Opt. Maths Book for Class-9

= 02 + –21 2 +  –1 2 +  3 2 + (–1)2
  2  2 

= 14 + 1 + 1 + 3 + 1
2 2 4

= 1 + 2 +3+4
4
10
= 4

= 25

4. Prove that:

a) cos  pc  + cos  3pc  + cos  5pc  + cos  7pc  = 0
8 8 8 8
Solution

LHS cos  pc  + cos  3pc  + cos  5pc  + cos  7pc 
8 8 8 8

= cos  pc  + cos  3pc  + cos p – 3pc  + cos pc – pc 
8 8 8  8

= cos  pc  + cos  3pc  – cos  3pc  – cos  pc 
8 8 8 8

= 0

 LHS = RHS proved

b) cos2  pc  + cos2  3pc  + cos2  5pc  + cos2  7pc  = 2
8 8 8 8
Solution

LHS cos2  pc  + cos2  3pc  + cos2  5pc  + cos2  7pc 
 8   8   8   8 

= cos2  pc  + cos2  3pc  + cos2 pc – 3pc  + cos2 pc – pc 
 8   8   8   8 

= cos2  pc  + cos2  3pc  + cos2  3pc  + cos2  pc 
 8   8   8   8 

= 2 cos2  pc  + cos2  3pc 
 8   8 

= 2 cos2  pc  + cos2  pc – pc 
 8   2 8 

= 2 cos2  pc  + sin2  pc 
 8   8 

= 2.1 ( sin2q + cos2q = 1)

= 2

 LHS = RHS proved

Vedanta Optional Mathematics Teacher's Guide ~ 9 151

c) sin2  7pc  + sin2  5pc  + sin2  3pc  + sin2  pc  = 2
8 8 8 8

Hint: Same as (b)

5. Find the value of q (0° ≤ q ≤ 90°)

a) sin2q = cosq

Solution
Here, sin2q = cosq
or, sin2q = sin(90° – q)
or, 2q = 90° – q
or, 3q = 90°
 q = 30°

b) tan2q = cotq

Solution
Here, tan2q = cotq
or, tan2q = tan(90° – q)
or, 2q = 90° – q
or, 3q = 90°
 q = 30°

6. Prove that:

sin30° – cos60° + tan45° + sin210° – cos240° + tan135° = 0

Solution
LHS sin30° – cos60° + tan45° + sin210° – cos240° + tan135°
= sin30° – cos60° + tan45° + sin(180° + 30°) – cos(180° + 60°) + tan(90° + 45°)

= 12 – 1 + 1 – 1 + 1 – 1
2 2 2
 sin(180° + 30°) = –sin30° = –21
= 0 cos(180° + 60°) = –cos60° = –21

 LHS = RHS proved

tan(90° + 45°) = –cot45° = –1

7. Prove that:

S olutsiionn(1s8in0q° – q) + tan(180° – q) + cos(180° – q) = –1
tanq cosq

LHS sin(180° – q) + tan(180° – q) + cos(180° – q)
sinq tanq cosq

= ssiinnqq + –tanq + –cosq
tanq cosq

= 1 – 1 – 1

152 Vedanta Optional Mathematics Teacher's Guide ~ 9

= –1
 LHS = RHS proved

8. Solve the following equation for x.
a) tan2225° – sin2120° = x . sin45° . cos135° . tan120°

Solution
Here, tan2225° – sin2120° = x . sin45° . cos135° . tan120°
or, tan2(180° + 45°) – sin2(180° – 60°) = x . sin45° . cos(180° – 45°) . tan(180° – 60°)
or, tan245° – sin260° = x . sin45° . (–cos45°) . tan (–tan60°)

or, 12 –  3 2 = x . 1 . – 1  . (– 3)
 2  2  2

or, 1 – 3 = x . 3
4 2

or, 41 = 3 x
2

 x = 1
23

b) 3 sin120° + x cos120° . tan135° = x . cot330° . tan150°

Solution
Here, 3 sin120° + x cos120° . tan135° = x . cot330° . tan150°
or, 3 sin(180° – 60°) + x cos(180° – 60°) . tan(180° – 45°)

= x . cot(360° – 30°) . tan(180° – 30°)
or, 3 sin60° + x(–cos60°) . (–tan45°) = x (–cot30°) . (–tan30°)

or, 3× 3 + x . 1 . 1 = x . 3× 1
2 2 3

or, 3 3 + x = x
2

 x = 3 3

c) x cot(270° + q) . cot(90° + q) = tan(180° – q) tan(360° – q) cosec(90° – q) cosec(90° + q)

Solution
Here, x cot(270° + q) . cot(90° + q) = tan(180° – q) . tan(360° – q)

. cosec(90° – q) . cosec(90° + q)
or, x(–tanq) . (–tanq) = –tanq . (–tanq) . secq . secq
or, x . tan2q = tan2q . sec2q
 x = sec2q

d) tan(90° + q) . cot(180° – q) + cosec(360° – q) . cosecq = x . cotq . tan(90° + q)

Solution
Here, tan(90° + q) . cot(180° – q) + cosec(360° – q) . cosecq = x . cotq . tan(90° + q)

Vedanta Optional Mathematics Teacher's Guide ~ 9 153

or, (–cotq) . (–cotq) + (–cosec2q) . cosecq = x cotq . (–cotq)
or, cot2q – cosecq = –x cot2q
or, x cot2q = cosec2q – cot2q
or, x cot2q = 1
 x = tan2q

e) sin2135° – x tan2150° = cos2150°

Solution
Here, sin2135° – x tan2150° = cos2150°
or, sin2(180° – 45°) – tan2(180° – 30°) = cos2(180° – 30°)
or, sin245° – x tan230° = cos230°

or,  1 2 – x .  1 2 =  3 2
 2 3 2 

or, 21 – x . 1 = 3
3 4

or, –3x = 3 – 1
4 2

or, –3x = 3 – 2
4

or, –3x = 1
4

 x = –43
9. If A, B and C are the angles of a triangle, prove the following:

a) cos(A + B) = –C

Solution ( sum of angles of a triangle)
Here, A + B + C = 180°
A + B = 180° – C
or, cos(A + B) = cos(180° – C)
 cos(A + B) = –cosC

b) sin  A + B  = cos C
2 2
Solution

Here, A + B + C = 180° ( sum of angles of a triangle)

A + B = 180° – C

or, A +2 B = 180° – C
2

or, A + B = sin 90° – C2 
 2  

 sin  A + B  = cos C
 2  2

154 Vedanta Optional Mathematics Teacher's Guide ~ 9

c) tan(A + B) + tanC = 0

Solution
Here, A + B + C = 180°
A + B = 180° – C
or, tan(A + B) = tan(180° – C)
or, tan(A + B) = –tanC
 tan(A + B) + tanC = 0

d) tan  A + B  tan C = 1
2 2
Solution

Here, A + B + C = 180°

A + B = 180° – C

or, A +2 B = 180° – C
2

or, tan A + B = tan 90° – C 
 2   2 

or, tan A + B = cot C
 2  2

or, tan A + B = 1 C
 2  tan 2

 tan A + B tan C = 1
 2  2

Compound Angles 3
2
1. Prove that: a) sin75° + cos75° =
Solution

LHS sin75° + cos75°

= sin(45° + 30°) + sin(45° + 30°)

= sin45° cos30° + sin30° cos45° + cos45° cos30° – sin45° sin30°

= 1 . 3 + 1 . 1+ 1. 3 – 1 . 1
2 2 2 2 2 2 2 2

= 3 + 3
22 22

= 2 3
22

= 3
2

 LHS = RHS proved

Vedanta Optional Mathematics Teacher's Guide ~ 9 155

b) tan15° + cot15° = 4

Solution
Here, tan15° + cot15°

= 1t+ant4a5n°4–5°tatnan303°0° + cot45° cot30° + 1
cot30° – cot45°

1– 1 +1. 3+1
3
=
1 3–1
1+1. 3

= 3 – 1 + 3 + 1
3+1 3–1

= ( 3 – 1)2 + ( 3 + 1)2
( 3 + 1) ( 3 – 1)

= 3 – 2 3 + 1 + 3 + 2 3 + 1

= 8 3–1
2

= 4

 LHS = RHS proved

c) sin75° – sin15° = 1

Solution 2

LHS sin75° – sin15°

= sin(45° + 30°) – sin(45° – 30°)

= sin45° cos30° + sin30° cos45° – sin45° cos30° + sin30° cos45°

= 2 sin30° . cos45°

= 2 . 1 . 1
2 2

= 1 2

 LHS = RHS proved

2. Prove that: a) tan53° – tan8° = 1 + tan53° . tan8°

Solution
Here, 45° = 53° – 45°
tan45° = tan(53° – 8°)

or, 1 = tan53° – tan8°
1 + tan53° tan8°

156 Vedanta Optional Mathematics Teacher's Guide ~ 9

 tan53° – tan8° = 1 + tan53° . tan8°
 LHS = RHS proved

b) tan8q – tan5q – tan3q = tan8q tan5q tan3q

Solution
Here, 8q = 5q + 3q
tan8q = tan(5q + 3q)

or, tan8q = tan5q + tan3q
1 – tan5q tan3q

or, tan8q – tan8q . tan5q . tan3q = tan5q + tan3q
 tan8q – tan5q – tan3q = tan8q tan5q tan3q
 LHS = RHS proved

c) tan(A – B) + tan(B – C) + tan(C – A) = tan(A – B) . tan(B – C) . tan(C – A)

Solution
We have, A – B + B – C + C – A = 0
(A – B) = –[(B – C) + (C – A)]

or, tan(A – B) = –{tan(B – C) + tan(C – A)}
1 – tan(B – C) tan(C – A)

or, tanA – B) – tan(A – B) . tan(B – C) . tan(C – A) = –tan(B – C) –tan(C – A)
 tan(A – B) + tan(B – C) + tan(C – A) = tan(A – B) . tan(B – C) . tan(C – A)
 LHS = RHS proved

3. Prove that: a) 2 sin(q + 45°) . sin(q – 45°) = sin2q – cos2q

Solution
LHS 2 sin(q + 45°) . sin(q – 45°)
= 2(sinq . cos45° + cosq . sin45°) . (sinq . cos45° – cosq . cos45°)

= 2  1 sinq + 1 cosq .  1 sinq – 1 cosq 
 2 2  2 2 

= 2 . 1 (sinq + cosq) (sinq – cosq)
2

= sin2q – cos2q
 LHS = RHS proved

b) sin(x + 2)q . cos(x + 1)q – cos(x + 2)q . sin(x + 1)q = sinq

Solution
LHS sin(x + 2)q . cos(x + 1)q – cos(x + 2)q . sin(x + 1)q
= sin{(x + 2)q – (x + 1)q}
= sin(x + 2 – x – 1)q
= sinq
 LHS = RHS proved

Vedanta Optional Mathematics Teacher's Guide ~ 9 157

c) sin2(q + 45°) + sin2(45° – q) = 1

Solution
LHS sin2(q + 45°) + sin2(45° – q)
= (sinq . cos45° + cosq . sin45°)2 + (sin45° cosq – cos45° sinq)2

= 21 (sinq + cosq)2 + 1 (cosq – sinq)2
2

= 12 [sin2q + 2 sinq . cosq + cos2q + cos2q – 2 sinq . cosq + sin2q]

= 21 . 2 . 1
= 1

 LHS = RHS proved

d) 1 – 1 = cosec2A
tanA tan2A
Solution

LHS 1 – 1
tanA tan2A

= csoinsAA – cos2A
sin2A

= sin2A . cosA – cos2A . sinA
sinA . sin2A

= ssiinnA(2.Asi–nA2A)

= sinAsi.nsAin2A

= cosec2A
 LHS = RHS proved

4. Prove that: a) cos9° + sin9° = tan54°
cos9° – sin9°
Solution

LHS cos9° + sin9°
cos9° – sin9°

= ccooss((5544°° – 45°) + sin(54° – 45°)
– 45°) – sin(54° – 45°)

= ccooss5544°° . cos45° + sin54° . sin45° + sin54° . cos45° – cos54° . sin45°
. cos45° + sin54° . sin45° – sin54° . cos45° + cos54° . sin45°

cos54° × 1 + sin54° × 1 + sin54° × 1 – cos54° × 1
= 2 2 2 2

cos54° × 1 + sin54° × 1 – sin54° × 1 + cos54° × 1
2 2 2 2

158 Vedanta Optional Mathematics Teacher's Guide ~ 9

2 sin54° . 1
= 2

2 cos54° . 1
2
= tan54°

 LHS = RHS proved

b) sin20° + cos20° = cot155°
sin20° – cos20°
Solution

LHS sin20° + cos20°
sin20° – cos20°

= ssiinn((4455°° – 25°) + cos(45° – 25°)
– 25°) – cos(45° – 25°)

= ssiinn4455°°..ccooss2255°°––ccooss4455°°..ssiinn2255°°+– cos45° . sin25° + sin45° . sin25°
cos45° . cos25° – sin45° . sin25°

1 . cos25° – 1 . sin25° + 1 . cos25° + 1 . sin25°
= 2 2 2 2

1 . cos25° – 1 . sin25° – 1 . cos25° – 1 . sin25°
2 2 2 2
1
2. 2 . cos25°
=
1
–2 . 2 . sin25°

= –cot25°
= –cot(180° – 155°)
= cot155°
 LHS = RHS proved

c) cos35° – sin35° = tan10°
cos35° + sin35°
Solution

LHS cos35° – sin35°
cos35° + sin35°

= ccooss((4455°° – 10°) – sin(45° – 10°)
– 10°) + sin(45° – 10°)

= ccooss4455°° . cos10° + sin45° . sin10° – sin45° . cos10° + cos45° . sin10°
. cos10° + sin45° . sin10° + sin45° . cos10° + cos45° . sin10°

1 . cos10° + 1 . sin10° – 1 . cos10° + 1 . sin10°
= 2 2 2 2

1 . cos10° + 1 . sin10° + 1 . cos10° – 1 . sin10°
2 2 2 2

Vedanta Optional Mathematics Teacher's Guide ~ 9 159

2. 1 . sin10°
= 2

2. 1 . cos10°
2

= tan10°
 LHS = RHS proved

d) cos20° – cos70° = 2 sin25°

Solution
LHS cos20° – cos70°
= cos(45° – 25°) – sin(45° + 25°)
= cos45° . cos25° + sin45° . sin25° – cos45° . cos25° + sin45° . sin25°

= 1 . cos25° + 1 . sin25° – 1 . cos25° + 1 . sin25°
2 22 2

= 2 . 1 . sin25°
2

= 2 sin25°
 LHS = RHS proved

5. Prove that: a) 2 tan50° = tan70° – tan20°

Solution
Here, 70° – 20° = 50°
tan(70° – 20°) = tan50°

or, 1t+ant7a0n°7–0°tatnan202°0° = tan50°

or, 1 + tan70° – tan20° = tan50°
tan(90° – 20°) tan20°

or, 1 t+anc7o0t2° 0–°t.atna2n02°0° = tan50°

or, tan701° – tan20° = tan50°
+ 1

or, tan70° – tan20° = 2 tan5°
 2 tan50° = tan70° – tan20°

b) 2 tan20° = tan55° – tan35°

Solution
Here, 20° = 55° – 35°
tan20° = tan(55° – 35°)

or, tan20° = tan55° – tan35°
1 + tan55° tan35°

160 Vedanta Optional Mathematics Teacher's Guide ~ 9

or, tan20° = 1 tan55° – tan35° [tan55° = tan(90° – 35° = cot35°]
+ cot35° . tan35°

or, tan20° = tan55° – tan35°
1 + 1

 2 tan20° = tan55° – tan35°

c) 2 tan70° = tan80° – tan10°

Solution
Here, 70° = 80° – 10°
tan70° = tan(80° – 10°)

or, tan70° = tan80° – tan10°
1 + tan80° tan10°

or, tan70° = tan80° – tan10°
1 + 1

 2 tan70° = tan80° – tan10°

6. 2 tan10° = tan50° – tan40°

Solution
Here, 10° = 50° – 40°
tan10° = tan(50° – 40°)

or, tan10° = tan50° – tan40°
1 + tan50° tan40°

or, tan10° = tan50° – tan14°
1 + 1

 2 tan10° = tan50° – tan40°

7. a) If tanq = 5 and tanf = 111, then show that: q + f = pc
6 4
Solution

Here, tan(q + f) = tanq + tanf
1 – tanq . tanf

= 5 + 1
6 11

1 – 5 × 1
6 11

55 + 6

= 66 = 61 =1
66 – 5 61

66

= tan pc
4

 q + f = pc
4

Vedanta Optional Mathematics Teacher's Guide ~ 9 161

b) If cosq = 4 and cosf = 7 , then prove that: q + f = pc
5 2 4
Solution 5

Here, cos(q + f) = cosq . cosf – sinq . sinf

= 4 × 7 – 1 – cos2q . 1 – cos2f
5
5 2

= 4 × 7 – 1 –  4 2 . 1 –  7 2
5  5   
5 2 5 2

= 28 – 3 × 1
25 2 5 52

= 28 – 3 = 25 = 1
25 2 25 2 2

 (q + f) = cos45°
q + f = 45°

c) If tana = m 1 and tanb = 1 1, then prove that: a + b = pc
m+ 2m + 4
Solution

Here, tan(a + b) = tana + tanb
1 – tana . anb

m + 1
2m +
= m+1 1

1 – m 1 × 1 1
m+ 2m +

2m2 + m + m + 1

= (m + 1) (2m + 1)
(m + 1) (2m + 1) – m

(m + 1) (2m + 1)

= 2m2 2m2 + 2m + 1 – m
+ 2m + m + 1

= 2m2 + 2m + 1
2m2 + 2m + 1

= 1

 tan(a + b) = tanp4c

a + b = pc
4

8. If a + b = p4c, then prove that: a) tana + tanb + tana . tanb = 1
Solution

Given, a + b = pc
4

162 Vedanta Optional Mathematics Teacher's Guide ~ 9

tan(a + b) = tanpc
4

or, 1ta–ntaan+attaannbb = 1

or, tana + tanb = 1 – tana . tanb
 tana + tanb + tana . tanb = 1

b) (1 + tana) (1 + tanb) = 2

Solution
From (a) it is proved that
tana + tanb + tana tanb = 1
or, 1 + tana + tanb(1 + tana) = 1 + 1
or, (1 + tana) + tanb(1 + tana) = 2
 (1 + tana) (1 + tanb) = 2

c) cotb(cota – 1) – cota = 1

Solution

Given, a + b = pc
4

cot(a + b) = cotpc
4
or, ccootbtbc+otcaot–a1 = 1

or, cotb . cota – 1 = cotb + cota
or, cotb . cota – cotb – cota = 1
 cotb(cota – 1) – cota = 1

d) (cota – 1) (cotb – 1) = 2

Solution
From above
cotb . cota – cotb – cota = 1
or, cotb(cota – 1) – cota + 1 = 1 + 1
or, cotb(cota – 1) – (cota – 1) = 2
 (cota – 1) (cotb – 1) = 2

9. Prove the following: a) sinA . sin(B – C) + sinB . sin(C – A) + sinC . sin(A – B) = 0

Solution
LHS sinA . sin(B – C) + sinB . sin(C – A) + sinC . sin(A – B)
= sinA(sinB . cosC – cosB . sinC) + sinB(sinC . cosA – cosC . sinA)

+ sinC(sinA . cosB – cosA . sinB)
= sinA . sinB . cosC – sinA . cosB . sinC + cosA . sinB . sinC

– sinA . sinB . cosC + sinA . cosB . sinC – cosA . sinB . sinC

Vedanta Optional Mathematics Teacher's Guide ~ 9 163

= 0
 LHS = RHS proved

b) sinA + sin(A + 120°) + sin(A – 120°) = 0

Solution
LHS sinA + sin(A + 120°) + sin(A – 120°)
= sinA + sinA . cos120° + cosA . sin120° + sinA . cos120° – cosA . sin120°

= sinA + 2 sinA . cos120° = sinA + 2sinA (– 1 )
= sinA – sinA 2

= 0

 LHS = RHS proved

c) cosA + cos(120° + A) + cos(120° – A) = 0

Solution
LHS cosA + cos(120° + A) + cos(120° – A)
= cosA + cos120° . cosA – sin120° . sinA + cos120° . cosA + sin120° . sinA
= cosA + 2 cos120° . cosA

= cosA + 2 . –12 . cosA

= cosA – cosA
= 0
 LHS = RHS proved

d) (1 + tan21°) (1 + tan28°) (1 + tan24°) (1 + tan17°) = 4

Solution
LHS (1 + tan21°) (1 + tan28°) (1 + tan24°) (1 + tan17°)
Here, tan(21° + 24°) = tan45°

1t–anta2n12° 1+° tan24° = 1
. tan24°

or, tan21° + tan24° = 1 – tan21° . tan24°

or, tan21° + tan24° + tan21° . an24° = 1

or, 1 + tan21° + tan24°(1 + tan21°) = 1 + 1

or, (1 + tan21°) (1 + tan24°) = 2 ..........(i)

Similarly we have, (1 + tan28°) (1 + tan17°) = 2........(ii)

Multiplying (i) and (ii), we get

(1 + tan21°) (1 + tan28°) (1 + tan24°) (1 + tan17°) = 4

 LHS = RHS proved

9. If A + B + C = pc and cosA = cosB . cosC, then prove that:
tanA = tanB + tanC

164 Vedanta Optional Mathematics Teacher's Guide ~ 9

Solution
Given, A + B + C = p
B + C = p – A
 cos(B + C) = cos(p – A) = –cosA
sin(B + C) = sin(p – A) = sinA
Now, RHS = tanB + tanC

= csoinsBB + sinC
cosC

= sinB . cosC + cosB . sinC
cosB . cosC

= csoinsB(B. + C)
cosC

= csoinsAA

= tanA
 tanA = tanB + tanC
 LHS = RHS proved

10 csoisnA(A. – B) + sin(B – C) + sin(C – A) =0
cosB cosB . cosC cosC . cosA
Solution

LHS sin(A – B) + sin(B – C) + sin(C – A)
cosA . cosB cosB . cosC cosC . cosA

= sinA . cosB – cosA . sinB + sinB . cosC – cosB . sinC + sinC . cosA – cosC . sinA
cosA . cosB cosB . cosC cosC . cosA

= csoinsAA . cosB – cosA . sinB + sinB . cosC – cosB . sinC
. cosB cosA . cosB cosB . cosC cosB . cosC

+ sinC . cosA – cosC . sinA
cosC . cosA cosC . cosA

= tanA – tanB + tanB – tanC + tanC – tanA
= 0
 LHS = RHS proved

11. Prove the following:
a) cos(A + B + C) = cosA . cosB . cosC(1 – tanA . tanB – tanB . tanC – tanC . tanA)

Solution
LHS cos(A + B + C)
= cos(A + B) . cosC – sin(A + B) . sinC
= (cosA . cosB – sinA . sinB) cosC – (sinA . cosB + cosA . sinB) sinC
= cosA . cosB . cosC – sinA . sinB . cosC – sinA . cosB . sinC – cosA . sinB . sinC

Vedanta Optional Mathematics Teacher's Guide ~ 9 165

= cosA . cosB . cosC 1 – sinA . sinB . cosC
cosA . cosB . cosC

– sinA . cosB . sinC – cosA . sinB . sinC
cosA . cosB . cosC cosA . cosB . cosC

= cosA . cosB . cosC(1 – tanA . tanB – tanB . tanC – tanC . tanA)
 LHS = RHS proved

b) tan(A + B + C) = tanA + tanB + tanC – tanA . tanB . tanC
1 – tanA . tanB – tanB . tanC – tanC . tanA
Solution

LHS tan(A + B + C)

= 1 t–anta(An(A+ B) + tanC
+ B) – tanC

tanA + tanB + tanC
= 1 – tanA . tanB

1 – tanA + tanB . tanC
1 – tanA . tanB

tanA + TanB + tanC – tanA . tanB . tanC

= 1 – tanA . tanB

1 – tanA . tanB – tanC . tanA – tanB . tanC

1 – tanA . tanB

= 1ta–ntAan+A tanB + tanC – tanA . tanB . tanC
. tanB – tanB . tanC – tanC . tanA

 LHS = RHS proved

12. If tan(a + b) = 1 and tan(a – b) = 25, then prove that: a) tan2a = 11
3 13
Solution

LHS tan2a = tan(a + b + a – b)

= tan(a + b) + tan(a – b)
1 – tan(a + b) . tan(a – b)

= 1 + 2
3 5

1 – 1 × 2
3 5

15 + 6 = 11
= 15 13

15 – 2
15

 LHS = RHS proved

166 Vedanta Optional Mathematics Teacher's Guide ~ 9

13. b) tan2b = –117
Solution
LHS tan2b = tan(a + b – (a – b))

= tan(a + b) – tan(a – b)
1 – tan(a + b) . tan(a – b)

= 1 – 2
3 5

1 + 1 × 2
3 5

= 5–6 = –117
15 + 2

 LHS = RHS proved

14. If an angle q is divided into two parts A and B. Such that tanA . tanB = x:y, then show that:

sin(A – B) = x–y sinq
x+y
Solution

LHS q is divided into A and B, than θ = A+B

 q = A + B

Given, tanA = x
tanB y
By componendo and dividendo

ttaannAA – tanB = x–y
+ tanB x+y

sinA – sinB = x–y
or, cosA cosB x+y
sinA
cosA + sinB
cosB

sinA . cosB – cosA . sinB

or, . cosA . cosB . sinB = x–y
sinA cosB + cosA x+y

cosA . cosB

or, ssiinn((AA – B) = x–y
+ B) x+y

or, sin(A – B) = x–y sin(A + B)
x+y

 sin(A – B) = x–y sinq
x+y

 LHS = RHS proved

Vedanta Optional Mathematics Teacher's Guide ~ 9 167

15. If tana = k tanb, then prove that:

sin(a – b) = k –1 sin(a + b)
k +1
Solution

LHS tana = k tanb

ttaannab = k
1
By componendo and dividendo

tana – tanb = k–1
tana + tanb k+1

sina – sinb = k–1
or, cosa cosb k+1
sina
cosa + sinb
cosb

sina . cosb – cosa . sinb

or, cosa . cosb = k– 1
sina . cosb + cosa . sinb k+ 1

cosa . cosb

or, ssiinn((aa – b) = k–1
+ b) k+1

 sin(a – b) = k– 1 sin(a + b)
k+ 1

 LHS = RHS proved

16. If 2 tanb + cotb = tana, then prove that: 2 tan(a – b) = cotb

Solution

Given 2 tanb + cotb = tana

Now, 2 tan(a – b) = 2 × 1 tana – tanb
+ tana . tanb

= 2 × (2 tanb + cotb – tanb)
1 + (2 tanb + cotb) tanb

= 2 × 1 + 2 tanb + cotb . tanb
tan2b + cotb
1
= 2× tanb + tanb

1 + 2 tan2b + 1

tan2b + 1
= 2 × tanb

= ta1nb2(1 + tan2b)

 cotb

 LHS = RHS proved

168 Vedanta Optional Mathematics Teacher's Guide ~ 9

17. If sina + sinb = m and cosa + cosb = n, then prove that:

cos(a – b) = 1 (m2 + n2 – 2)
2
Solution

RHS 1 (m2 + n2 – 2)
2

= 1 [(sina + sinb)2 + (cosa + cosb)2 – 2]
2

= 21 [sin2a + 2 sina . sinb + sin2b + cos2a + 2 cosa . cosb + cos2b – 2]

= 12 [sin2a + cos2a + sin2b + cos2b + 2(sina . sinb + cosa . cosb) – 2]

= 21 [1 + 1 + 2 cos(a – b) – 2]

= 21 × 2 cos(a – b)

= cos(a – b)
 LHS = RHS proved

18. If sin(q + f) = 2 sin(q – f), prove that: tanq = 3 tanf

Solution
Given, sin(q + f) = 2 sin(q – f)
or sinq . cosf + cosq . sinf = 2(sinq . cosf – cosq . sinf)
Dividing both sides by cosq . cosf

csoinsqq . cosf + cosq . sinf – 2 sinq . cosf – cosq . sinf
. cosf cosq cosf cosq . cosf cosq cosf

or, tanq + tanf = 2[tanq – tanf]
or, tanq + tanf = 2 tanq – 2 tanf
 tanq = 3 tanf
 LHS = RHS proved

Vedanta Optional Mathematics Teacher's Guide ~ 9 169

Questions for practice

1. Prove that:

(a) sin(90°– θ ) . cos(90°– θ ) . tan(90°– θ) = 1
sinθ cosθ cot (90°–θ)

(b) sin2θ ) + sin2(90°– θ ) = 1
cot2 (90°–θ cot2θ

2. Find the value of a from the given equation :
(a) xcotθ . tan(90°+θ) = tan(90°+θ). cos(90°–θ)
(b) 2cos120°–xsin120°.cos210°= xtan150°

3. Prove that:
(a) cos(α–β) – cos(α+β)= 2sinα .sinβ
(b) cos (α+β) + cos (α–β) = 2cosα .cosβ
(c) sin (α+β) –sin(α–β) = 2cosα .sinβ

4. Prove that following:
(a) 1+tanθ .tan2θ = sec2θ
(b) sinθ +sin(θ+120°)+ sin(θ–120°) = 0

5. Prove that:
(a) 2sin(45°+θ) . sin(45°–θ) = cos2θ – sin2θ
(b) 2cos(45°+θ) . cos(45°–θ) = cos2θ – sin2θ

6. Prove that:
(a) tan10°+tan135°+tan10°.tan35° = 1
(b) tan5A–tan3A–tan2A = tan5A.tan3A.tan2A

7. Prove that:

πc πc
tan ( 4 +θ) + tan ( 4
(a) cosec2 = –θ)

tan s(eπ4cc+θπ4)c - tan (π4c–θ)
– sec =
(b) sec πc + θ . θ . 2secθ
4 2 2

8. If tanα = ktanβ , then prove that
(k+1).sin(α–β) = (k–1) sin(α+β).

170 Vedanta Optional Mathematics Teacher's Guide ~ 9

UNIT

twelve Vectors

Estimated teaching hours : 12

1. Objectives

S.N. Level Objectives

To define vectors.

i. Knowledge (K) To define different type of vectors (row, column, position,
unit, zero, equal, negative vectors, etc).

To define like and unlike vectors.

To classify the given physical quantities into vectors or scalars.
To multiply a vector by a scalar.

To calculate magnitude and direction of a vectors.
ii. Understanding (U) To state triangle law, parallelogram law and polygon laws of

vector addition.

To find unit vector along a given vector.

To find sum and difference of two vectors.

iii. Application (A) To apply triangle law of vector addition, parallelogram law of vector
addition, polygon law of vector addition in solving problems.

iv. Higher Ability To solve problems by using vector geometry.
(HA)

2. Teaching Materials
Graph papers.

Chart paper with figures showing

Triangle law of vector addition

– parallelogram law of vector addition.

– polygon law of vector addition.

3. Teaching Learning Strategies
– Give some physical quantities and let the students to classify them into scalars
and vectors.
– Define vectors and scalars with appropriate examples.
– Giving examples define different types of vectors (row, column, position, zero /
null, unit, negative, like and unlike vectors.
– Define magnitude and direction of a vector with diagrams.
– Discuss addition and subtraction of vectors.
– State the triangle law of vector addition and the parallelogram law of vector addition.
– Solve some problems from the exercises as examples and motivate the students
giving required guidance.
– State and prove the mid point theorem in vector geometry.

Vedanta Optional Mathematics Manual ~ 9 171

Notes :

1. The point (x, y) represents the position vector of P with reference point origin O. Then,

OP =  x r,owinvceocltuomr n vector
= (x, y), y
in

magnitude of OP ,|OP |= x2 + y2
y y
Direction of OP tanθ = x ⇒ θ = tan – 1 x

2. The points P (x1, y1) and Q(x2, y2) represent the position vector of P and Q with reference
point origin 0.Then

=OQ – OP =  x2  –  x1 =  x2 – x1 
y2 y1 y2 – y1

magnitude of PQ ,|PQ |= (x2 – x1)2 + (y2 –y1)2

direction of PQ , tanθ = xy22 – yx11 ⇒ θ = tan – 1 xy22 – xy11
– –

3. Two vectors AB =  a  and CD =  p  are said to be equal if and only if their
b q

corresponding components are equal. (i.e. a = p, b = q ). It also can be said that if|AB|=

bq
|PQ |and tan – 1 a = tan – 1 p , then AB and PQ are equal vectors.

4.Unit vector along OP = (x,y) is given by OP = OP | = (x, y)
= x,y |OP x2 + y2

x2 + y2 x2 + y2

5.Unit vector along x – axis is i =(1,0) and along y – axis j =(0,1) respectively.

6. If a = k b , where k is a scalor .Then
i)two vectors a and b are called like parallel vectors if k is positive.
ii)two vectors a and b are called unlike parallel vectors if k is negative.

7. Triangle law of vector addition. C
If two sides of a triangle are taken in order represent two
vectors, then their resultant is given by the third side of
the triangle whose initial point is the initial point of the
first vector and the terminal point is the terminal point of
the second vector.

In ∆ ABC, AB + BC = AC
Here, AC is called the resultant vector of AB and BC .

A B

172 Vedanta Optional Mathematics Manual ~ 9

8. Parallelogram law of vector addition: C B
If two adjacent sides of a parallelogram represent c b
two co-initial vectors, then the diagonal of the A
parallelogram passing through the same point gives D
their resultant.
C
In the parallelogram ∆ABC, OA + AB = OB O a
Also, a + b = c B

9. Polygon law of vector addition : E
Polygon law of vector addition is the
generalized triangle law of vector addition.

In the figure, ABCEF is a hexagon.Then
AB + BC +CD + DE + EF = AF
AB + BC +CD + DE + EF + FA = 0

F

10.Addition of vectors :

If a=  x1  b =  x2  , then sum of
a and  y1   y2 

b

x1 x2 x1 + x2 A
y1 y2 y1 + y2
is a +b =   +   =  
   

11. Difference of two vectors :

If a =  x1  and b =  x2  , then the difference of a and b
y1 y2

is a – b =  x1  –  x2  =  x1 – x2 
 y1   y2   y1 – y2

Some solved problems

1. MN displaces M to N, where M(4, 6) and N (8, 10) find the magnitude and direction of MN

Solution :

Here, M(4, 6) and N (8, 10) , MN is given by
x2 – x1
MN =  y2 – y1 

Vedanta Optional Mathematics Manual ~ 9 173

=  8 – 4  =  4 
10 – 6 4

Magnitude of MN =|MN |= (x – component)2 + (y – comonemt)2

= 42 + 42

= 16 + 16

= 32 = 4 2

Direction of MN , tanθ = y – component
x – component
4
= 4 =1

= tan 45˚

θ = 45˚

MN makes an angle 45 with x – axis in positive direction.

2. Let A(–2, 3), B(3, 5) and C(x + 1, 4) and D (3, – 1) be four points. If | AB | = | CD |,
find the value of x.

Solution

For AB , A(–2, 3) and B(3, 5)
x2 – x1
AB =  y2 – y1  =  3 + 2  =  5 
5 – 3 2

for CD, C(x + 1, 4) and D (3, – 1) is given by
x2 – x1
=  y2 – y1  =  3–x–1  =  2–x 
   –1–4   –5 

Now, |AB | =tan θ|CD |= (x – component)2 + (y – comonemt)2

ie. 52 + 22 = (2 – x)2 + (– 5)2 -
or, 29 = 4 – 4x + x2 + 25
or, x2 – 4x = 0
or, x (x – 4) =
Either x = 0 or x – 4 = 0 →x –4= 0 ⇒ x = 4

x = 0, 4

3. If PQ displaces a point P (2, 3) to Q (5, 7), find PQ , express it in the form of xi + yj
4=, also find the unit vector along PQ .

Solution
Here, P (2, 3) and Q (5, 7) are given two points.

PQ =  x2 – x1 
y2 – y1

174 Vedanta Optional Mathematics Manual ~ 9

=  5 – 2  =  3 
7 – 3 4

Also, PQ =(3, 4) = 3i + 4j ,|PQ |= 32 + 42 =5
unit vector along is given by,

P^Q PQ = (3,4)
= 5

|PQ |

= 3 , 4
5 5

4. Find the values of p when the pair of vectors are parallel,

AB =  3p  and CD =  – 6 
4  – 8 

Solution 3p – 6
4 – 8
Here, AB =   and CD =  

since AB and CD are parallel, we can write

AB = k CD , where k is a scalar
Equ34apting=thke co––rr68esp=ond––in86gkkcomponents, we get,

4 = 8k ⇒ k = – 1
2
and 3p = – 6k

→3p = – 6 × – 1
2
⇒3p = 3

p=1

Alternative method
3p4parallaenl,dtheCirDdi=recti––on86
Here, AthBey=are  also same.
Since are

For AB , let its direction be 1, tan 1 = y – component = 4
x – component 3p
For CD , let its direction be 2,
tan 2 =
–8 = 4
–6 3
As AB and CD are parallel

tan 1==34tan 2
4
⇒
3p

p = 1

Vedanta Optional Mathematics Manual ~ 9 175

4. If a =  2  and b =  4  , then find | 2a – 3b | .
1 3

Solution 2 4
1 3
Here, a =   and b = 
 

2a = 2  2  =  4
 1   2

3 b =3  4  =  12 
3 9

| 2a – 3b |=  4  –  12  =  –8 
Now, 2 9  –7

2 a – 3b = (– 8)2 + (– 7)2 = 64 + 49 = 113 units.

5. In the given figure OA = a , and OB = b are co – initial vectors, then draw arrow diagram
for each of the following vectors.

(a) a + b B

(b) b – a

(c) a – b b

(d) 2a + b

(e) 2a – b A
Oa

(Hints: use concept of negative vector and parallelogram law of vector addition,

(a) a + b C
B

a+b

Oa A

Here,
OC = OA + OB

176 Vedanta Optional Mathematics Manual ~ 9

(b) b – a B
C

a–bb
b–a
A' –a Oa A
here,

OC = OA′ + OB = – a + b = b – a

Note:
OA = – a is negative vector of OA .

2a means two times in length of but direction is same as that of a .

(c) a – b B a A
Here, OB′ = – b
OC =OA + OB′ = – a – b b
O

(d) 2a + b –b C
B
B'
C

b 2a + b

a
O A a A'
Here, OA1 =OA + OA1 where, (OA = OA1 )
= a+a

Vedanta Optional Mathematics Manual ~ 9 177

=2 a
Now,
OC =OA′ + OB

= 2a + b
(e) 2a – b
B

b A A'
O a

a

B' C

Here, OB1 = – b
OC = OA1 + OB1 )
=2a – b

6. PQRS is a quadrilateral. If PR = p , SQ = q , QR = r , QP = n , SP = m , SR =
s , then express the following vectors as a single vector.

(a) n – m (b) r – s (c) p – q – s (d) p – s + m
Hints : use triangle law of vector addition

Solution = – PS ) Pm S
(a)Here, n – m nq s
= QP – SP Qr
= QP + PS ( SP p
= SP (From ∆QPS) R

= SQ = – q

(b) r – s = – RS )
= QR – SR
= QR + RS ( SR
= QS (From ∆QRS)

=– q

178 Vedanta Optional Mathematics Manual ~ 9

(c) p + q – s = – PS )
= PR + SQ – SR
= PR + SQ + RS ( SP
= ( PR + RS )+ SQ
= PS + SQ (From ∆PRS)

= PQ = – QP (From ∆PQS)

=– n

(d) p – s + m A
= PR – SR + SP
= PR + RS + SP
= PS + SP
= PS – PS = 0

7. In a regular hexagon ABCDEF of AB = a , BC = b , then R Q
(a) Express AC , AD and AE in terms of a and b . Also

show that AB + BC CD + DE + EF + FA = 0

(b) Express the remaining sides in terms of a and b taken B P C
in order. D

Solution E

(a) In a regular hexagon ABCDEF, F C

AB = ED =a , BC = FE = b b
AD = 2 BC B
Now,

AD = 2 BC = 2b

AC = AB + BC (From ∆ABC)
=a +b

AE = AD + DE (From ∆ABC) A

= 2b + (– a ) ( DE = – ED = – a ) a

= 2b – a FA = 0
Again, to prove

AB + BC + CD + DE + EF

LHS = ( AB + BC ) + CD + DE + EF + FA = 0

Vedanta Optional Mathematics Manual ~ 9 179

= ( AC + CD ) + DE + EF + FA = 0 ( AB + BC = AC )
= ( AD + DE ) + EF + FA = 0 ( AC + CD = AD )
= ( AE + EF ) + FA = 0 ( AD + DE = AF )
= ( AE + EF ) + FA
= AF + FA

= AF – AF
= 0 proved

(b) CD =CA + AD
= – AC + AD
= – ( a + b ) + 2b
= – a + b = – (b – a )
=a –b
EF = CB = – b
FA = DC
= DA + AC =– 2b + a + b = a – b

8. In the adjoining figure PQRS is a parallelogram two diagonals PR and QS bisect at T
fand 0 is any point then prove that

OP + OQ + OR + OS =4 OT

Solution
We prove gives problem by using mid point theorem.

In a parallelogram, the diagonals PR and QS bisect each other. So T is the mid point of PR and QS.
1
Now, OT = 2 (OQ + OS ), (T is the mid point of QS )

⇒ 2OT = OQ + OS .............(i) S R
OT
Similarly, T is the mid point of PR,

OT = 1 (OP + OR )
2
⇒ 2OT = OP + OR .............(ii)

adding (i) and (ii), we get

OP + OQ + OR + OS = 4 OT proved.

Alternative Method: PQ
By using triangle law of vector addition,

RHS = 4 OT

= OT + OT + OT + OT

180 Vedanta Optional Mathematics Manual ~ 9

=OS + ST + OQ + QT + OR + RT + OP + PT
But ST = TQ , PT = TR (diagonals of parm bisect each other )
(OP + OQ + OR +OS ) + (ST + QT ) + (PT + RT )
= OP + OQ + OR +OS + ( 0 + 0)
= OP + OQ + OR +OS LHS proved

9. In the adjoining figure P, Q, R are the mid points of BC, CA and AB respectively. then

prove that AP + BQ + CR = 0

Solution A
By using the mid point theorem

AP = 1 (AB + AC )
2

BQ = 1 (BA + BC ) RQ
2

CR = 1 (CB + CA ) BP C
2

LHS = AP + BQ + CR
1
AP = 2 (AB + AC + BA + BC + CA + CB )

= 1 (AB + BA + AC + CA + AC + CA )
2

= 1 (0 + 0 + 0) (  AB + BA = AB – AB =0)
2

=0

= RHS proved

Alternative method.
By using triangle law of vector addition.

LHS = AP + BQ + CR

=( AB + BP ) + ( BC + CQ ) + ( CA + AR )

=( AB + BC + CA ) + ( BP + CQ + AR )

(In ∆ABC, AB + BC + CA = 0)
1 1 1
= 2 BC + 2 CA + 2 AB

= 1 ( AB + BC + CA )
2

Vedanta Optional Mathematics Manual ~ 9 181

= 1 . 0
2

=0
=RHS Proved

Questions for practice

1.If the vector PQ displaces the point P(6,9) to Q (10,13), find PQ , its magnitude and
direction.

2.If the vector AB displaces A(5,2) to B(2,5), then find AB as a column vector. Express
AB in the form of x i + y j .

3.If p = (3,4) and q = (6,7), then find p - q and |2p - 3q |.

4.In the goven figure , C is the mid point of AB, then prove O
that OA + OB = 2 OC

5.In the figure, PQRSTU is a regular hexagon, then prove A CB
that: T S
PQ + PR + PS + TP + UP = 4PQ
R
U

6.In the given triangle ABC, P, Q and R the mid points of sides P Q
BC, CA and AB respectively, then prove that: GA + GB + Q C
GC = 0, where G is the point of intersection of the medians.
P
A G
RB

182 Vedanta Optional Mathematics Manual ~ 9

UNIT

thirteen Transformation

Estimated teaching periods : 15

1. Objectives

S.N. Level Objectives
(i) Knowledge(K)
To define a transformation

To define a isometric and non-isometric transformation

To define reflection, rotation, translation and enlargement
and formula of these transformations

To define invariant points in transformation

To find the image of a given point when it is reflected on
x=0 (or y-axis), y=0 (or x-axis), y=x, y=-x

To find the image of a given point when it is rotated

(ii) Understanding(U) -through +90°, -90°, 180°, +270°, -270° about origin

To ternalnasrlgaeteangiovbenjecptoainbotsuut scienngtrteraantstlhateioonrigvienctaonrdTs=cale a 
To b 

factor k.

(iii) Application(A) To find image vertices of given geometrical figures and plot
their graphs under reflections, rotations, translation and
enlargement.

To derive formula of the following transformation

(iv) Higher -reflection on lines x=h and y=k
Ability(H.A.)
-rotation of +90°, -90° and 180° about centre (a,b)

To solve problems in transformation using these formula.

2. Teaching Materials
– Compass
– Ruler
– set squares
– Graph papers
– mirrors, etc.

3. Teaching learning strategies.
Giving examples of reflection – natural phenomena image of animals and plants on the
surface of water, ( in river, lake, ponds etc. ) – discuss about reflections.

– Discuss about the of reflection on x = 0 y – axis, y = 0, x – axis, y = x, y = – x and
formula of them, explain how these formula works also graphically.

– Derive the formula of reflection on lines x = h and y = k .

– Give concept of rotation about a centre and through given angles, positive rotation

Vedanta Optional Mathematics Manual ~ 9 183

negative rotation.

Discuss about + 90°, – 90°, 180°, +270°, 270° rotations about origin and their formula
giving examples with graphs.

–Review the construction of angles 30°, 60°, 90°, 180°.Derive the formulas of rotation of
+90°, – 90°, 180°, about centre (a, b) other than the origin.

–Discuss about translation. Use set squares to translate given geometrical objects.

–Explain about use of coordinates in translation with use of graphs.

–Giving examples of photographs, globe, circles etc., clear the concept of enlargement and
reduction.

– Discuss about use of coordinates in enlargement and reduction with using of graphs.

– Derive the formula of enlargement with scale factor k and centre (a, b), other than origin.

P (x, y) E[(a, b) k] P1 P(k (x – a) + a, k (y – b ) + b)

In each case of above formula discussed, give at least two examples of them and solve
some corresponding problems from exercises giving necessary guidance.

Notes :-A. Reflection: P′(x, –y )
x – axis P′(–x, y )
P′(y, x )
i) P (x, y) P′(– y, –x )
y – axis P′(2h – x, y )
P′(x, 2k –y )
ii) P (x, y)
y=x

iii) P (x, y)
y = –x

iv) P (x, y)
v) P (x, y) x = h
vi) P (x, y) y = k

Some solved problems

1. Find the image of P (4, – 5) when its reflected on

i) x – axis ii) y – axis iii) y = x iv) y = – x

Solution x – axis P′(x, –y )
i) we have, P (x, y)

P (4, 5) x – axis P′(4, – 5 )

y – axis P′(– x, y )
ii) P (x, y)

y – axis P′(– 4, 5 )
P (4, 5)

184 Vedanta Optional Mathematics Manual ~ 9

iii) We have, P (x, y) y = x P′(y, x )

y=x P′(5, 4 )
P (4, 5)

y = –x P′(– y, –x )
iv) P (x, y)

y = –x P′(– 5, – 4 )
P (4, 5)

2. Reflect P(– 3, – 5) and Q(5, 6) under the following axis of reflections.

(a) y = 3 (b) x = 3

Solution y=k P′ (x, 2k – y )
(a) we have P (x, y)

P (–3,–5) y=3 P′(– 3, 2 × 3 + 5 ) = P′(– 3, 11 )
Q (5, 6) y=3 Q′(5, 2 × 3 – 6 ) = Q′(5, 0)

x=h P′(2h – x, y )
(a) we have P (x, y)

x=3 P′(2 × 3 + 3, – 5 ) = P′(9, – 5 )
P (– 3, – 5) Q′(2 × 3 – 5, 6 ) = Q′(1, 6 )

x=3
Q (5, 6)

3. Determine the vertices of image ∆A'B'C' formed when∆ABC with the vertices A(-1, 8 ),
B(7,6) and C(4,1) is reflected on the line x+2=0. Also draw triangles on the same graph.

Solution
Given vertices of ∆ABC are A(-1, 8 ), B(7,6) and C(4,1). ∆ABC is reflected on line x = – 2

Now, P′(2h – x. y )
x= h A′(2 × – 2 + 1 , 8 ) = A′( –3, 8 )

P (x, y) B′(2 × – 2 – 7, 6 ) = A′( –11, 6 )
x= – 2 C′(2 × – 2 – 4, 1) = C′( –8, 1 )

A (– 1, 8)
x= – 2

B (7, 6)
x= –2

C (4, 1)

∆ABC and its image ∆A'B'C' are plotted on the same graph.

Vedanta Optional Mathematics Manual ~ 9 185

4. The image of P(3 –m, 4–n) when reflected on the line y = – x is P' (2n – 6, 2m – 5).Find
the values of m and n.

Solution y=x P' (y, x )
we have,
P (x, y)

y=–x P′(–4+n, –3+m)
Now, P(3–m, 4–n)

But p′(2n – 6, 2m – 5)

(–4+n, –3+m ) = (2n–6, 2m–5)

⇒ –4+n= 2n–6 ⇒ n = 2

and –3+m = 2m–5 ⇒ m = 2

m = 2 and n = 2

b. Rotation P' (– y, x)
R[(0, 0), 90° ] P' (y, – x)
P'(– x, – y)
i) P (x, y) P' (– y + a + b, x – a + b)
ii) P (x, y) R[(0, 0), – 90° ] P' (y + a – b, – x + a + b)
P' (2a – x, 2b – y)
R[(a, b), ±180°]
iii) P (x, y)

E[(a, b), + 90° ]
iv) P (x, y)
v) P (x, y) E[(a, b), – 90° ]

E[(a, b), 180°]
vi) P (x, y)

vii) 90° positive rotation is equivalent to – 270° about the same centre.

viii) – 90° negative rotation is equvalent to + 270° about the same centre.

ix) In case of 180° rotation positive or negative is same, but better to rotate in positive
direction .

x) Clockwise direction is taken as negative and anticlockwise direction is taken as positive.

5. Find the image of P (4, 5) when rotated through + 90° about the centre (1, 2).

Solution P' (– y + a + b, x – a + b)
we have, P' (– 5 + 1 + 2, 4 –1 + 2) = P' (– 2, 5)

R[(a, b), + 90° ]
P (x, y)

R[(1, 2)]
P (4, 5)

6. If P (2, 4) is mapped into P' (– 4, 2) under a rotation about origin. what is the image of
Q(1, 2) under the same rotation ?

186 Vedanta Optional Mathematics Manual ~ 9

Solution : Here, P (x, y) P' (– 4, 2)

It seem same as, P' (– y , x )
R[(0, 0), + 90° ] P' (– 6 , 3) .

P (x, y)
R[(0, 0), + 90°]

Q(3, 6)

7. A triangle with vertices A (3, 4), B (3, – 4) and C (1, 1) through an angle θ˚about (p, q). to
get the image A' (– 4, 3), B' (4, 3) and C' (– 1, 1).Find the centre of rotation using graph.

Solution

y

A'(–4,3) 5 A(3,4)
B(4,3)
4
3

2
C'(–1,1) 1 C(1,1)
x' x

–5 –4 –3 –2 –1 o –11 2 3 4 5

–2
–3

–4 B(3,–4)
–5

y'

– Draw ∆ABC and its image ∆A'B'C' on the same graph.
– Join AA', BB' and CC' .
– Draw perpendicular bisectors of AA', BB' and CC' (two perpendicular bisectors are
sufficient).
– The perpendicular bisectors are produced on the side where they become closer.
– The point of intersection of the perpendicular bisectors(say P) will be the centre of
rotation (in the figure P = 0).
– Measure the APA' and check the direction of rotation positive or negative.
– In the given problem, the point of intersection of the perpendicular bisector is at the
origin and the angle of rotation is +90° .

Centre of rotation = O(0,0)
Angle of rotation = +90°

8. Rotate the triangle ABC with vertices A (4, 4), B (5,3) and C (2,1) through 180° about
centre (1,1) to get ∆A'B'C' . Draw both the triangles on the same graph.

Solution
Given vertices of ∆ABC are A (4, 4), B (5,3) and C (2,1). The ∆ABC is rotated through about
180° centre (1,1).

Vedanta Optional Mathematics Manual ~ 9 187

We have. P' (2a – x, 2b – y )
R[(a, b), + 180° ] A' (2.1 – 4, 2.2 – 4 ) = A' (– 2, 0)
B' (2.1 – 5, 2.2 – 3 ) = B' (– 3, 1)
P (x, y) C' (2.1 – 2, 2.2 – 1 ) = C' (0, 3)

R[(1, 2), 180° ]
A (4, 4)

B (5,3) R[(1, 2), 180° ]
C (2,1) R[(1, 2), 180° ]

∆ABC and ∆A'B'C' both are plotted on the same graph. image ∆A'B'C' is shaded.
y

5 A(4,4)

A'(–3,1) C'(0,3) B(5,3)

x' C(2,1) x

–5 B(–2,0)o 5

–5
y'

c. Translation a
b
T =  
 
i) P (x, y) P' (x + a, y +b )

P' = x+a  =(x + a, y +b )
y + b
x2 – x1
ii) If A(x1, y1) and B(x2, y2), then AB =  y2 – y1  .

9. If A(1, 2), B (4,5) and C (– 3,4). are the vertices of ∆ABC . Find the images of A, B and C
4
under the translation. T =  6  -

Solution
Here, A(1, 2), B (4,5) and C (– 3,4). are the vertices of ∆ABC .

translation. vector T =  4 
6

T =  a 
 b 
we have, P (x, y) P' (x + a, y +b )

Now,

188 Vedanta Optional Mathematics Manual ~ 9

A (1, 2) T =  4  A' (1 + 4, 2 +6 ) = A' (5, 8 )
6 B' (4 +4, 5 +6 ) = B' (8, 11)
C' (– 3 +4, 4 +6 ) = C' (1, 10)
B(4, 5) T =  4 
6

C(–3, 4) T =  4 
6

10. If a point M(6, 7) is translated by a translation vector to M'(2,3), what will be the image
of N (5, 2) by the same translation.

Solution a
b
Let T =   the required translation vector.
P' (x
Then, we have, P (x, y) T =  a  + a, y +b )
b

T =  a 
 b 
M(6, 7) M' (6 + a, 7+b )

But M' (2, 3 )

2=6+a⇒a=–4

and 3 = 7 + b ⇒ b= 3 – 7 = – 4

T =  –4 
– 4
Again,

T =  – 4 
 – 4 
N(5, 2) N' (5 – 4, 2 – 4 ) = N' (1, – 2 )

11. If a point P(p, q) under the translated vector T=  2  is p' (2p – 4, 2q – 3),find the
value of p and q.  3

Solution 2
3
Here, translated vector T=  
Now,

Then, we have, P (p, q) T=  2  P' (p + 2, q +3 )
3

But P' (p + 2, q +3 )

p' (2p – 4, 2q – 3) = P' (p + 2, q +3 )

Equating the corresponding components , we get

⇒ 2p – 4 = p + 2 ⇒ p = 6

and 2q – 3 = q + 3 ⇒ q= 6

p = 6 and q = 6

Vedanta Optional Mathematics Manual ~ 9 189

12. A parallelogram having vertices P (3, 2) , Q(11, 2), R(14, 7) and S(6,7) is translated by its

diagonal PR in its magnitude and direction. Find the coordinates of image parallelogram.
Draw both parallelogram in the same graph.

Solution

Here, to find PR , we have

P(x1, =y1)=xy22P––(3yx,11 2) and R(14, 7) = R(x2, y2)
PR
 =  14 – 3  =  11  .
7–2 5

Again,

T=  11 
 5
P (3, 2) P' (3 + 11, 2 +5 ) = P' (14, 7 )

T =  11 
5
(11, 2) Q' (11 + 11, 2 + 5 ) = Q' (22, 7 )

T=  11 
 5
R(14, 7) R' (14 +11, 7+ 5 ) = R' (25, 12 )

T =  11 
5
S(6,7) S' (6 + 11 , 7 + 5 ) = S' (17, 12 )

Both of the parallelogram are plotted in the graph.

y

12 R'(17,12) R'(24,12)

10 P' Q'(2,7)
8 R(6,7) R(14,7)
6

4

x' 2 P(3,2) Q(11,2) x

–4 –2 o 2 4 6 8 10 12 14 16 18 20 22 24

–2

–4
y'

190 Vedanta Optional Mathematics Manual ~ 9

D. Enlargement P' (kx, ky )
E[(0, 0), k ]

i) P (x, y)

E[(a, b), k ] P' (k(x – a) + a, k(y – b) + b )
ii) P (x, y)

iii) Scale factor k) = OP' = distance of image point from centre
OP distance of object point from centre

where, P = object point, P' = image point

13. Find the image of R( 2, 4) under E[(1, 2), 3].

Solution
Here, given point is R ( 2, 4).

we have, P' (k(x – a) + a, k(y – b) + b )
E[(a, b), k ]

P (x, y)

Now, P' (3(2 – 1) + 1, 3(4 – 2) +2 ) = P' (4, 8)
E[(1, 2), 3 ]

R( 2, 4)

14. A triangle PQR with vertices P (3, 0), Q (0, 2) and R (3, 2) is enlarged about centre origin
O and scale factor – 2. Find the vertices of image triangle P'Q'R'. Plot both the object
triangle and the image triangle on the same graph.

Solution
Here, P (3, 0), Q (0, 2) and R (3, 2) are the vertices of ∆PQR. we have to enlarge ∆PQR
about origin with scale factor – 2.

we have, P' (kx , ky )
P (x, y) E[(0, 0), k ] P' (– 2 × 3, – 2 × 0 ) =P' (– 6, 0 )
P (3, 0) E[(0, 0), – 2 ] Q' (– 2 × 0, – 2 ×2 ) =Q' (0, – 4 )
P' (– 2 × 3, – 2 × 2 ) =R' (– 6, – 4 )
E[(0, 0), – 2 ]
Q (0, 2)

E[(0, 0), – 2 ]
R (3,2)

Vedanta Optional Mathematics Manual ~ 9 191

y
5

Q(0,2) R(3,2)

x' P'(–6,0) x

–6 –4 –2 o P(3,0) 6

–2

R'(–6,–4) Q'(0,–4) –Q
–5

y'

∆PQR and ∆P'Q'R' are plotted on the graph below.
Q.11 (a) p. 304 textbook to be corrected.

15. A triangle ABC with vertices A (4, 2), B (– 3, 1), and C (0, 8) is enlarged to ∆A'B'C' where
A' (– 8, – 4), B' (6, – 2), and C' (0, – 16). Find the centre and scale factor.

Solution
Let (a, b) be the centre and k scale factor of the enlargement. Then, we have

E[(a, b), k ] P' (k(x – a) + a, k(y – b) + b )
P (x, y)

Now, A' (k(4 – a) + a, k(2 – b) +b) -
E[(1, 2), 3 ]

A (4, 2)

But A' (– 8, – 4)

– 8 = 4k – ka +a............(i)

– 4 = 2k – kb + b..........(ii)

Again, B (– 3,1)

E[(a, b), k ] B' (k(– 3 – a) + a, k(1 – b) +b )
B (– 3, 1)

But B' (6, – 2)

6 = – 3k – ka + a..........(iii)

– 1 = k – kb + b...............(iv)

and

E[(a, b), k ] B' (k(0 – a) + a, k(8 – b) + b)
C (0, 8)

But B' 0, – 16)

0 = – ka + a..........(v)

– 16 = 8k – kb + b...............(vi)

Solving equation (i) and (ii), we get,

192 Vedanta Optional Mathematics Manual ~ 9

k = – 2 and (a,b)= (0,0)
Q. N. 15

Alternative Method
The above problem can also be solved graphically. ∆ABC and ∆A'B'C' are plotted
on the same graph.

i) Join AA', BB' and CC' be produced if necessary, their point of intersection will be the
centre of enlargement. Let it be P.

ii) Scale factor k = PA' = PB' = PC'
PA PB PC

iii) If object point and image points are on the opposite sides of the centre. Then scale
factor will be negative otherwise positive.

y
8C(0,8)

6

4

2 A(4,2)

x' R(–3,1) x

–8 –6 –4 –2 o 2 4 6 8 10

–2 B'(6,–2)

A"(8,4) –4
–6

–8
–10
C"(0,–10)
y'

16. An enlargement mapped A to A' . Find the centre and the scale factor of the enlargement.

A(– 1,2) → A(– 4,3), B(2, – 3) → B'(2, – 7)

Hints It can be solved the following two methods.

Method :(I) Taking centre (a,b) and scale factor k.

use formula, P' (k(x – a) + a, k(y – b) + b)
P (x, y) E[(a, b), k ]

Four equations will be formed using the given points and their vertices solving we get the
values of a, b and k.

Method : (II) The points A, B, A' and B' are plotted in a graph . Join AA' and BB' and

Vedanta Optional Mathematics Manual ~ 9 193

produced if necessary, then their point of intersection will be the centre of enlargement.
Let it be p .

k= PA' = PB'
PA PB

k is positive if the image and object points are on the same side of the centre otherwise
negative.

Questions for practice

1. Find the image of point A (6, 7) when it is reflected on i) x – axis ii) y – axis
iii) y = x lines, line y = – x iv) line x = 2 v) line y = 4

2. Find the image of point P (2, 3) when it is rotated through

i) + 90° about origin

ii) – 90° about origin

iii) 180° about origin

iv) + 90° about (1,2)

v) + 180° about (2, 3)

3. Find the image of MP(7(2,5,6))wwhheennititisisetnralanrsgleadteudnbdyerTt=hefoballow. ing stated conditions.
4. Find the image of

i) E[(0,0), –3] ii) E[(1,2), 2] iii) E[(2,4), 3]

5. Let A(5,3), B(9,3), C(9,7) and D(5,7) be the vertices of square ABCD. Find the mage
of the square when it is reflected on

i) y = x line ii) x = –2 line

plot all the squares on the same graph.

6. A(3,3), B(6,3), C(9,5) and D(2,5) are the vertices of a quadrilateral. Find its image
under the enlargement with centre (3,–1) and scale factor 2. Plot the quadrilateral and its
image on the same graph.

7. A(–4,6), B(0,3) and C(5,8) are the vertices of ∆ABC. Translate the triangle by
a23nd.CP(l1o,t1b) oatrhe
translation vector T =  the triangles on the same graph.
 the vertices of ∆ABC to ∆A'B'C' under
8. A(–3,–3), B(–2,3) translation

vector AC . Plot ∆ABC and ∆A'B'C' on the same graph.

9. A triangle PQR with vertices P(4,0), Q(4,2) and R(1,3) is enlarged to trianble P'Q'R'
where P'(6,–1), Q'(6,3) and R'(0,5) respectively. Find the centre of enlargement and scale
factor. Draw both triangles on the same graph paper.

194 Vedanta Optional Mathematics Manual ~ 9

UNIT

fourteen Statistics: Partition values

Estimated Teaching hours : 5

1. Objectives

S.N. Level Objectives

(i) Knowledge(K) –To define partition values.

– To define quartiles, deciles and percentiles.

– To tell formula to calculate Q1, Q3, D1, D2, .......... D9
P1, P2, ......... P99, etc.

(ii) Understanding(U) To calculate quartiles, deciles and percentiles of given
individual data.

(iii) Application(A) To calculate quartiles, deciles and percentiles for given
discrete sires.

(iv) Higher Ability (HA) To make frequency table of given raw data and compute
partition values from it.

2. Teaching materials
Formula chart of quartiles, percentiles and deciles.

3. Teaching Learning Strategies
Review concept of mean, median and mode of individual and discrete series.
Give an examples of set of individual series with odd and even number of observation,
then discuss how to calculate quartiles, deciles and percentiles with formula.
Also explain importance of writing ascending or descending order to calculate the partition
values.
Give an examples discrete series and discuss to calculate the quartiles, deciles and
percentiles with formula.
Discuss about units of partition values.

Notes : b. For discrete Series
1. For individual Series

Md = value of n+1 th Md = value of N+1 th
2 2
item item

Q1 = value of (n + 1) th Q1 = value of (N + 1) th
4 4
item item

Q3 = value of 3(n + 1) th Q3 = value of 3(N + 1) th
4 4
item item

Vedanta Optional Mathematics Manual ~ 9 195

Qi = value of 3(n + 1) th Qi = value of i(N + 1) th
4 4
item, item

i = 1, 2, 3. n+1 th i = 1, 2, 3. N+1 th
D1 = value of 10 D1 = value of 10
item item

D2 = value of 2(n + 1) th D2 = value of 2(N + 1) th
10 10
item item

Di = value of i(n + 1) th Di = value of i(N + 1) th
10 10
item, item,

i = 1, 2, 3, ..........9. i = 1, 2, 3, ..........9.

P1 = value of n+1 th P1 = value of N+1 th
100 100
item item

Pi = value of i(n + 1) th Pi = value of i(N + 1) th
100 100
item, item,

i = 1, 2, 3, ..........99. i = 1, 2, 3, ..........99.

Some solved problems

1. Calculate Q1 and Q3 from given data.
20, 25, 15, 10, 30, 45, 22, 35

Solution
writing the given data in ascending order

10, 15, 20, 22, 25, 30, 35, 45

n=8 n+1 th 8+1 th
Q1 = value of 4 4
item = item = 2.25 th item

= 2nd item + (3rd item – 2nd item) 0.25

= 15 + (20 – 15) × 0.25

= 16.25

Q3 = value of 3(n + 1) th 3×9 th
4 4
item = item = 6.75 th item

= 6 th item + (7 th item – 6th item) 0.75
= 30 + (35 – 30) × 0.75
= 33.75

196 Vedanta Optional Mathematics Manual ~ 9

2. Find D3, D7 and D8 from the following data :
22, 45, 23, 67, 18, 70, 80, 75

Solution
writing the given data in ascending order.

18, 22, 23, 45, 67, 70, 75, 80

Now,

D3 = value of 3(n + 1) th 3×9 th
10 10
item = item = 2.7 th item

= 2nd item + (3rd item – 2nd item) 0.7

= 22 + (23 – 22) × 0.7

= 22.7 7(n + 1) th
D7 = value of 10
item

= 7×9 th
10
= 6.3 th item

= 6th item + (7th item – 6th item)

= 70 + (75 – 70) × 0.3

= 71.5

D8 = 8(n + 1) th
10
item

= 8×9 th
10
item = 7.2 th item

= 7th item + (8th item – 7th item) × 0.2
= 75 + (80 – 75) × 0.2
= 76

3. Find P25, P50 and P75 from the following data :
12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40

Solution
Given data is in ascending order.

n = 15

P25= 25(n + 1) th 25 × 16 th
100 100
item = item

= 4 th item = 18

P50 = value of 50(n + 1) th
100
item

= 50 × 16 th
100
= 8 th item = 26

P75 = 75(n + 1) th
100
item

Vedanta Optional Mathematics Manual ~ 9 197

= 75 × 16 th
100
item = 12 th item = 34

4. Find Q1 and Q3 of the following data. 22
Size(inch) 11 13 15 17 19 21 5
No. of Boxes 2 9 20 25 24 15

Solution

Size x (inch) No. of boxes c.f.
11 2 2

13 9 11

15 20 31

17 25 56

19 24 80

21 15 95

22 5 100

N = 100

N = 100

N+1 th 101 th
4 4
Q1= item = item = 25.25 th item

c.f. just greater 25.25 is 31 whose corresponding value is 15

Q1 = 15 inch

Q3 = 3(N + 1) th 3 × 101 th 101 th item = 75.75 th item .
4 4 4
item = item =

c.f. first greater 75.75 is 80 whose corresponding value is 19
Q3 = 19 inch

5. Find D4 , D6 and D9 from given data.

Marks 10 20 30 40 50 70 80
4
No. of Boxes 4 6 8 16 8 6

Solution c.f.
Marks (x) No. of boxes 4
10 4 10
20 6 18
30 8 34
40 16 42
50 8 48
60 6 52
70 4

198 Vedanta Optional Mathematics Manual ~ 9

N = 52

Here, N = 52

D4 = 4(N + 1) th
10
item

= 4×(52 + 1) th
10
item

=(21.5)th item

c.f. just greater than 21.5 is 34 whose corresponding value is 40.

D4 = 40 marks th

D6 = 6(N + 1) item
10

= 6×(52 + 1) th
10
item

=(31.8)th item
c.f. just greater than 31.8 is 34 whose corresponding value is 40.

D6 = 40 marks

D9 = 9(N + 1) th
10
item

= 9× 53 th
10
item

= 47.7

c.f. just greater than 47.7 is 48 whose corresponding value is 70.

D9 = 70 marks

6. Construct discrete frequency table and find 40th and 80th percentile(daily income of 23
labourers in Rs)

200, 100, 150, 300, 350, 250, 100, 300, 150, 100, 200, 300, 200, 250, 200, 300, 350, 150,
200, 400, 200, 350, 400

Solution
Writing the given data in discrete frequency table.

Daily wages (Rs.) Number of laborers (f) c.f.
100 3 3
150 3 6
200 6 12
250 2 14
300 4 18

Vedanta Optional Mathematics Manual ~ 9 199

350 3 21

400 2 23

Here, N = 23, N = 23

P40 = 40(N + 1) th
100
item

= 40× 24 th
100
item

= 9.6

c.f. just greater than 9.6 is 12. whose corresponding value is 200.

P40 = Rs. 200

P80 = 80(N + 1) th
100
item

= 80 × 24 th
100
item

=(19.2)th item
c.f. just greater than 19.2 is 21 whose corresponding value is 350.

P80 = Rs 350

Questions for practice

1. Find Q1, Q2 and Q3 from given data
Height of plants (cm) : 10, 12, 14, 6, 8, 15, 17, 20

2. Find D6, D8 and D9 from given data
(Weight in kg) : 22, 24, 28, 30, 20, 27, 36, 40, 45, 41, 45

3. Find P20, P60 and P9 from given data
marks : 40, 60, 70, 35, 75, 80, 75, 72, 80, 90, 49, 64, 75, 80

4. Compute Q1 and Q3 from given data

Marks 40 45 60 70 80 90 95
8 4 21
No. of students 2 4 6
50 55
5. Calculate P40 , P60 , D5, D8 from given table 10 7

Height(inch) 40 44 46 47 60 65 72
63 2
No. of students 2 4 68

200 Vedanta Optional Mathematics Manual ~ 9