°=180°CEG=50°
ABCD and EG is the transversal.
EGF=CEG=50°   Alternate Interior AnglesEGF=50°

Page No 168:

Question 15:

Match the following columns:

Column I Column II
(a) An angle is 10° more than its complement. The measure of the angle is ...... . (p) 160°
(b) If in ABC, ∠A = 65° and ∠B − ∠C = 25°, then ∠B = ...... (q) 50°
(c) In ABC, ∠A = 40° and ∠B = ∠C. If EF || BC, then ∠EFC = ...... .
(r) 70°
(d) If the angles around a point are 2x°, 3x°, 5x° and 40°, the measure of the largest angle is ...... (s) 110°
(a) ......
(b) ......
(c) ......
(d) ......

Answer:

(a)
Let the required angle be x°.
Then, measure of its complement = 90-x°.
x=90-x+102x=100°x=50°
Hence, (a)(q).

(b)
We have:
A+B+C=180°   Sum of the angles of a triangle65°+B+C=180°B+C=115°  ...iAlso, B-C=25°  ...ii   Given
Adding i and ii, we get:  2B=140°B=70°
Hence, (b)(r).

(c)
In ABC, we have:
A+B+C=180°   Sum of the angles of a triangle40°+B+B=180°   B=C40°+2B=180°2B=140°B=70°
EFBC and AEBis the transversal.
ABC=AEF=70°  Corresponding Angles 
Now, side AF of triangle AEF is produced to C.
EFC=EAF+AEFEFC=40°+70°EFC=110°
Hence, (c)(s).

(d)
We know that the sum of the angles around a point is 360°.
2x+3x+5x+40°=360°10x=320°x=32°
Hence, the angles are
232°=64°, 332°=96°, 532°=160° and 40°
Therefore, the largest angle is 160°.
Hence, (d)(p).



Page No 169:

Question 16:

(i) In the given figure, lines AB and CD intersect at O, such that AOD + ∠BOD + ∠BOC = 300°. Find ∠AOD.

(ii) In the given figure, AB || CD, ∠APQ = 50° and ∠PRD = 120°. Find ∠QPR.

Answer:

(i)
We know that the sum of the angles around a point is 360°.
AOD+BOD+BOC+AOC=360°AOD+BOD+BOC+AOC=360°300°+AOC=360°AOC=60°
Also,
AOC=BOD=60°   Vertically-Opposite Angles
Now,
AOD+BOD=180°   AOB is a straight lineAOD+60°=180°AOD=120°

(ii)
ABCD and PQ is the transversal.
APQ=PQR=50°   Alternate Interior Angles
Side QR of triangle PQR is produced to D.
PRD=QPR+PQR120°=QPR+50°QPR=70°

Page No 169:

Question 17:

In the given figure, BE is the bisector of B and CE is the bisector of ∠ACD. Prove that BEC=12A.

Answer:

Side BC of triangle ABC is produced to D.
ACD=B+A12ACD=12B+12AECD=12B+12A   ...i
Also, side BC of triangle EBC is produced to D.
ECD=CBE+BECECD=12B+BEC   ...ii
From (i) and (ii), we get:
12B+BEC=12B+12ABEC=12A



Page No 170:

Question 18:

In ABC, sides AB and AC are produced to D and E, respectively. BO and CO are the bisectors of ∠CBD and ∠BCE, respectively. Prove that BOC=90°-12A.

Answer:

Let
OBC=OBD=1 andOCB=OCE=2
We have:
ABC+CBD=180°   Linear pair12ABC+12CBD=90°12ABC+1=90°   12CBD=OBC=OBD1=90°-12ABC   ...i

Also,
ACB+BCE=180°   Linear pair12ACB+12BCE=90°   12BCE=OCB=OCE12ACB+2=90°2=90°-12ACB   ...ii
Now, in OBC, we have:
OBC+OCB+BOC=180°1+2+BOC=180°90°-12ABC +90°-12ACB+BOC=180°   Using i and ii

BOC=12ABC+ACBBOC=12A+ABC+ACB-12ABOC=12180°-12A   Sum of the angles of a triangle is 180°BOC=90°-12A

Page No 170:

Question 19:

Of the three angles of a triangle, one is twice the smallest and another is thrice the smallest. Find the angles.

Answer:

Let ABC be a triangle and let C be the smallest angle.
Then, let A=2C and B=3C
Now,
A+B+C=180°   Sum of the angles of a triangle2C+3C+C=180°6C=180°C=30°A=230°=60°and B=330°=90°

Page No 170:

Question 20:

In ABC, ∠B = 90° and BDAC. Prove that ∠ABD = ∠ACB.

Answer:

Side AD of triangle ABD is produced to C.
BDC=BAD+ABD   ...i
From the right ABC, we have:
ABC=BAC+ACB90°=BAD+ACB90°=BDC-ABD+ACB   Using i90°=90°-ABD+ACB   BDACABD=ACB



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