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Page No 252:

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 252:

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 252:

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 252:

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 252:

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 252:

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 252:

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 252:

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 252:

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 252:

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 252:

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 252:

Question 12:

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 252:

Question 13:

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 252:

Question 14:

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 252:

Question 15:

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 253:

Question 16:

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 253:

Question 17:

In the figure given alongside, AB || CD, EF || BC, ∠BAC = 60º and ∠DHF = 50º. Find ∠GCH and ∠AGH. 

Answer:


In the given figure, AB || CD and AC is the transversal.

∴ ∠ACD = ∠BAC = 60º     (Pair of alternate angles)

Or ∠GCH = 60º

Now, ∠GHC = ∠DHF = 50º      (Vertically opposite angles)

In ∆GCH,

∠AGH = ∠GCH + ∠GHC       (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)

⇒ ∠AGH = 60º + 50º = 110º
 

Page No 253:

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 254:

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 254:

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 254:

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 254:

Question 22:

In the given figure, AMBC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, then ∠MAN = ?

Answer:

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 254:

Question 23:

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

Answer:


In the given figure, EF || BD and CE is the transversal.

∴ ∠CAD = ∠AEF         (Pair of corresponding angles)

⇒ ∠CAD = 55°

In ∆ABC,

∠CAD = ∠ABC + ∠ACB       (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)

⇒ 55° = ∠ABC + 25°

⇒ ∠ABC = 55° − 25° = 30°

Thus, the measure of ∠ABC is 30°.



Page No 255:

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 255:

Question 25:

In the given figure, AB || DE and BD || FG such that ∠ABC = 50° and ∠FGH = 120°. Find the values of x and y.

Answer:


We have, ∠FGE + ∠FGH = 180°       (Linear pair of angles)

y + 120° = 180°

⇒ y = 180° − 120° = 60°

Now, AB || DF and BD is the transversal.

∴ ∠ABD = ∠BDF           (Pair of alternate angles)

⇒ ∠BDF = 50°

Also, BD || FG and DF is the transversal.

∴ ∠BDF = ∠DFG           (Pair of alternate angles)

⇒ ∠DFG = 50°       .....(1)

In ∆EFG,

∠FGH = ∠EFG + ∠FEG       (Exterior angle of a triangle is equal to the sum of the two opposite interior angles)

⇒ 120° = 50° + x                   [Using (1)]

⇒ x = 120° − 50° = 70°

Thus, the values of x and y are 70° and 60°, respectively.

Page No 255:

Question 26:

In the given figure, AB || CD and EF is a transversal. If ∠AEF = 65°, ∠DFG = 30°, ∠EGF = 90° and ∠GEF = x°, find the value of x.

Answer:


It is given that, AB || CD and EF is a transversal.

∴ ∠EFD = ∠AEF          (Pair of alternate angles)

⇒ ∠EFD = 65°

⇒ ∠EFG + ∠GFD = 65° 

⇒ ∠EFG + 30° = 65°

⇒ ∠EFG = 65° − 30° = 35°

In ∆EFG, 

∠EFG + ∠GEF + ∠EGF = 180°           (Angle sum property)

⇒ 35° + x + 90° = 180°

⇒ 125° + x = 180°

⇒ x = 180° − 125° = 55°

Thus, the value of x is 55°.

Page No 255:

Question 27:

In the given figure, AB || CD, ∠BAE = 65° and ∠OEC = 20°. Find ∠ECO.

Answer:


In the given figure, AB || CD and AE is the transversal.

∴ ∠DOE = ∠BAE         (Pair of corresponding angles)

⇒ ∠DOE = 65°

In ∆COE,

∠DOE = ∠OEC + ∠ECO          (Exterior angle of a triangle is equal to the sum of two opposite interior angles)

⇒ 65° = 20° + ∠ECO

⇒ ∠ECO = 65° − 20° = 45°

Thus, the measure of ∠ECO is 45°.

Page No 255:

Question 28:

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

Answer:


In the given figure, AB || CD and EF is a transversal.

∴ ∠PHQ = ∠EGB          (Pair of alternate exterior angles)

⇒ ∠PHQ = 35°

In ∆PHQ,

∠PHQ + ∠QPH + ∠PQH = 180°     (Angle sum property)

⇒ 35° + 90° + x = 180°

⇒ 125° + x = 180°

x = 180° − 125° = 55°

Thus, the measure of ∠PQH is 55°.

Page No 255:

Question 29:

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

Answer:


In the given figure, AB || CD and GE is the transversal.

∴ ∠GED + ∠EGF = 180°       (Sum of adjacent interior angles on the same side of the transversal is supplementary)

⇒ 130° + ∠EGF = 180°

⇒ ∠EGF = 180° − 130° = 50°

Thus, the measure of ∠EGF is 50°.



Page No 257:

Question 1:

In ∆ABC, 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:


LCM of 3, 4 and 6 = 12

3∠A = 4∠B = 6∠C      (Given)

Dividing throughout by 12, we get

3A12=4B12=6C12

A4=B3=C2

Let A4=B3=C2=k, where k is some constant

Then, ∠A = 4k,  ∠B = 3k, ∠C = 2k

∴ ∠A : ∠B : ∠C = 4k : 3k : 2k = 4 : 3 : 2

Hence, the correct answer is option (b).



Page No 258:

Question 2:

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

Figure

Answer:

(c) 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 258:

Question 3:

In ∆ABC, side BC is produced to D. If ∠ABC = 50° and ∠ACD = 110° then A = ?
(a) 160°
(b) 60°
(c) 80°
(d) 30°

Answer:

A+B=ACDA+50°=110°A=60°
Hence, the correct answer is option (b).

Page No 258:

Question 4:

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

Answer:

(d) 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 258:

Question 5:

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

Answer:

(a) 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 258:

Question 6:

The sides BC, CA and AB of ∆ABC have been produced to D, E and F, respectively. ∠BAE + ∠CBF + ACD = ?
(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 258:

Question 7:

The the given figure, EAD  BCD. Ray FAC cuts ray EAD at a point A such that ∠EAF = 30°. Also, in ∆BAC, BAC = x° and ∠ABC = (x + 10)°. Then, the value of x is
(a) 20
(b) 25
(c) 30
(d) 35

Answer:


In the given figure, ∠CAD = ∠EAF            (Vertically opposite angles)

∴ ∠CAD = 30°        

In ∆ABD,

∠ABD + ∠BAD + ∠ADB = 180°          (Angle sum property)

⇒ (x + 10)° + (x° + 30°) + 90° = 180°

⇒ 2x° + 130° = 180°

⇒ 2x° = 180° − 130° = 50°

⇒ x = 25

Thus, the value of x is 25.

Hence, the correct answer is option (b).



Page No 259:

Question 8:

In the given figure, two rays BD and CE intersect at a point A. The side BC of ∆ABC have been produced on both sides to points F and G respectively. If ABF = x°, ∠ACG = y° and ∠DAE = z° then z = ?
(a) x + y – 180
(b) x + y + 180
(c) 180 – (x + y)
(d) x + y + 360°

Answer:


In the given figure, ∠ABF + ∠ABC = 180°         (Linear pair of angles)

∴ x° + ∠ABC = 180°

⇒ ∠ABC = 180° − x°        .....(1)

Also, ∠ACG + ∠ACB = 180°         (Linear pair of angles)

∴ y° + ∠ACB = 180°

⇒ ∠ACB = 180° − y°        .....(2)

Also, ∠BAC = ∠DAE = z°       .....(3)         (Vertically opposite angles)

In ∆ABC,

∠BAC + ∠ABC + ∠ACB = 180°       (Angle sum property)

∴ z° + 180° − x° + 180° − y° = 180°        [Using (1), (2) and (3)]

⇒ zxy − 180

Hence, the correct answer is option (a).

Page No 259:

Question 9:

In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and F respectively such that ∠OAE = x° and ∠DBF = y°.

If ∠OCA = 80°, ∠COA = 40° and ∠BDO = 70° then x° + y° = ?
(a) 190°
(b) 230°
(c) 210°
(d) 270°

Answer:


In the given figure, ∠BOD = ∠COA          (Vertically opposite angles)

∴ ∠BOD = 40°        .....(1)

In ∆ACO,

∠OAE = ∠OCA + ∠COA        (Exterior angle of a triangle is equal to the sum of two opposite interior angles)

⇒ x° = 80° + 40° = 120°           .....(2)

In ∆BDO,

∠DBF = ∠BDO + ∠BOD        (Exterior angle of a triangle is equal to the sum of two opposite interior angles)

⇒ y° = 70° + 40° = 110°           [Using (1)]         .....(3)

Adding (2) and (3), we get

x° + y° = 120° + 110° = 230°

Hence, the correct answer is option (b).

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Question 10:

In a ∆ABC, it is given that ∠A : ∠B : ∠C = 3 : 2 : 1 and ∠ACD = 90°. If BC is produced to E then ∠ECD = ?
(a) 60°
(b) 50°
(c) 40°
(d) 25°

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°
Hence, the correct answer is option (a).

Page No 259:

Question 11:

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 260:

Question 12:

In the given figure, side BC of ∆ABC has been produced to a point D. If ∠A = 3y°, ∠B = x°, ∠C = 5y° and ∠CBD = 7y°. Then, the value of x is

(a) 60
(b) 50
(c) 45
(d) 35

Answer:

Disclaimer: In the question ACD should be 7y°.

In the given figure, ∠ACB + ∠ACD = 180°         (Linear pair of angles)

∴ 5y° + 7y° = 180°

⇒ 12y° = 180°

⇒ y = 15         .....(1)

In ∆ABC,

∠A + ∠B + ∠ACB = 180°       (Angle sum property)

∴ 3y° + x° + 5y° = 180°

⇒ x° + 8y° = 180°

⇒ x° + 8 × 15° = 180°              [Using (1)]

⇒ x° + 120° = 180°

⇒ x° = 180° − 120° = 60°

Thus, the value of x is 60.

Hence, the correct answer is option (a).



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