RS Aggarwal Solutions: Triangles | Mathematics (Maths) Class 9 PDF Download

 Page 1


Q u e s t i o n : 1
In ?ABC, if ?B = 76° and ?C = 48°, find ?A.
S o l u t i o n :
In ? ABC, ?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? ?A +76°+48° = 180° ? ?A +124° = 180° ? ?A = 56°
Q u e s t i o n : 2
The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles.
S o l u t i o n :
Let the angles of the given triangle measure (2x)°, (3x)° and (4x)°
, respectively.
Then,
2x +3x +4x = 180°   [Sum of the angles of a triangle] ? 9x = 180° ? x = 20°
Hence, the measures of the angles are 2 ×20° = 40°, 3 ×20° = 60° and 4 ×20°=80°
.
Q u e s t i o n : 3
In ?ABC, if 3 ?A = 4 ?B = 6 ?C, calculate ?A, ?B and ?C.
S o l u t i o n :
Let 3 ?A = 4 ?B = 6 ?C = x°
.
Then,
?A =
x
3
°
, ?B =
x
4
°
 and ?C =
x
6
°
? 
x
3
+
x
4
+
x
6
= 180°   [Sum of the angles of a triangle] ? 4x +3x +2x = 2160° ? 9x = 2160° ? x = 240°
Therefore,
?A =
240
3
°
= 80°, ?B =
240
4
°
= 60° and ?C =
240
6
°
= 40°
Q u e s t i o n : 4
In ?ABC, if ?A + ?B = 108° and ?B + ?C = 130°, find ?A, ?B and ?C.
S o l u t i o n :
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°
Q u e s t i o n : 5
In ?ABC, ?A + ?B = 125° and ?A + ?C = 113°. Find ?A, ?B and ?C.
S o l u t i o n :
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°
Q u e s t i o n : 6
In ?PQR, if ?P - ?Q = 42° and ?Q - ?R = 21°, find ?P, ?Q and ?R.
S o l u t i o n :
 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 triangle] ? 3 ?Q = 159° ? ?Q = 53°
? ?P = 42°+ ?Q
         = (42 +53)° = 95°
? ?R = ?Q-21°
         = (53 -21)° = 32°
Q u e s t i o n : 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.
( ) ( ) ( )
( ) ( ) ( )
Page 2


Q u e s t i o n : 1
In ?ABC, if ?B = 76° and ?C = 48°, find ?A.
S o l u t i o n :
In ? ABC, ?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? ?A +76°+48° = 180° ? ?A +124° = 180° ? ?A = 56°
Q u e s t i o n : 2
The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles.
S o l u t i o n :
Let the angles of the given triangle measure (2x)°, (3x)° and (4x)°
, respectively.
Then,
2x +3x +4x = 180°   [Sum of the angles of a triangle] ? 9x = 180° ? x = 20°
Hence, the measures of the angles are 2 ×20° = 40°, 3 ×20° = 60° and 4 ×20°=80°
.
Q u e s t i o n : 3
In ?ABC, if 3 ?A = 4 ?B = 6 ?C, calculate ?A, ?B and ?C.
S o l u t i o n :
Let 3 ?A = 4 ?B = 6 ?C = x°
.
Then,
?A =
x
3
°
, ?B =
x
4
°
 and ?C =
x
6
°
? 
x
3
+
x
4
+
x
6
= 180°   [Sum of the angles of a triangle] ? 4x +3x +2x = 2160° ? 9x = 2160° ? x = 240°
Therefore,
?A =
240
3
°
= 80°, ?B =
240
4
°
= 60° and ?C =
240
6
°
= 40°
Q u e s t i o n : 4
In ?ABC, if ?A + ?B = 108° and ?B + ?C = 130°, find ?A, ?B and ?C.
S o l u t i o n :
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°
Q u e s t i o n : 5
In ?ABC, ?A + ?B = 125° and ?A + ?C = 113°. Find ?A, ?B and ?C.
S o l u t i o n :
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°
Q u e s t i o n : 6
In ?PQR, if ?P - ?Q = 42° and ?Q - ?R = 21°, find ?P, ?Q and ?R.
S o l u t i o n :
 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 triangle] ? 3 ?Q = 159° ? ?Q = 53°
? ?P = 42°+ ?Q
         = (42 +53)° = 95°
? ?R = ?Q-21°
         = (53 -21)° = 32°
Q u e s t i o n : 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.
( ) ( ) ( )
( ) ( ) ( )
S o l u t i o n :
Let ?A + ?B = 116° and ?A - ?B = 24°
Then,
? ?A + ?B + ?A - ?B = (116 +24)° ? 2 ?A = 140° ? ?A = 7 0 °
? ?B = 116°- ?A
          = (116 -70)° = 4 6 °
Also, in ? ABC:
?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? 70°+46°+ ?C = 180° ? ?C = 6 4 °
Q u e s t i o n : 8
Two angles of a triangle are equal and the third angle is greater than each one of them by 18°. Find the angles.
S o l u t i o n :
Let ?A = ?B and ?C = ?A +18°
.
Then,
?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ?A + ?A + ?A +18° = 180° ? 3 ?A = 162° ? ?A = 54°
Since,
?A = ?B ? ?B = 54° ? ?C = ?A +18°
        = (54 +18)° = 72°
Q u e s t i o n : 9
Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.
S o l u t i o n :
Let the smallest angle of the triangle be ?C
and let ?A = 2 ?C and ?B = 3 ?C
.
Then,
?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? 2 ?C +3 ?C + ?C = 180° ? 6 ? = 180° ? ?C = 30°
? ?A = 2 ?C
         = 2(30)° = 60°
Also,
?B = 3 ?C     = 3(30)°     = 90°
Q u e s t i o n : 1 0
In a right-angled triangle, one of the acute angles measures 53°. Find the measure of each angle of the triangle.
S o l u t i o n :
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 triangle] ? 53°+90°+ ?C = 180° ? ?C = 37°
Hence, ?A = 53°, ?B = 90° and ?C = 37°
.
Q u e s t i o n : 1 1
If one angle of a triangle is equal to the sum of the other two, show that the triangle is right-angled.
S o l u t i o n :
Let ABC be a triangle.
Then, ?A = ?B + ?C
? ?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? ?B + ?C + ?B + ?C = 180° ? 2 ?B + ?C = 180° ? ?B + ?C = 90° ? ?A = 9 0 °   [ ? ?A = ?B + ?C]
This implies that the triangle is right-angled at A.
Q u e s t i o n : 1 2
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute-angled.
S o l u t i o n :
Let ABC be the triangle.
Let ?A < ?B + ?C
Then,
2 ?A < ?A + ?B + ?C   [Adding ?A to both sides] ? 2 ?A < 180°   [ ? ?A + ?B + ?C = 180°] ? ?A < 9 0 °
Also, let ?B < ?A + ?C
Then,
2 ?B < ?A + ?B + ?C   [Adding ?B to both sides] ? 2 ?B < 180°   [ ? ?A + ?B + ?C = 180°] ? ?B < 9 0 °
And let ?C < ?A + ?B
Then,
2 ?C < ?A + ?B + ?C   [Adding ?C to both sides] ? 2 ?C < 180°   [ ? ?A + ?B + ?C = 180°] ? ?C < 9 0 °
Hence, each angle of the triangle is less than 90°
Page 3


Q u e s t i o n : 1
In ?ABC, if ?B = 76° and ?C = 48°, find ?A.
S o l u t i o n :
In ? ABC, ?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? ?A +76°+48° = 180° ? ?A +124° = 180° ? ?A = 56°
Q u e s t i o n : 2
The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles.
S o l u t i o n :
Let the angles of the given triangle measure (2x)°, (3x)° and (4x)°
, respectively.
Then,
2x +3x +4x = 180°   [Sum of the angles of a triangle] ? 9x = 180° ? x = 20°
Hence, the measures of the angles are 2 ×20° = 40°, 3 ×20° = 60° and 4 ×20°=80°
.
Q u e s t i o n : 3
In ?ABC, if 3 ?A = 4 ?B = 6 ?C, calculate ?A, ?B and ?C.
S o l u t i o n :
Let 3 ?A = 4 ?B = 6 ?C = x°
.
Then,
?A =
x
3
°
, ?B =
x
4
°
 and ?C =
x
6
°
? 
x
3
+
x
4
+
x
6
= 180°   [Sum of the angles of a triangle] ? 4x +3x +2x = 2160° ? 9x = 2160° ? x = 240°
Therefore,
?A =
240
3
°
= 80°, ?B =
240
4
°
= 60° and ?C =
240
6
°
= 40°
Q u e s t i o n : 4
In ?ABC, if ?A + ?B = 108° and ?B + ?C = 130°, find ?A, ?B and ?C.
S o l u t i o n :
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°
Q u e s t i o n : 5
In ?ABC, ?A + ?B = 125° and ?A + ?C = 113°. Find ?A, ?B and ?C.
S o l u t i o n :
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°
Q u e s t i o n : 6
In ?PQR, if ?P - ?Q = 42° and ?Q - ?R = 21°, find ?P, ?Q and ?R.
S o l u t i o n :
 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 triangle] ? 3 ?Q = 159° ? ?Q = 53°
? ?P = 42°+ ?Q
         = (42 +53)° = 95°
? ?R = ?Q-21°
         = (53 -21)° = 32°
Q u e s t i o n : 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.
( ) ( ) ( )
( ) ( ) ( )
S o l u t i o n :
Let ?A + ?B = 116° and ?A - ?B = 24°
Then,
? ?A + ?B + ?A - ?B = (116 +24)° ? 2 ?A = 140° ? ?A = 7 0 °
? ?B = 116°- ?A
          = (116 -70)° = 4 6 °
Also, in ? ABC:
?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? 70°+46°+ ?C = 180° ? ?C = 6 4 °
Q u e s t i o n : 8
Two angles of a triangle are equal and the third angle is greater than each one of them by 18°. Find the angles.
S o l u t i o n :
Let ?A = ?B and ?C = ?A +18°
.
Then,
?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ?A + ?A + ?A +18° = 180° ? 3 ?A = 162° ? ?A = 54°
Since,
?A = ?B ? ?B = 54° ? ?C = ?A +18°
        = (54 +18)° = 72°
Q u e s t i o n : 9
Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.
S o l u t i o n :
Let the smallest angle of the triangle be ?C
and let ?A = 2 ?C and ?B = 3 ?C
.
Then,
?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? 2 ?C +3 ?C + ?C = 180° ? 6 ? = 180° ? ?C = 30°
? ?A = 2 ?C
         = 2(30)° = 60°
Also,
?B = 3 ?C     = 3(30)°     = 90°
Q u e s t i o n : 1 0
In a right-angled triangle, one of the acute angles measures 53°. Find the measure of each angle of the triangle.
S o l u t i o n :
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 triangle] ? 53°+90°+ ?C = 180° ? ?C = 37°
Hence, ?A = 53°, ?B = 90° and ?C = 37°
.
Q u e s t i o n : 1 1
If one angle of a triangle is equal to the sum of the other two, show that the triangle is right-angled.
S o l u t i o n :
Let ABC be a triangle.
Then, ?A = ?B + ?C
? ?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? ?B + ?C + ?B + ?C = 180° ? 2 ?B + ?C = 180° ? ?B + ?C = 90° ? ?A = 9 0 °   [ ? ?A = ?B + ?C]
This implies that the triangle is right-angled at A.
Q u e s t i o n : 1 2
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute-angled.
S o l u t i o n :
Let ABC be the triangle.
Let ?A < ?B + ?C
Then,
2 ?A < ?A + ?B + ?C   [Adding ?A to both sides] ? 2 ?A < 180°   [ ? ?A + ?B + ?C = 180°] ? ?A < 9 0 °
Also, let ?B < ?A + ?C
Then,
2 ?B < ?A + ?B + ?C   [Adding ?B to both sides] ? 2 ?B < 180°   [ ? ?A + ?B + ?C = 180°] ? ?B < 9 0 °
And let ?C < ?A + ?B
Then,
2 ?C < ?A + ?B + ?C   [Adding ?C to both sides] ? 2 ?C < 180°   [ ? ?A + ?B + ?C = 180°] ? ?C < 9 0 °
Hence, each angle of the triangle is less than 90°
.
Therefore, the triangle is acute-angled.
Q u e s t i o n : 1 3
If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse-angled.
S o l u t i o n :
Let ABC be a triangle and let ?C > ?A + ?B
.
Then, we have:
2 ?C > ?A + ?B + ?C   [Adding ?C to both sides] ? 2 ?C > 180°   [ ? ?A + ?B + ?C = 180°] ? ?C > 9 0 °
Since one of the angles of the triangle is greater than 90°
, the triangle is obtuse-angled.
Q u e s t i o n : 1 4
In the given figure, side BC of ?ABC is produced to D. If ?ACD = 128° and ?ABC = 43°, find ?BAC and ?ACB.
S o l u t i o n :
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 triangle] ? 85°+43°+ ?ACB = 180° ? 128°+ ?ACB = 180° ? ?ACB = 52°
Q u e s t i o n : 1 5
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.
S o l u t i o n :
Side BC of triangle ABC is produced to D.
? ?ABC = ?A + ?C ? 106° = ?A + ?C   . . . (i)
Also, side BC of triangle ABC is produced to E.
?ACE = ?A + ?B ? 118° = ?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 (ii)] ? ?B = (118 -44)° ? ?B = 74°
And,
?C = 106°- ?A   [Using (i)] ? ?C = (106 -44)° ? ?C = 62°
Q u e s t i o n : 1 6
Calculate the value of x in each of the following figures.
S o l u t i o n :
Page 4


Q u e s t i o n : 1
In ?ABC, if ?B = 76° and ?C = 48°, find ?A.
S o l u t i o n :
In ? ABC, ?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? ?A +76°+48° = 180° ? ?A +124° = 180° ? ?A = 56°
Q u e s t i o n : 2
The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles.
S o l u t i o n :
Let the angles of the given triangle measure (2x)°, (3x)° and (4x)°
, respectively.
Then,
2x +3x +4x = 180°   [Sum of the angles of a triangle] ? 9x = 180° ? x = 20°
Hence, the measures of the angles are 2 ×20° = 40°, 3 ×20° = 60° and 4 ×20°=80°
.
Q u e s t i o n : 3
In ?ABC, if 3 ?A = 4 ?B = 6 ?C, calculate ?A, ?B and ?C.
S o l u t i o n :
Let 3 ?A = 4 ?B = 6 ?C = x°
.
Then,
?A =
x
3
°
, ?B =
x
4
°
 and ?C =
x
6
°
? 
x
3
+
x
4
+
x
6
= 180°   [Sum of the angles of a triangle] ? 4x +3x +2x = 2160° ? 9x = 2160° ? x = 240°
Therefore,
?A =
240
3
°
= 80°, ?B =
240
4
°
= 60° and ?C =
240
6
°
= 40°
Q u e s t i o n : 4
In ?ABC, if ?A + ?B = 108° and ?B + ?C = 130°, find ?A, ?B and ?C.
S o l u t i o n :
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°
Q u e s t i o n : 5
In ?ABC, ?A + ?B = 125° and ?A + ?C = 113°. Find ?A, ?B and ?C.
S o l u t i o n :
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°
Q u e s t i o n : 6
In ?PQR, if ?P - ?Q = 42° and ?Q - ?R = 21°, find ?P, ?Q and ?R.
S o l u t i o n :
 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 triangle] ? 3 ?Q = 159° ? ?Q = 53°
? ?P = 42°+ ?Q
         = (42 +53)° = 95°
? ?R = ?Q-21°
         = (53 -21)° = 32°
Q u e s t i o n : 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.
( ) ( ) ( )
( ) ( ) ( )
S o l u t i o n :
Let ?A + ?B = 116° and ?A - ?B = 24°
Then,
? ?A + ?B + ?A - ?B = (116 +24)° ? 2 ?A = 140° ? ?A = 7 0 °
? ?B = 116°- ?A
          = (116 -70)° = 4 6 °
Also, in ? ABC:
?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? 70°+46°+ ?C = 180° ? ?C = 6 4 °
Q u e s t i o n : 8
Two angles of a triangle are equal and the third angle is greater than each one of them by 18°. Find the angles.
S o l u t i o n :
Let ?A = ?B and ?C = ?A +18°
.
Then,
?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ?A + ?A + ?A +18° = 180° ? 3 ?A = 162° ? ?A = 54°
Since,
?A = ?B ? ?B = 54° ? ?C = ?A +18°
        = (54 +18)° = 72°
Q u e s t i o n : 9
Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.
S o l u t i o n :
Let the smallest angle of the triangle be ?C
and let ?A = 2 ?C and ?B = 3 ?C
.
Then,
?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? 2 ?C +3 ?C + ?C = 180° ? 6 ? = 180° ? ?C = 30°
? ?A = 2 ?C
         = 2(30)° = 60°
Also,
?B = 3 ?C     = 3(30)°     = 90°
Q u e s t i o n : 1 0
In a right-angled triangle, one of the acute angles measures 53°. Find the measure of each angle of the triangle.
S o l u t i o n :
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 triangle] ? 53°+90°+ ?C = 180° ? ?C = 37°
Hence, ?A = 53°, ?B = 90° and ?C = 37°
.
Q u e s t i o n : 1 1
If one angle of a triangle is equal to the sum of the other two, show that the triangle is right-angled.
S o l u t i o n :
Let ABC be a triangle.
Then, ?A = ?B + ?C
? ?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? ?B + ?C + ?B + ?C = 180° ? 2 ?B + ?C = 180° ? ?B + ?C = 90° ? ?A = 9 0 °   [ ? ?A = ?B + ?C]
This implies that the triangle is right-angled at A.
Q u e s t i o n : 1 2
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute-angled.
S o l u t i o n :
Let ABC be the triangle.
Let ?A < ?B + ?C
Then,
2 ?A < ?A + ?B + ?C   [Adding ?A to both sides] ? 2 ?A < 180°   [ ? ?A + ?B + ?C = 180°] ? ?A < 9 0 °
Also, let ?B < ?A + ?C
Then,
2 ?B < ?A + ?B + ?C   [Adding ?B to both sides] ? 2 ?B < 180°   [ ? ?A + ?B + ?C = 180°] ? ?B < 9 0 °
And let ?C < ?A + ?B
Then,
2 ?C < ?A + ?B + ?C   [Adding ?C to both sides] ? 2 ?C < 180°   [ ? ?A + ?B + ?C = 180°] ? ?C < 9 0 °
Hence, each angle of the triangle is less than 90°
.
Therefore, the triangle is acute-angled.
Q u e s t i o n : 1 3
If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse-angled.
S o l u t i o n :
Let ABC be a triangle and let ?C > ?A + ?B
.
Then, we have:
2 ?C > ?A + ?B + ?C   [Adding ?C to both sides] ? 2 ?C > 180°   [ ? ?A + ?B + ?C = 180°] ? ?C > 9 0 °
Since one of the angles of the triangle is greater than 90°
, the triangle is obtuse-angled.
Q u e s t i o n : 1 4
In the given figure, side BC of ?ABC is produced to D. If ?ACD = 128° and ?ABC = 43°, find ?BAC and ?ACB.
S o l u t i o n :
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 triangle] ? 85°+43°+ ?ACB = 180° ? 128°+ ?ACB = 180° ? ?ACB = 52°
Q u e s t i o n : 1 5
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.
S o l u t i o n :
Side BC of triangle ABC is produced to D.
? ?ABC = ?A + ?C ? 106° = ?A + ?C   . . . (i)
Also, side BC of triangle ABC is produced to E.
?ACE = ?A + ?B ? 118° = ?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 (ii)] ? ?B = (118 -44)° ? ?B = 74°
And,
?C = 106°- ?A   [Using (i)] ? ?C = (106 -44)° ? ?C = 62°
Q u e s t i o n : 1 6
Calculate the value of x in each of the following figures.
S o l u t i o n :
i
Side AC of triangle ABC is produced to E.
? ?EAB = ?B + ?C ? 110° = x + ?C   . . . (i)
Also,
?ACD + ?ACB = 180°   [linear pair] ? 120°+ ?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 triangle] ? 30°+40°+ ?C = 180° ? ?C = 110° ? ?ACB = 110°
Also,
?ECB + ?ECD = 180°   [linear pair] ? 110°+ ?ECD = 180° ? ?ECD = 70°Now, in ? ECD, ? ?AED = ?ECD + ?EDC     [exterior angle property ] ? x = 70°+50° ? x = 120°
iii
?ACB + ?ACD = 180°   [linear pair] ? ?ACB +115° = 180° ? ?ACB = 65°
Also,
?EAF = ?BAC   [Vertically-opposite angles] ? ?BAC = 60° ? ?BAC + ?ABC + ?ACB = 180°   [Sum of the angles of a triangle] ? 60°+x +65° = 180° ? x = 55°
iv
?BAE = ?CDE   [Alternate angles] ? ?CDE = 60° ? ?ECD + ?CDE + ?CED = 180°     [Sum of the angles of a triangle] ? 45°+60°+x = 180° ? x = 75°
v
From ? ABC
, we have:
?BAC + ?ABC + ?ACB = 180°   [Sum of the angles of a triangle] ? 40°+ ?ABC +90° = 180° ? ?ABC = 50°
Also, from ? EBD
, we have:
?BED + ?EBD + ?BDE = 180°   [Sum of the angles of a triangle] ? 100°+50°+x = 180°   [ ? ?ABC = ?EBD] ? x = 30°
vi
From ? ABE
, we have:
?BAE + ?ABE + ?AEB = 180°   [Sum of the angles of a triangle] ? 75°+65°+ ?AEB = 180° ? ?AEB = 40° ? ?AEB = ?CED   [Vertically-opposite angles] ? ?CED = 40°
Also, From ? CDE
, we have
?ECD + ?CDE + ?CED = 180°   [Sum of the angles of a triangle] ? 110°+x +40° = 180° ? x = 30°
Q u e s t i o n : 1 7
In the figure given alongside, AB || CD, EF || BC, ?BAC = 60º and ?DHF = 50º. Find ?GCH and ?AGH. 
S o l u t i o n :
In the given figure, AB || CD and AC is the transversal.
? ?ACD = ?BAC = 60º     Pairofalternateangles
Or ?GCH = 60º
Now, ?GHC = ?DHF = 50º      Verticallyoppositeangles
In ?GCH,
?AGH = ?GCH + ?GHC       Exteriorangleofatriangleisequaltothesumofthetwointerioroppositeangles
? ?AGH = 60º + 50º = 110º
 
Q u e s t i o n : 1 8
Calculate the value of x in the given figure.
Page 5


Q u e s t i o n : 1
In ?ABC, if ?B = 76° and ?C = 48°, find ?A.
S o l u t i o n :
In ? ABC, ?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? ?A +76°+48° = 180° ? ?A +124° = 180° ? ?A = 56°
Q u e s t i o n : 2
The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles.
S o l u t i o n :
Let the angles of the given triangle measure (2x)°, (3x)° and (4x)°
, respectively.
Then,
2x +3x +4x = 180°   [Sum of the angles of a triangle] ? 9x = 180° ? x = 20°
Hence, the measures of the angles are 2 ×20° = 40°, 3 ×20° = 60° and 4 ×20°=80°
.
Q u e s t i o n : 3
In ?ABC, if 3 ?A = 4 ?B = 6 ?C, calculate ?A, ?B and ?C.
S o l u t i o n :
Let 3 ?A = 4 ?B = 6 ?C = x°
.
Then,
?A =
x
3
°
, ?B =
x
4
°
 and ?C =
x
6
°
? 
x
3
+
x
4
+
x
6
= 180°   [Sum of the angles of a triangle] ? 4x +3x +2x = 2160° ? 9x = 2160° ? x = 240°
Therefore,
?A =
240
3
°
= 80°, ?B =
240
4
°
= 60° and ?C =
240
6
°
= 40°
Q u e s t i o n : 4
In ?ABC, if ?A + ?B = 108° and ?B + ?C = 130°, find ?A, ?B and ?C.
S o l u t i o n :
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°
Q u e s t i o n : 5
In ?ABC, ?A + ?B = 125° and ?A + ?C = 113°. Find ?A, ?B and ?C.
S o l u t i o n :
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°
Q u e s t i o n : 6
In ?PQR, if ?P - ?Q = 42° and ?Q - ?R = 21°, find ?P, ?Q and ?R.
S o l u t i o n :
 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 triangle] ? 3 ?Q = 159° ? ?Q = 53°
? ?P = 42°+ ?Q
         = (42 +53)° = 95°
? ?R = ?Q-21°
         = (53 -21)° = 32°
Q u e s t i o n : 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.
( ) ( ) ( )
( ) ( ) ( )
S o l u t i o n :
Let ?A + ?B = 116° and ?A - ?B = 24°
Then,
? ?A + ?B + ?A - ?B = (116 +24)° ? 2 ?A = 140° ? ?A = 7 0 °
? ?B = 116°- ?A
          = (116 -70)° = 4 6 °
Also, in ? ABC:
?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? 70°+46°+ ?C = 180° ? ?C = 6 4 °
Q u e s t i o n : 8
Two angles of a triangle are equal and the third angle is greater than each one of them by 18°. Find the angles.
S o l u t i o n :
Let ?A = ?B and ?C = ?A +18°
.
Then,
?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ?A + ?A + ?A +18° = 180° ? 3 ?A = 162° ? ?A = 54°
Since,
?A = ?B ? ?B = 54° ? ?C = ?A +18°
        = (54 +18)° = 72°
Q u e s t i o n : 9
Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.
S o l u t i o n :
Let the smallest angle of the triangle be ?C
and let ?A = 2 ?C and ?B = 3 ?C
.
Then,
?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? 2 ?C +3 ?C + ?C = 180° ? 6 ? = 180° ? ?C = 30°
? ?A = 2 ?C
         = 2(30)° = 60°
Also,
?B = 3 ?C     = 3(30)°     = 90°
Q u e s t i o n : 1 0
In a right-angled triangle, one of the acute angles measures 53°. Find the measure of each angle of the triangle.
S o l u t i o n :
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 triangle] ? 53°+90°+ ?C = 180° ? ?C = 37°
Hence, ?A = 53°, ?B = 90° and ?C = 37°
.
Q u e s t i o n : 1 1
If one angle of a triangle is equal to the sum of the other two, show that the triangle is right-angled.
S o l u t i o n :
Let ABC be a triangle.
Then, ?A = ?B + ?C
? ?A + ?B + ?C = 180°   [Sum of the angles of a triangle] ? ?B + ?C + ?B + ?C = 180° ? 2 ?B + ?C = 180° ? ?B + ?C = 90° ? ?A = 9 0 °   [ ? ?A = ?B + ?C]
This implies that the triangle is right-angled at A.
Q u e s t i o n : 1 2
If each angle of a triangle is less than the sum of the other two, show that the triangle is acute-angled.
S o l u t i o n :
Let ABC be the triangle.
Let ?A < ?B + ?C
Then,
2 ?A < ?A + ?B + ?C   [Adding ?A to both sides] ? 2 ?A < 180°   [ ? ?A + ?B + ?C = 180°] ? ?A < 9 0 °
Also, let ?B < ?A + ?C
Then,
2 ?B < ?A + ?B + ?C   [Adding ?B to both sides] ? 2 ?B < 180°   [ ? ?A + ?B + ?C = 180°] ? ?B < 9 0 °
And let ?C < ?A + ?B
Then,
2 ?C < ?A + ?B + ?C   [Adding ?C to both sides] ? 2 ?C < 180°   [ ? ?A + ?B + ?C = 180°] ? ?C < 9 0 °
Hence, each angle of the triangle is less than 90°
.
Therefore, the triangle is acute-angled.
Q u e s t i o n : 1 3
If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse-angled.
S o l u t i o n :
Let ABC be a triangle and let ?C > ?A + ?B
.
Then, we have:
2 ?C > ?A + ?B + ?C   [Adding ?C to both sides] ? 2 ?C > 180°   [ ? ?A + ?B + ?C = 180°] ? ?C > 9 0 °
Since one of the angles of the triangle is greater than 90°
, the triangle is obtuse-angled.
Q u e s t i o n : 1 4
In the given figure, side BC of ?ABC is produced to D. If ?ACD = 128° and ?ABC = 43°, find ?BAC and ?ACB.
S o l u t i o n :
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 triangle] ? 85°+43°+ ?ACB = 180° ? 128°+ ?ACB = 180° ? ?ACB = 52°
Q u e s t i o n : 1 5
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.
S o l u t i o n :
Side BC of triangle ABC is produced to D.
? ?ABC = ?A + ?C ? 106° = ?A + ?C   . . . (i)
Also, side BC of triangle ABC is produced to E.
?ACE = ?A + ?B ? 118° = ?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 (ii)] ? ?B = (118 -44)° ? ?B = 74°
And,
?C = 106°- ?A   [Using (i)] ? ?C = (106 -44)° ? ?C = 62°
Q u e s t i o n : 1 6
Calculate the value of x in each of the following figures.
S o l u t i o n :
i
Side AC of triangle ABC is produced to E.
? ?EAB = ?B + ?C ? 110° = x + ?C   . . . (i)
Also,
?ACD + ?ACB = 180°   [linear pair] ? 120°+ ?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 triangle] ? 30°+40°+ ?C = 180° ? ?C = 110° ? ?ACB = 110°
Also,
?ECB + ?ECD = 180°   [linear pair] ? 110°+ ?ECD = 180° ? ?ECD = 70°Now, in ? ECD, ? ?AED = ?ECD + ?EDC     [exterior angle property ] ? x = 70°+50° ? x = 120°
iii
?ACB + ?ACD = 180°   [linear pair] ? ?ACB +115° = 180° ? ?ACB = 65°
Also,
?EAF = ?BAC   [Vertically-opposite angles] ? ?BAC = 60° ? ?BAC + ?ABC + ?ACB = 180°   [Sum of the angles of a triangle] ? 60°+x +65° = 180° ? x = 55°
iv
?BAE = ?CDE   [Alternate angles] ? ?CDE = 60° ? ?ECD + ?CDE + ?CED = 180°     [Sum of the angles of a triangle] ? 45°+60°+x = 180° ? x = 75°
v
From ? ABC
, we have:
?BAC + ?ABC + ?ACB = 180°   [Sum of the angles of a triangle] ? 40°+ ?ABC +90° = 180° ? ?ABC = 50°
Also, from ? EBD
, we have:
?BED + ?EBD + ?BDE = 180°   [Sum of the angles of a triangle] ? 100°+50°+x = 180°   [ ? ?ABC = ?EBD] ? x = 30°
vi
From ? ABE
, we have:
?BAE + ?ABE + ?AEB = 180°   [Sum of the angles of a triangle] ? 75°+65°+ ?AEB = 180° ? ?AEB = 40° ? ?AEB = ?CED   [Vertically-opposite angles] ? ?CED = 40°
Also, From ? CDE
, we have
?ECD + ?CDE + ?CED = 180°   [Sum of the angles of a triangle] ? 110°+x +40° = 180° ? x = 30°
Q u e s t i o n : 1 7
In the figure given alongside, AB || CD, EF || BC, ?BAC = 60º and ?DHF = 50º. Find ?GCH and ?AGH. 
S o l u t i o n :
In the given figure, AB || CD and AC is the transversal.
? ?ACD = ?BAC = 60º     Pairofalternateangles
Or ?GCH = 60º
Now, ?GHC = ?DHF = 50º      Verticallyoppositeangles
In ?GCH,
?AGH = ?GCH + ?GHC       Exteriorangleofatriangleisequaltothesumofthetwointerioroppositeangles
? ?AGH = 60º + 50º = 110º
 
Q u e s t i o n : 1 8
Calculate the value of x in the given figure.
S o l u t i o n :
Join A and D to produce AD to E.
Then,
?CAD + ?DAB = 55° and ?CDE + ?EDB = x°
Side AD of triangle ACD is produced to E.
? ?CDE = ?CAD + ?ACD   . . . (i)
   Exteriorangleproperty
Side AD of triangle ABD is produced to E.
? ?EDB = ?DAB + ?ABD   . . . (ii)
   Exteriorangleproperty
Adding (i) and (ii)we get, ?CDE + ?EDB = ?CAD + ?ACD + ?DAB + ?ABD
 
                 ? x° = ( ?CAD + ?DAB)+30°+45° ? x° = 55°+30°+45° ? x° = 130° ? x = 130
Q u e s t i o n : 1 9
In the given figure, AD divides ?BAC in the ratio 1 : 3 and AD = DB. Determine the value of x.
S o l u t i o n :
?BAC + ?CAE = 180°   [ ? BE is a straight line] ? ?BAC +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 18
o
 and 54
o
? ?BAD = 18° and ?DAC = 54°
Given,
AD = DB ? ?DAB = ?DBA = 18°
In ? ABC
, we have:
?BAC + ?ABC + ?ACB = 180°   [Sum of the angles of a triangle] ? 72°+18°+x° = 180° ? x° = 90° ? x = 90
Q u e s t i o n : 2 0
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.
S o l u t i o n :
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 = 2( ?A + ?B + ?C)
                               = 2(180)° = 360° = 4 ×90° = 4 right angles
Hence, the sum of the exterior angles so formed is equal to four right angles.
Q u e s t i o n : 2 1
In the adjoining figure, show that
?A + ?B + ?C + ?D + ?E + ?F = 360°.