Vedanta Optional Math Teacher's guide

b. sin2A – sin2B + sin2C = 4cosA . sinB . cosC

Solution

LHS = sin2A – sin2B + sin2C

= 2cos 2A + 2B . sin 2A – 2B + 2sinC . cosC
2 2

= 2cos(A + B) . sin(A – B) + 2sinC . cosC
= –2cosC . sin(A – B) + 2sinC cos C
= 2cosC[–sin(A – B) + sinC]
= 2cosC[sin(A + B) – sin(A – B)]
= 2cosC . 2sinB . cosA
= 4cosA . sinB . cosC
= RHS proved.

6. If A + B + C = 180°, then prove that:

a. cosA – cos B + cosC = 4cosA2 . sinB2 . cosC2 – 1

Solution

LHS = cosA – cos B + 2cos2 C – 1
2

= 2sin A +B . sin B–A + 2cos2 C – 1
2 2 2

= 2cosC2 . sin B–A + 2cos2C2 – 1  (A + B) = C
2 sin 2 cos 2

= 2cosC2 sinB – A  + cosC2 –1
2

= 2cosC2 sinB – A  + sinA + B  –1
2 2

= 2cosC2 2sinB2 . cosA2
= 4cosA2 . sinB2 . cosC2 – 1

= RHS proved.

b. cosA – cosB – cosC = 1 – 4sinA2 . cosB2 . cosC2
Solution

LHS = cosA – cosB – cosC

= 1 – 2sin2A2 – [cosB + cosC]

= 1– 2sin2A2 – 2cos B+C . cos B–C
2 2

Vedanta Optional Mathematics Teacher's Guide ~ 10 301

= 1 – 2sin2A2 – 2sinA2 . cos B–C
2

= 1 – 2sinA2 sinA2 – cos B – C
2 

= 1 – 2sinA2 cos B + C  + cos B – C
2 2 

= 1 – 2sinA2 . 2cosB2 . cosC2

= 1 – 4sinA2 . cosB2 . cosC2

= RHS proved.

7. If A + B + C = pc, then prove that:
a. sin2A – sin2B – sin2C = –2cosA . sinB . sinC

Solution
Here, A + B + C = pc
A + B = pc – C
∴ sin(A + B) = sin(pc – C) = sinC
and cos(A + B) = cos(pc – C) = –cosC

LHS = sin2A – sin2B – sin2C

= 1 [2sin2A – 2sin2B] – sin2C
2

= 12[1 – cos2A – 1 + cos2B] – sin2C

= –21[cos2A – cos2B] – sin2C

= –1 2sin 2A – 2B  . sin2B – 2A – sin2C
2 2 2 

= –sin(A + B) . sin(B – A) – sin2C

= –sinC[sin(B – A) + sinC]
= –sinC[sin(B – A) + sin(A + B)]
= –sinC . 2sinB . cosA
= –2 cosA . sinB . sinC
= RHS proved.

b. sin2 A – sin2 B – sin2 C = 2sinA2 . cosB2 . cosC2 – 1
2 2 2

Solution
LHS = sin2A2 – sin2B2 – sin2C2

302 Vedanta Optional Mathematics Teacher's Guide ~ 10

= 1 2sin2A2 – 2sin2B2 – 1 + cos2C2
2

= 12[1 – cosA – 1 + cosB] – 1 + cos2 C
2

= 12[cosB – cosA] + cos2C – 1
2sinB + A A – B
= 1 . 2  . sin 2  + cos2C2 – 1
2
A – B
= cosC2 . sin 2  + cos2C2 – 1

= cosC2 sinA – B + sin A + B –1
2 2
= cosC2 . 2sinA2 . cosB2 – 1
= 2sinA2 . cosB2 . cosC2 – 1

= RHS proved.

c. cos2A2 – cos2B2 + cos2C2 = 2cosA2 . sinB2 . cosC2

Solution
LHS = cos2A2 – cos2B2 + cos2C2

= 1 2cos2A2 – 2cos2B2 + cos2C2
2

= 12[1 + cosA – 1 – cosB] + cos2C2

= 12[cosA – cosB] + cos2C2
A + B B – A
= 1 . 2sin 2  . sin 2  + cos2C2
2 

= cosC2 sinB – A + cosC2
2

= cosC2 sinB – A + sin A + B 
2  2

= cosC2 –sinA – B + sin A + B 
2  2 

= cosC2 . 2sinB2 . cosA2
= 2cosA2 . sinB2 . cosC2

= RHS proved.

d. cos2A2 – cos2B2 – cos2C2 = –2sinA2 . cosB2 . cosC2
Solution

Here, A + B + C = pc

Vedanta Optional Mathematics Teacher's Guide ~ 10 303

A + B = pc – C

∴ sin(A + B) = sin(pc – C) = sinC

and cos(A + B) = cos(pc – C) = –cosC
LHS = cos2A2 – cos2B2 – cos2C2

= 1 2cos2A2 – 2cos2B2 – cos2 C
2 2

= 21[1 + cosA – 1 – cosB] – cos2 C
2

= 12[cosA – cosB] – cos2 C
2
A + B B – A
= 1 . 2sin 2  . sin 2  – cos2C2
2
sinB – A
= cosC2 . 2 – cos2C2

= cosC2 sin B – C  – cosC2
2 

= cosC2 sinB – A – sin A + B
2  2 

= cosC2 2cosB – A + A + B . sin B – A – A – B 
4 4

= cosC2 2cosB2 . sin –A 
2 
= –2sinA2 . cosB2 . cosC2

= RHS proved.

8. If A + B + C = pc, prove that

a. sinA2 + sinB2 + sinC2 = 1 + 4sin p – A  . sin p – B  . sin p – C 
4 4 4

= 1 + 4sinB + C . sin C + A  . sinA + B
4 4  4

Solution

LHS = sinA2 + sinB2 + sinC2 + 1 – 1

= 1 + sinA2 + sinB2  + sinC – sin p
2

= 1 + 2sin A+B . cos A–B + 2cos C+p . sin C–p
4 4 4 4

= 1 + 2sin p – C  cos A – B  – 2cos p + C  . sin p – C 
4 4 4 4

= 1 + 2sin p – C  cos A – B  – cos p + C 
4 4  4 

= 1 + 2sin p – C  . 2sin A – B + p + C . sin p + C – A + B 
4  8  8 

304 Vedanta Optional Mathematics Teacher's Guide ~ 10

= 1 + 4sin p – C  . sin A + C – B + p . sin B + C – A + p 
4  8  8 

= 1 + 4sin p – C  . sin  p – B –B + p  . sin p – A –A + p 
4 8 8

= 1 + 4sin p – C  . sin p – B  . sin p – A 
4 4 4

= 1 + 4sin p – A  . sin p – B  . sin p – C  = MS
4 4 4

= 1 + 4sinB + C . sin C + A . sinA + B 
4  4  4

= RHS proved.

b. cosA + cosB + cosC = 1 + 4cos p – A  . cos  p – B  . cos  p – C 
 2   2   2 

= 1 + 4cosB + C . sin C + A  . sin A + B 
Solution 2  2  2 

LHS = cosA + cosB + cosC

= cosA + cosB + cosC + cosp + 1 ( cosp = –1)

= 1 + (cosA + cosB) + (cosC + cosp)

= 1 + 2cos A + B . cos A – B  + 2 cos p + C  cos  p – C 
2  2  2   2 

= 1 + 2cos p – C  . cos A–B + 2cos p + C  . cos p – C 
2  2 2  2 

= 1 + 2cos p – C  cos A–B + cos p + C 
2 2 2 

= 1 + 2cos p – C  2cos A – B + p + C . cos A – B– p – C 
2  4  4

= 1 + 4cos p – C  . cosA + C – B + p  . cosA – p +(B + C)
2 4 4 

= 1 + 4cos p – C  . cosp – B – B + p . cos A – p –p + A 
2 4 4

= 1 + 4cos p – C  . cos p – B  . cos A – p 
2  2  2 

= 1 + 4cos p – A  . cos p – B  . cos p – C 
2 2 2

= 1 + 4cosB + C . sin C + A . sin A + B
2 2 2

= RHS proved.

9. If A + B + C = pc, prove that
sin(B + C – A) + sin(C + A – B) + sin(A + B – C) = 4 sinA . sinB. sinC

Vedanta Optional Mathematics Teacher's Guide ~ 10 305

Solution
Here, A + B + C = p
A + B = p – C
∴ sin(A + B) = sin(p – C) = sinC
and cos(A + B) = cos(p – C) = –cosC

LHS = sin(B + C – A) + sin(C + A – B) + sin(A + B – C)

= sin(p – 2A) + sin(p – 2B) + sin(p – 2C)

= sin2A + sin2B + sin2C

= 2sin2A + 2B . cos 2A – 2B  + sinC cosC
2  2

= 2sin(A + B) . cos(A – B) + 2sinC . cos C

= 2sinC . cos(A – B) + 2sinC . cosC

= 2sinC[cos(A – B) + cosC]

= 2sinC[cos(A – B) – cos(A + B)]

= 2sinC . 2sinA . sinB

= 4sinA . sinB . sinC

= RHS proved.

10. If A + B + C = pc, prove that

cos(B + 2C) + cos(C + 2A) + cos(A + 2B) = 1 – 4cos A – B . cosB – C . cos C – A 
2  2  2 

Solution

Here, A + B + C = p

A + B = p – C

∴ sin(A + B) = sin(p – C) = sinC

and cos(A + B) = cos(p – C) = –cosC

LHS = cos(B + 2C) + cos(C + 2A) + cos(A + 2B)

= cos(A + C + C) + cos(C + A + A) + cos(A + B + B)

= cos(p – A + C) + cos(p – B + A) + cos(p – C + B)

= cos{p – (A – C)} + ocs{p – (B – A)} + cos{p – (C – B)}
= cos(A – C) – cos(B – A) – cos(C – B)

= –[cos(A – C) + cos(B – A)] – cos(C – B)

= –2cos A – C + B – A . cosA – C – B + A – 2cos2 B – C + 1
2 2 2

= –2cos B – C . cos 2A –C – B – 2cos2B – C + 1
2  2  2 

= 1 – 2cosB – C cos 2A –C – B + cos B – C
2  2  2 

= 1 – 2cosB – C 2cos 2A – C – B + B – C  + cos2A – C – B – B + C
2 4 4

306 Vedanta Optional Mathematics Teacher's Guide ~ 10

= 1 – 4cosB – C . cos A – C . cos A – B 
2  2  2 

= 1 – 4cosA – B . cos B – C . cosC – A 
2  2  2 
= RHS proved.

11. If α + β + γ = pc , prove that
2
cos(α – β – γ) + cos(β – γ – α) + cos(γ – α – β) = 4cosα . cosβ . cosγ

Solution

Here, α + β + γp2=c –p2γc
α + β =

∴ sin(α + β) = sin pc – γ = cosγ
2
and cos(α + β) = sinγ

LHS = cos(α – β – γ) + cos(β – γ – α) + cos(γ – α – β)

= 2cosα – β – γ + β – γ – α  . cos α – β – γ –β + γ + α  . cos γ – p + γ 
2  2  2 

= 2cosγ . cos(α – β) + sin2γ

= 2cosγ . cos(α – β) + 2sinγ . cosγ

= 2cosγ [cos(α – β) + sinγ]

= 2cosγ [cos(α – β) + cos (α + β)]

= 2cosγ . 2cosα . cosβ

= 2cosα . cosβ . cosγ

= RHS proved.

12. If A + B + C = pc, prove that

a. sin2A + sin2B + sin2C = 8sinA2 . sinB2 . sinC
4cosA . cosB . cosC
222

Hints :
Numerator = sin2A + 2sinB + 2sinC
It gives 4sinA . sinB . sinC
and 4sinA . sinB . sinC
= 16sinA2 . sinB2 . sinC2 . cosA2 . cosB . cosC2
b. cosA . cosB . sinC + cosB . sinC . sinA + cosC . sinA . sinB = 1 + cosA . cosB . cosC

Solution
Here, A + B + C = pc
A + B = pc – C
∴ sin(A + B) = sin(pc – C) = sinC

Vedanta Optional Mathematics Teacher's Guide ~ 10 307

and cos(A + B) = cos(pc – C) = –cosC
LHS = cosA . cosB . sinC + cosB . sinC . sinA + cosC . sinA . sinB
= sinC(cosA . sinB + cosB . sinA) + cosC . sinA . sinB
= sinC . sin(A + B) + cosC . sinA . sinB
= sinC . sinC + cosC . sinA . sinB
= 1 – cos2C + cosC . sinA . sinB
= 1 – cosC [cosC – sinA . sinB]
= 1 – cosC [–cos(A + B) – sinA . sinB]
= 1 + cosC [cos(A + B) + sinA . sinB]
= 1 + cosC [cosA . cosB – sinA . sinB + sinA . sinB]
= 1 + cosA . cosB . cosC
= RHS proved.

Questions for practice



If A + B + C = p, prove the following:
1. tan2A + tan2B + tan2C = tan2A tan2B tan2C

2. sin2A + sin2B + sin2C = 4sinA sinB sinC

3. cos2A + cos2B – cos2C = 1 – 4sinA sinB sinC

4. cos2A – cos2B – cos2C = 4cosA sinB sinC – 1

5. sinA + sinB + sinC = 4cosA2 . cosB2 . cosC2

6. sin2A + sin2B + sin2C = 2 + 2cosA . cosB . cosC

7. sin2A2 + sin2B2 – sin2 C = 1 – 2cosA2 . cosB2 . sinC2
2

8. cos(B + C – A) + cos(C + A – B) + cos(A + B – C) = 1 + 4 cosA . cosB . cosC

9. sinA2 + sinB2 + sinC2 = 1 + 4sinp – A . sin p – B  . sin p – C
4  2  2 

10. tanA2 + tan B + C = secA2 . secB + C
2  2 

11. cotA2 + cot B + C  = cosecA2 . cosecB + C
2  2

12. sinB + sinC – sinA = tanB2 . tanC2
sinA + sinB + sinC

308 Vedanta Optional Mathematics Teacher's Guide ~ 10

Trigonometric Equations

Estimated Periods : 4

1. Objectives

Knowledge (K) To define trigonometric equation.
To define principal solutions of trigonometric equations.
To say limitations of variables in trigonometric equations.

To explain the rule of 'CAST'
Understanding (U) To find the values of unknown angles from trigonometric equations.

Skill/Application To solve simple trigonometric equations.
(S/A)
To solve harder trigonometric equations.
Higher Ability To check solutions trigonometric equations.
(HA) To check solutions are true or false when equations are solved
squaring on both sides of equations.

2. Teaching Materials
Chart paper with rule of 'CAST' and table of trigonometric values of standard angles.

3. Teaching Learning Strategies
→ Review the concept of equations and identities with some appropriate examples.

→ Explain the rule of 'CAST' with examples

Give some basic ideas for solution of equations, define principal solutions.
→ Ask the solution of simple equations like sinθ = 23, tan2θ = 1, 0°  θ  180°

→ Discuss how equation in the form of a sin2θ + b sinθ + c = 0 can be solved, when it can
be factorized. Also state limitations of values of trigonometric equations.

Example: 2 sin2θ – 3 sinθ + 1 = 0, 0°  θ  360°.

Give idea how to check roots of equations are true or false as in algebra solving
simultanesous equations in two variables.

→ Solve an equation, for example sinx + cosx = 1, 0°  θ  360°. Solve it in two ways:

i) dividing by 12 + 12 = 2

ii) squaring on both sides.

Suggest how to check roots so obtained are true or false.

→ Give ideas to solve some harder questions. For example:

2 tan3x cos2x + 1 = tan3x + 2cosx, 0°  θ  360°.

Vedanta Optional Mathematics Teacher's Guide ~ 10 309

Rule of CAST (From Book)

Y

(S) (A) (all ratios are positive
(Sine and cosec are positive) θ and (360° + θ), θ is acute angle

XO X′
(T) (C)

(Tan and cot are positive) (Cos and sec are positive)

Y′
Some Basic Ideas (From Book page 245)

Working rules for solutions of trigonometric equations (From Book page 246)

Some special cases (From Book page 297)

Some solved problems

1. Solve 0°  θ  180°
a. 2cos θ + 1 = 0

Solution

Here, 2cos θ + 1 = 0
–1
or, cos θ = 2

Since cosθ is negatyive, θ lies on the second and third quadrant. But 0°  θ  180°.

cosθ = cos 120°

 θ = 120°.

b. 3 tanθ – 1 = 0

Solution

Here, 3 tanθ – 1 = 0

or, 3 tanθ = 1

310 Vedanta Optional Mathematics Teacher's Guide ~ 10

or, tanθ = 1
3

Since tanθ is positive, it lies on the first quadrant.

tanθ = tan30°

 θ = 30°

2. Find the value of θ, 0°  θ  90°
a. tanθ = cotθ

Solution tanθ = cotθ
Here, tanθ = tan(90° – θ)
or, θ = 90° – θ
 2θ = 90°
or, θ = 45°


b. sin4θ = cos2θ

Solution
Here, sin4θ = cos2θ
or, sin4θ = sin(90° – 2θ)
 4θ = 90° – 2θ
or, 6θ = 90°
 θ = 15°

3. Solve: (0°  θ  360°)
a. 3tan2θ = 1

Solution 3tan2θ = 1
Here,
tan2θ = 1
or, 3

 tanθ = ± 1
3

Taking positive sign, we get.

tanθ = 1
3

or, tanθ = tan30°, tan(180° + 30°)

 θ = 30°, 210°

Taking negative sign, we get.

Vedanta Optional Mathematics Teacher's Guide ~ 10 311

tanθ = – 1
3

or, tanθ = tan150°, tan(360° – 30°)

 θ = 150°, 330°

b. sec2θ = 2tan2θ

Solution

Here, sec2θ = 2tan2θ
or, 1 + tan2θ = 2 tan2θ
or, tan2θ = 1
 tanθ = ±1
Taking positive sign, we get.
tanθ = 1
or, tanθ = tan45°, tan(180° + 45°)
 θ = 45°, 225°
Taking negative sign, we get.
tanθ = –1
or, tanθ = tan135°, tan(360° – 45°)
 θ = 135°, 315°
Hence the required values of θ are 45°, 135°, 225°, 315°.
4. Solve : (0°  θ  180°)
a. sin4θ = cosθ – cos7θ

Solution

Here, sin4θ = cosθ – cos7θ

or, sin4θ = 2sin θ + 7θ  . sin 7θ2– θ 
2

or, sin4θ = 2sin4θ . sin3θ = 0

or, sin4θ (1 – 2sin3θ) = 0

Either, sin4θ = 0 ... ... ... (i)

 1 – 2sin3θ = 0 ... ... ... (ii)

From equation (i) sin4θ = 0

sin4θ = sin 0°, sin180°, sin360°, sin540°

 4θ = 0°, 180°, 360°, 540°

 θ = 0°, 45°, 90°, 135°

From equation (ii), we get

312 Vedanta Optional Mathematics Teacher's Guide ~ 10

2sin3θ = 1

or, sin3θ = 1
2

or, sin3θ = sin30˚, sin150˚, sin(360˚+30˚)

⸫ 3θ = 30˚, 150˚, 390˚

⇒ θ = 10˚, 50˚, 130˚

b. sinθ + sin2θ + sin3θ = 0
Solution
Here, sinθ + sin2θ + sin3θ = 0

or, (sinθ + sin3θ) + sin2θ = 0

or, 2sin θ + 3θ  . cos θ – 3θ  + sin2θ = 0
2  2

or, 2sin2θ . cosθ + sin2θ = 0

or, 2sin2θ (cosθ + 1) = 0

or, sin2θ (cosθ + 1) = 0

Either, sin2θ = 0 ... ... ... (i)

 cosθ + 1 = 0 ... ... ... (ii)

From equation (i) sin2θ = 0

sin2θ = sin0°, sin180°, sin360°

 θ = 0°, 90°, 180°

From equation (ii) cosθ = –1

cosθ = cos180°

 θ = 180°

Hence the required values of θ are 0°, 180°.

c. 3cotθ – tanθ = 2

Solution

Here, 3cotθ – tanθ = 2

or, 3 – tanθ =2
tanθ

or, 3 – tan2θ = 2 tanθ

or, tan2θ + 2 tanθ – 3 = 0

or, tan2θ + 3 tanθ – tanθ – 3 = 0

or, tanθ(tanθ + 3) – 1(tanθ + 3) = 0

or, (tanθ + 3) (tanθ – 1) = 0

Vedanta Optional Mathematics Teacher's Guide ~ 10 313

Either, tanθ – 1 = 0 ... ... ... (i)
 tanθ + 3 = 0 ... ... ... (ii)
From equation (i) tanθ = 1
tanθ = tan45°
 θ = 45°
From equation (ii) tanθ = –3
tanθ = tan108.43°
 θ = 108.43°
Hence the required values of θ are 45°, 108.43°.

5. Solve : (0° ≤ θ ≤ 360°)
a. 2cos2θ – 3sinθ = 0

Solution
Here, 2cos2θ – 3sinθ = 0

or, 2 – 2 sin2θ – 3 sinθ = 0

or, 2 sin2θ + 3 sinθ – 2 = 0

or, 2 sin2θ + 4 sinθ – sinθ – 2 = 0

or, 2 sinθ(sinθ + 2) – 1(sinθ + 2) = 0

or, (sinθ + 2) (2 sinθ – 1) = 0

Either, sinθ + 2 = 0 ... ... ... (i)

 2 sinθ – 1 = 0 ... ... ... (ii)

From equation (i) sinθ = –2

It has no solution as –1 ≤ sinθ ≤ 1

From equation (ii) sinθ = 1
2
sinθ = sin30°, sin150°

 θ = 30°, 150°

Hence the required values of θ are 30°, 150°.

b. 4 cos2θ + 4 sinθ = 5
Solution

Here, 4 cos2θ + 4 sinθ = 5
or, 4 – 4 sin2θ + 4 sinθ – 5 = 0
or, 4 sin2θ – 4 sinθ + 1 = 0
or, (2 sinθ – 1)2 = 0

314 Vedanta Optional Mathematics Teacher's Guide ~ 10

or, 2 sinθ – 1 = 0

or, sinθ = 1
2
or, sinθ = sin30°, sin150°

 θ = 30°, 150°

c. 3 – 2 sin2θ = 3 cosθ

Solution

Here, 3 – 2 sin2θ = 3 cosθ

or, 3 – 2 + 2 cos2θ = 3 cosθ

or, 2 cos2θ – 3 cosθ + 1 = 0

or, 2 cos2θ – 2 cosθ – cosθ + 1 = 0

or, 2 cosθ(cosθ – 1) – 1(cosθ – 1) = 0

or, (cosθ – 1) (2 cosθ – 1) = 0

Either, cosθ – 1 = 0 ... ... ... (i)

 2 cosθ – 1 = 0 ... ... ... (ii)

From equation (i) cosθ = 1

cosθ = cos0°, cos360°

 θ = 0°, 360°
1
From equation (ii) cosθ = 2

cosθ = cos60°, cos300°

 θ = 60°, 300°

Hence the required values of θ are 0°, 60°, 300°, 360°

d. tan2θ + (1 – 3) tan θ = 3
Solution
Here, tan2θ + (1 – 3) tan θ = 3
or, tan2θ + tan θ – 3 tan θ – 3 = 0
or, tanθ(tan θ + 1) – 3(tan θ + 1) = 0
or, (tan θ + 1) (tan θ – 3) = 0
Either, tanθ + 1 = 0 ... ... ... (i)
 tanθ – 3 = 0 ... ... ... (ii)
From equation (i) tanθ = –1
tanθ = tan135°, tan315°
 θ = 135°, 315°

Vedanta Optional Mathematics Teacher's Guide ~ 10 315

From equation (ii) tanθ = 3
tanθ = tan60°, tan240°
 θ = 60°, 240°
Hence the required values of θ are 60°, 135°, 240°, 315°

e. cot2θ +  3+ 1 cotθ + 1 = 0
3

Solution

Here, cot2θ +  3+ 1  cotθ + 1 = 0
3

or, 3 cot2θ + 3 cotθ + cotθ + 3 = 0

or, 3 cotθ(cotθ + 3) + 1(cotθ + 3) = 0

or, (cotθ + 3) ( 3 cotθ + 1) = 0

Either, cotθ + 3 = 0 ... ... ... (i)

 3 cotθ + 1 = 0 ... ... ... (ii)

From equation (i) cotθ = – 3

cotθ = cot150°, cot330°

 θ = 150°, 330°

From equation (ii) cotθ = 1
3
cotθ = cot120°, cot300°

 θ = 120°, 300°

Hence the required values of θ are 120°, 150°, 300°, 330°

6. Solve (0°  x  360°)
a. 3 cosx + sinx = 3

It can be solved in two ways:

First Method 3 cosx + sinx = 3
Here,

It is in the form of a sinθ + b cosθ = c

Where, a = 1, b = 3

a2 + b2 = 1 + 3 = 4 = 2

Dividing the given equation on both sides by '2', we get,

316 Vedanta Optional Mathematics Teacher's Guide ~ 10

1 sinx + 3 cosx = 3
2 2 2

or, cosx cos30° + sinx sin30° = 3
2

or, cos(x – 30°) = cos30°. cos(360° – 30°)

or, cos(x – 30°) = cos 30°, cos 330°

or, x – 30° = 30°, 330°
 x = 60°, 360°

Second Method
Here, 3 cosx + sinx = 3

32 + 12 = 2

Dividing the given equation by '2', on both sides, we get,

3 cosx + sinx = 3
2 2 2

or, sinx cos60° + cosx sin60° = 3
2
or, sin(x + 60°) = sin60°, sin120°, sin420°

or, x + 60° = 60°, 120°, 420°

 x = 0°, 60°, 360°

Hence, the required values of x are 0°, 60°, 360°.

Alternative Method
In this method, we square on both sides, the roots so obtained must be checked whether it
is true or false, only true values are accepted.

Here, 3 cosx + sinx = 3

3 cosx = 3 – sinx

Squaring on both sides, we get

( 3 cosx)2 = ( 3 – sinx)2
or, 3 cos2x = 3 – 2 3 sinx + sin2x

or, 3 – 3 sin2x = 3 – 2 3 sinx + sin2x

or, –4 sin2x + 2 3 sinx = 0

or, 2 sin2x – 3 sinx = 0

 sinx(2 sinx – 3) = 0
Either, sinx = 0 ... ... ... (i)

Vedanta Optional Mathematics Teacher's Guide ~ 10 317

 2 sinx – 3 = 0 ... ... ... (ii)

From equation (i) sinx = 0

sinx = sin0°, sin180°, sin360°

 x = 0°, 180°, 360°

From equation (ii) sinx = 3
2
sinx = sin60°, sin120°

 x = 60°, 120°

In checking for x = 0°, 120°, 180°, 360°

For x = 0° 3 . 1 + 0 = 3 (true)

For x = 120° 3 . –21 + 3 = 3 0= 3 (false)
2
For x = 60°
For x = 180° 3 . 1 + 3 = 3 3 = 3 (true)
2 2 – 3 = 3 (false)

3 . (–1) + 0 = 3

For x = 360° 3 . 1 + 0 = 3 (true)

Hence, the required values of x are 0°, 60°, 360°.

b. 3 sinx + cosx = 2, 0˚≤ x ≤ 360˚

Solution

Here 3 sinx + cosx = 2

( 3)2 + 1 = 2

Dividing on both sides by 2, we get,

3 sinx + 1 cosx = 1
2 2
or, cosx . cos60° + sinx . sin60° = 1

or, cos(x – 60°) = cos0°

or, x – 60° = 0°

 x = 60°

Alternatively,
we can solve it by squaring on both sides,
3 sinx = 2 – cosx
Squaring on both sides, we get,
3 sin2x = 4 – 4 cosx + cos2x
or, 3 – 3 cos2x = 4 – 4 cosx + cos2x

318 Vedanta Optional Mathematics Teacher's Guide ~ 10

or, –4 cos2x + 4 cosx – 1 = 0

or, 4 cos2x – 4 cosx + 1 = 0

or, (2 cosx – 1)2 = 0

or, 2 cosx – 1 = 0

or, cosx = 1
2

or, cosx = cos60°, cos300°

 x = 60°, 300°

On checking,

When x = 60 3 . 3 + 1 = 2 3 + 1 = 2 2 = 2 (true)
2 2 2 4 2 –1 = 2 (false)

When x = 300 3 . – 3 + 1 = 2 –3 + 1 = 2
2 2 2

Hence, the required value of x is 60°.

7. Solve for x, (0°  x  360°)

a. 3 + 1 = 0
sin2x cos2x

Solution

Here, 3 + 1 = 0
sin2x cos2x

or, 3 cos2x + sin2x = 0

or, 3 cos2x + 1 sin2x = 0
2 2
or, cos2x . cos30° + sin2x . sin30° = 0

or, cos(2x – 30°) = cos90°, cos270°

or, 2x – 30° = 90°, 270°

or, 2x = 120°, 300°

 x = 60°, 150°

b. 3 + 1 = 4
sin2x cos2x

Solution

Here, 3 + 1 = 4
sin2x cos2x

or, 3 cos2x + sin2x = 4 sin2x . cos2x

or, 3 cos2x + 1 sin2x = 2 sin2x . cos2x
2 2

Vedanta Optional Mathematics Teacher's Guide ~ 10 319

or, cos2x . sin60° + sin2x . cos60° = sin4x

or, cos(2x + 60°) = sin4x, sin(180° – 4x)

 2x + 60° = 4x, 180° – 4x

2x = 60° or, x = 30°

and 2x + 60° = 180° – 4x

or, x = 20°

 x = 20°, 30°

c. 2 secx + tanx = 1

Solution
Here, 2 secx + tanx = 1

or, 2 + sinx = 1
cosx cosx

or, 2 + sinx = cosx

or, 2 = cosx – sinx

Squaring on both sides, we get,

2 = cos2x – 2 cosx sinx + sin2x

or, 2 = (sin2x + cos2x) – sin2x

or, 2 = 1 – sin2x

or, sin2x = –1

or, sin2x = sin270°, sin(360° + 270°)

or, 2x = 270°, 630°

 x = 135°, 315°

8. Solve : (0° ≤ x ≤ 360°)

a. 2 tan3x . cos2x + 1 = tan3x + 2 cos2x

Solution
Here, 2 tan3x . cos2x + 1 = tan3x + 2 cos2x

or, 2 tan3x . cos2x – tan3x + 1 – 2 cos2x = 0

or, tan3x(2 cos2x – 1) – 1(2 cos2x – 1) = 0

 (tan3x – 1) (2 cos2x – 1) = 0

Either, tan3x – 1 = 0 ... ... ... (i)

 2 cos2x – 1 = 0 ... ... ... (ii)

From equation (i) tan3x = 1

tan3x = tan45°, tan225°, tan405°, tan765°, tan945°

320 Vedanta Optional Mathematics Teacher's Guide ~ 10

or, 3x = 45°, 225°, 405°, 765°, 945°

 x = 15°, 75°, 135°, 225°, 315°

From equation (ii) 2 cos2x = 1

cos2x = 1
2

or, cos2x = cos60°, cos300°, cos420°, cos660°

or, 2x = 60°, 300°, 420°, 660°

 x = 30°, 150°, 210°, 330°

Hence, the required values of x are 15°, 30°, 75°, 135°, 150°, 210°, 225°, 315°, 330°

b. sin2x . tanx + 1 = sin2x + tanx

Solution
Here, sin2x . tanx + 1 = sin2x + tanx

or, sin2x . tanx – sin2x + 1 – tanx = 0

or, sin2x(tanx – 1) – 1(tanx – 1) = 0

 (tanx – 1) (sin2x – 1) = 0

Either, tanx – 1 = 0 ... ... ... (i)

 sin2x – 1 = 0 ... ... ... (ii)

From equation (i) tanx = 1

tanx = tan45°, tan225°

 x = 45°, 225°

From equation (ii) sin2x = 1

sin2x = sin90°, sin450°
or, 2x = 90°, 450°
 x = 45°, 225°
Hence, the required values of x are 45°, 225°

9. Solve for θ, (0°  θ  360°)
a. tanθ – 3 cotθ = 2 tan3θ

Solution
Here, tanθ – 3 cotθ = 2 tan3θ

or, tanθ – 3 = 2 . 3 tanθ – tan3θ
tanθ 1–3 tan2θ

or, tan2θ – 3 = 6 tanθ – 2 tan3θ
tanθ 1 – 3 tan2θ

or, tan2θ – 3 – 3 tan4θ + 9 tan2θ = 6 tan2θ – 2 tan4θ

Vedanta Optional Mathematics Teacher's Guide ~ 10 321

or, –tan4θ + 4 tan2θ – 3 = 0
or, tan4θ – 4 tan2θ + 3 = 0
or, tan4θ – 3 tan2θ – tan2θ + 3 = 0
or, tan2θ(tan2θ – 3) – 1(tan2θ – 3) = 0
or, (tan2θ – 3) (tan2θ – 1) = 0
Either, tan2θ – 3 = 0 ... ... ... (i)
 tan2θ – 1 = 0 ... ... ... (ii)
From equation (i) tan2θ = 3
tan2θ = ( 3)2
 tanθ = ± 3
Taking positive sign,
tanθ = 3
or, tanθ = tan60°, tan240°
 θ = 60°, 240°
Taking negative sign,
tanθ = – 3
or, tanθ = tan120°, tan300°
 θ = 120°, 300°
From equation (ii)
tanθ = ±1
Taking positive sign,
tanθ = 1
or, tanθ = tan45°, tan225°
 θ = 45°, 225°
Taking negative sign,
tanθ = –1
or, tanθ = tan135°, tan315°
 θ = 135°, 315°
Hence, the required values of θ are 45°, 60°, 120°, 135°, 225°, 240°, 300°, 315°

322 Vedanta Optional Mathematics Teacher's Guide ~ 10

Height and Distance

1. Objectives

Knowledge (K) To define angle of depression and angle of eleration.

Understanding (U) To draw figures o show angle of eleration and angle of depression.

Skill/Application To solve simple problems on height and distance with drawing
(S/A) diagrams.

Higher Ability To solve harder problems on height and distance with verbal
(HA) expressions.

2. Teaching Materials Required
Chart paper with illustration of angle of elevation and angle of depression, theodolite,

sextant, clinometor, hypsometre etc.

3. Teaching Activities:
→ Discuss about the components of a triangle as review and fundamental trigonometric ratios.
→ Review on solution of a right angled triangle.
→ Define angle of eleration and angle of depression with draw for illustration.
→ Give problems from exercise of the text book Q.N. 2 and 3.
→ The teacher solves some verbal problems as examples.

Some solved problems

1. Find the values of x and y from the given figure. P
y
Solution Q
Here, from right angled triangle PQR,

tan45° = y
x

or, 1 = y
x
45°
 x = y ... ... ... (i) S 30° Rx
Again from right angled triangled PQR 20m

tan30° = PQ
or, SQ

1 = y x
3 20 +

 20 + x = 3y ... ... ... (ii)
Put the value of x from (i) in (ii), we get

Vedanta Optional Mathematics Teacher's Guide ~ 10 323

20 + y = 3y

or, y(1 – 3) = –20

or, y= 20 1 × 3+1
3– 3+1

or, y = 20( 3 + 1)
2

or, y = 10 × 2.732

 y = 27.32m

b.

Solution
Here, UPS = PSQ = 60°, PU//QS, corresponding angles.

UPR = PRT = 30°, PU//TR, corresponding angles.

From right angled ∆PRT, P U
R
tan30° = PT 30° 60°
or, TR
20m
1 = 20 ( TR = QS = x)
3 x

 x = 20 3m T
Again from right angled ∆PQR,

tan60° = PQ y
or, RS

3 = 20 + y
x

 3x = 20 + y ... ... ... (ii) Q S
x
Put the value of x in eqn (ii), from (i)

3 . 20 3 = 20 + y

or, 60 = 20 + y

 y = 40m

Hence, x = 20 3m and y = 40m.

2. The angle of elevation of the top of a tower from a point on ground was observed to be
45° on walking 30m away from that point it was found to be 30°. Find the height of the
house.

Solution
Let, PS be the height of the house and R the point of observer.

324 Vedanta Optional Mathematics Teacher's Guide ~ 10

PRS = 45°, PQS = 30°, QR = 30m

Let, PS = xm, RS = ym

From right angled ∆PQS, P
x
tan30° = PS S
or, QS
1
3 = 30 x y
+

 30 + y = 3x ... ... ... (i)

Again from right angled triangle PRS, 30° 45°
Q 30m Ry
tan45° = PS
RS
or, 1 = xy

 x = y ... ... ... (ii)

Put the value of x from eqn (ii) in (i), we get

30 + y = 3y
or,
y( 3 – 1) = 30
or,
y= 30 1 × 3+1
3– 3+1

or, y = 30( 3 + 1)
3–1

or, y = 15( 3 + 1)

or, y = 15 × 2.732
 y = 40.98m
Put the value of y in eqn (i)
x = 40.98m
Hence the height of the house was 40.98m.

3. From the top of a tower of 200m the angle of depression of two boats which are in
a straight line on the same side of the tower are to be 30° and 45°. Find the distance
between the boats.

Solution
Let, PQ = 200m, the height of the tower,
R and S are the positions of the boats such that
UPS = PSQ = 30°, UP//SQ, the corresponding angles
UPR = PRQ = 45°

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From right angled ∆PRQ, UP
45° 30°
tan45° = PQ
RQ

or, 1 = 2R0Q0 200m

 RQ = 200m

Again from right angled triangle PQR,

tan30° = PQ 30° 45°
or, SQ S R

1 = 200 Q
3 SR + RQ

or, SR + RQ = 200 3
or, SR = 200 = 200 3
or, SR = 200( 3 – 1)
or, SR = 200(1.732 – 1)
or, SR = 200 × 0.732
 SR = 146.4m
Hence the distance between the two boats is 146.4m.

4. From a helicopter flying vertically above a straight road, the angles of depression of
two consecutive kilometer stone on the same side are found to be 45° and 60°. Find the
height of the helicopter.

Solution
Let, R be the position of the helicopter and RS the height of it from the ground.

Let, P and Q be the positions of two stones on the ground such that PQ = 1km = 1000m

RPS = 45°, RQS = 60° R
Let, QS = ym, RS = xm

From right angled ∆RQS,

tan60° = RS
QS

or, 3 = RS
QS
x
x
or, 3 = y

 x = 3y ... ... ... (i)

Again, from right angled triangle RPS, 45° 60°
P 1km Q
tan45° = RS y S
PS

326 Vedanta Optional Mathematics Teacher's Guide ~ 10

or, 1 = y + x1000

 x = y + 1000 ... ... ... (ii)

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

y + 1000 = 3y

or, 1000 = ( 3 – 1)y

or, y= 1000 × 3+1
3–1 3+1

or, y = 1000( 3 + 1)
3–1

or, y = 1000(2.732)
2

or, y = 500 × 2.732

 y = 1366m

Put the value of y in eqn (ii)

x = 1366 + 1000
= 2366m.

Hence the height of the helicopter was 2366m.

5. From the top of 21m high cliff; the angles of depression of top and bottom of a towers
are observed to be 45° and 60° respectively. Find the height of the tower.

Solution
Let, MN and PQ be the heights of cliff and the tower respectively and MN = 21m

Then, draw SM//PR and PR//QN

SMP = MPR = 45°, SMQ = MQN = 60°, MN = 21m

From right angled ∆MQN, S M
60° 45°
tan60° = MN
or, QN
21
3 = QN

 QN = 7 3m

Again from right angled triangle MPR, 45°
P
tan45° = MR R
PR

But PR = QN = 7 3m

or, 1 = MR 60°
73 Q
N
 MR = 7 3m

Vedanta Optional Mathematics Teacher's Guide ~ 10 327

Hence the height of the tower is PQ = RN
= MN – MR
= 21 – 7 3
= 21 – 7 × 1.732 = 8.87m

6. A flagstaff is placed at one corner of a rectangular 40m long and 30m wide. If the angle of
elevation of the top of the flagstaff from the opposite corner is 30°. Find the height of the flagstaff.

Solution

Let, ABCD be rectangular garden of length 40m and breadth 30m and PD be the height

of the flagstaff. P
The angle of elevation of the flagstaff PD is 30°

Diagonel BD is drawn. By using pythagoras theorem, A D
BD = BC2 + CD2

= 1600 + 900

= 2500 30m 30°

= 50m

From right angled ∆PBD,

tan30° = PD B 40m C
or, BD

1 = PD
3 50

PD = 28.86m

9. From the top and bottom of a tower, the angle of depression of the top of the house and
angle of elevation of the house are found to be 60° and 30° respectively. If the height of
the building is 20m, find the height of the tower.

Solution EA
In the figure, EA//CF//DB 60°

AB = the height of the tower

CD = 20m, the height of the house = FB

EAC = ACF = 60°, DBC = 30°

EA//CF the corresponding angles.

From right angled ∆CDB, 60°
30°
tan30° = CD C F
DB 20m B

or, 1 = 20 D
3 DB

 DB = 20 3

328 Vedanta Optional Mathematics Teacher's Guide ~ 10

Again, from right angled triangle ACF,

tan60° = AF
CF

or, 3 = AF ( CF = DB)
20 3

 AF = 60m
Hence, the height of the tower = AB
= AF + FB
= 60m + 20m
= 80m.

10. A ladder of 18m reaches a point 18m below the top of a vertical flagstaff. From the foot
of the ladder the angle of elevation of the flagstaff is 60°. Find the height of the flagstaff.

Solution P
Let, SR be the ladder of length 18 and PQ the height of the flagstaff.
R
SR = 18m, PR = 18m Q

PSR = 60°. ∆PRS is an isoceles triangles.

 PSR = ∆RPS = 30°

RSQ = 60° – 30° = 30°

Now, from the right angled triangle RSQ, 18m

sin30° = RQ
SR

or, 1= RQ 18m
2 18 60°

 RQ = 9m

Hence, the height of the flagstaff is 18m + 9m = 27m. S

11. AB is a vertical pole with its foot B on a level of ground. A point C on AB divides such
that AC:CD = 3:2. If the parts AC and CB subtand equal angles at a point on the ground
which is at a distance of 20m from the foot of the pole. Find the height of the pole.

Solution
In the figure, AC:CB = 3:2.
Let, AC = 3x, BC = 2x
D is a point 20m away from the foot of the pole B. DB = 20m
BDC = θ, ADC = θ, ADB= 20θ
From the right angled triangle CDB,

tanθ = CB
DB

Vedanta Optional Mathematics Teacher's Guide ~ 10 329

or, tanθ = 2x = x A
20 10
Again, from right angled triangle ADB,

tan2θ = 3x + 2x 3x
20 C
2x
or, 2 tanθ = 5x
1 – tan2θ 20 B

2. x
10 x
or, 1 – x2 = 4

1 100 1 60°
5 100 4 D 20m
or, × 100 – x2 =

or, 80 = 100 – x2

or, x2 = 20

 x = 2 5m

 AB = 5x = 5.2 5 = 10 5 = 22.36m

12. A man 1.75m stands at a distance of 8.5m from a lamp post and it is observed that his
shadow 3.5m long. Find the height of the lamp post.

Solution
Let, PQ the height of the lamp post and MN the height of the man.

MN = 1.75m

RS = 3.5m, the shadow of the man. R

From the right angled triangle MRN,

Let, θ = MRN

tanθ = MN = 1.75 = 1 M
RN 3.5 2

Again, from right angled triangle PRQ,

tanθ = PR 1.75m
RQ N 8.5m

or, 1 = PQ Rθ Q
2 3.5 + 8.5
3.5m
1 PQ
or, 2 = 12

 PQ = 6m

13. The angle of elevation of the top of a tower is 45° from a point 10m above the water
level of a lake. The angle of depression of its image in the lake from the same point is
60°. Find the height of the tower above the water level.

330 Vedanta Optional Mathematics Teacher's Guide ~ 10

Solution
Let, MN be the height of the tower from the water level of the lake.

P is the position of the observer which is 10m above the water level.

Let, M' be the image of M in water of the lake

PQNR is a rectangle. M

PQ = RN = 10m xm
R
Let, MR = xm 10m
N
NM' = ym
ym
From the right angled triangle MPR, P 45° M'
10m 60°
tan45° = MR
PR Q

or, MR = PR

or, PR = x R
Again, from right angled triangle PRM',

tan60° = RM'
PR

or, 3 = 10 + y
x

or, 3x = 10 + y ... ... ... (i)

But by definition of reflection, MN = NM'

i.e. image distance = object distance

So, we can write x + 10 = y ... ... ... (ii)

From the equation (i) and (ii), we get

x + 10 = 3x – 10

or, 20 = ( 3 – 1)x

or, x= 20 1 × 3+1 = 20( 3 + 1) = 10 × 2.732 = 27.32m
3– 3+1 2

Now, the height of the tower above the level of water

= MN

= x + 10

= 27.32 + 10

= 37.32m

14. The angle of elevation of an aeroplane from a point in the ground is 45°. After 15 seconds
of flight the angle of elevation changes to 30°. If the aeroplane is flying horizontally at
a height of 4000m, in the same direction, find the speed of aeroplane.

Vedanta Optional Mathematics Teacher's Guide ~ 10 331

Solution
Let, PQ = RS = 4000m, the constant height of the aeroplane.

P the starting position of the aeroplane when the observer saw it.

PTQ = 45°

Let, R be the position of the aeroplane after 15 second RTS = 30°

Now, from the right angled triangle PTQ, P R
4000m 4000m
tan45° = PQ S
TQ

or, 1 = PQ
TQ

or, PQ = TQ = 4000m

Again, from right angled triangle RTS, 45°
30°
tan30° = RS T
TS
4000 Q
or, 1 = TQ + QS
3

or, 4000 3 = 4000 + QS

or, 4000( 3 – 1) = QS

or, QS = 4000( 3 – 1)

= 4000(1.732 – 1)

= 4000 × 0.732

= 2928m
But PR = QS = 2928m
The distance covered by the aeroplane in 15 second is 2928m.

Hence, speed of the aeroplane (v) = distance
time

= 2928m
15s

= 195.2ms–1

15. From the foot of mountain, the elevation of its summit is 45°. After going up at a distance
of 1km towards the top of the mountain at an angle of 30°, the elevation changes to 60°.
Find the height of the mountain.

Hints:

Solution
PQ = 1km, initially climbed part.

Again, from right angled triangle PQR,

tan30° = QR
PQ

332 Vedanta Optional Mathematics Teacher's Guide ~ 10

M

or, 1 = QR
2 1
x
 QR = 0.5km S
N
QRNS is a rectangle  QR = SN = 0.5

But MPN = 45°, ∆MPN is an isoceles right angled triangle Q 60°
y
MN = PN
y
Also, cos30° = PR 1 km R
PQ

3 45° 30°
or, PR = 2 = 0.8660 km. P
x
tan60° = y

or, x = 3y = 1.7321y

Again from right angled trinalge PMN

tan45° = MN
PN
or, PN = MN

or, PR + RN = MS + SN

or, 0.8660 + y = x + 0.5

or, 0.8660 + y = 1.7321y + 0.5

or, y = 0.3660 = 0.5
0.7321

and x = 1.7321 × 0.5 = 0.866

Total height of the mountain = 0.5 + 0.866 = 1.366km = 1366m.

Questions for practice

1. A chimney is 10 3m. high. Find the angle of elevation of its top from a point 100m away from its foot.
2. From the top of a building 45m high, a man observes the angle of depression of a stationary bus is

30˚. Find the distance of the bus from the building.
3. The angle of elevation of a tower was observed to be 60˚ from a point. on walking 200m away

from the point, it was found to be 30˚. Find the height of the tower.
4. A boy standing between two pillars of equal height observes the angle of elevation of the top of a

pillar to be 30˚. Approaching 15m, towards anyone of the pillars the angle of elevation is 45˚. Find
the height of the pillars and distance between them.
5. A cloud is observed above a river from opposite banks at angles of elevation are found to be 45˚
and 60˚. The cloud is vertically above the line joining the points of observation and the river is
80m wide. Find the height of the cloud.
7. The angles of elevation of the top of a tower from two points 'a' and 'b' m from the base in the
name straight line with it are complementary. Prove that the height of the tower is ab m.
8. The angles of depression and elevation of the top and the bottom of a tele-communication
tower are observed respectively 45˚ and 30˚ from the root of the house. The height of the
house is 40m. Find the height of the tower and house and tower are on the same plans.

Vedanta Optional Mathematics Teacher's Guide ~ 10 333

UNIT

ten Vectors

Scalar Product of two vectors

Estimated Teaching Periods : 5

1. Objectives

Level Objectives
Knowledge (K)
To define dot product of two vectors.
Understanding (U) Tell meaning of →a.→b = |→a| |→b| cosθ

Skill/Application (S/A) To establish relation cosθ = →a.→b
Higher Ability (HA) |→a| |→b|

To state conditions of perpendicularity and parallelism of two vectors.

To solve problems involving dot product of two vectors.

To solve harder problems of dot product of vectors.

2. Teaching Materials
Graph papers, list of formula of scaler product of two vectors.

3. Teaching Learning Strategies
→ Review the concept of vectors and scalars.
→ Take two position vectors O→A = (4, 5) and O→B = (–5, 4) plot them in a graph papers.

Find angle between them. Multiply (4, 5) and (–5, 4) to get 4 × –5 + 5 × 4 = 0. Draw the
conclusion from it.
→ Define dot product of two vectors →a and →b as →a.→b = |→a| |→b| cosθ with figure.

→ Dcoisscθu=ss|t→ao→a|f.i|→bn→bd|angle between two vectors by using formula

→ If →a = x1→i + y1→j and →b = x2→i + y2→j and show that →a.→b = x1x2 + y1y2.

→ Discuss the conditions of tpheerpmeneadnicinuglaorift→ya.a→bn=d p0aarnaldle→alis=mko→bf.two vectors by using above
formula and also discuss
→ Also show →i .→j = 0 and →i .→i = 1 as review.

→ Discuss the some properties of dot product of two vectors.

List of Formula
1. →a.→b = |→a| |→b| cosθ
→a.→b
or, cosθ = |→a| |→b|

2. If →a = x1→i + y1→j and →b = x2→i + y2→j then →a.→b = x1x2 + y1y2.

334 Vedanta Optional Mathematics Teacher's Guide ~ 10

3 4.. iII.ffe→→aa. .=→bco=ms9→b00,°toh=re→bn|→=at→ah|.ke|→b→a→bv,e|wcth⇒oerrs→ae→.a→bma=nadn0d→b karaerepsecrpaleanrds,icthuelanr to each other. and →b are parallel to
each other. two vectors →a

5. (→a + →b)2 = a2 + 2→a.→b + b2 = a2 + b2 + ab cosθ.

Some solved problems

1. Find the dot product of :

a. →a = 3 and →b = 2
4 1

Solution

Here, →a = 3 and →b = 2
4 1

→a . →b = 3 . 2
4 1

= (3, 4) . (2, 1)

=3×2+4×1
= 6 + 4 = 10

b. →a = →i + 2→j and →b = 3→i – →j

Solution
Here, →a = →i + 2→j and →b = 3→i – →j

→a . →b = (→i + 2→j ) . (3→i – →j )

= 1 × 3 + 2 × (–1)

=3–2 =1

2. If |→a| = 2, |→b| = 3, angle between them is 45°, find →a . →b.

Solution
Here, |→a| = 2, |→b| = 3, θ = 45°
→a . →b = |→a| |→b| cosθ

= 2 × 3 × cos45°

=6× 1
2

= 6 =3× 2× 2 =3 2
2 2

Vedanta Optional Mathematics Teacher's Guide ~ 10 335

3. Show that →p = 3 and →q = –4 are perpendicular to each other.
4 3
Solution

Here, →p = 3 and →q = –4
4 3

→p . →q = 3 . –4
4 3

= 3 × (–4) + 4 × 3

= –12 + 12 = 0

Since →p . →q = 0, hence →p and →q are perpendicular to each other. Proved

Alternatively
Let, θ be the angle between →p and →q, then

cosθ =|→p→p|.|→q→q|

→p . →q = 3 . –4
4 3

= –12 + 12 = 0

|→p| = 32 + 42 = 5

|→q| = (–4)2 + 32 = 5

Now, cosθ = 0 =0 = cos90°
5×5

 θ = 90°

This shows that →p and →q are perpendicular to each other.

4. Find the value of k, if the →a = 4→i + k→j and →b = 3→i – 6→j are perpendicular to each other.

Solution
Here, →a = 4→i + k→j , →b = 3→i – 6→j
Since →a and →b are perpendicular to each other, their dot product is zero.
i.e. →a . →b = 0
or, (4→i + k→j ) . (3→i – 6→j ) = 0
or, 4 × 3 + k(–6) = 0
or, 12 – 6k = 0
 k = 2

5. In ∆PQR, if →PQ = 4→j – 3→i , →PR = →j – 7→i , prove that ∆PQR is an isoceles right angled triangle.

336 Vedanta Optional Mathematics Teacher's Guide ~ 10

Solution Q –3, 4)
Let, P be taken as origin,
then O→Q = →PQ = 4→j – 3→i = (–3, 4)
O→R = →PR = →j – 7→i = (–7, 1)

Q→R = xy22 – xy11 4j→– 3i→
–

= –7 + 3
1–4

= –4 = (–4, –3) P j→– 7→i R(–7, 2)
–3

Now, Q→R . →PQ = (–4, –3) . (–3, 4) = 12 – 12 = 0

 PQR = 90° 16 + 9 = 5
Also, |→PQ| = (–3)2 + 42 = 5
|Q→R| = (–4)2 + (–3)2 =
 |→PQ| = |Q→R|

and PQR = 90°

Hence ∆PQR is an isoceles right angled triangle.

6. In ∆PQR, if →PQ = 5→i – 9→j and Q→R = 4→i + 14→j , prove that ∆PQR is an isoceles right

angled triangle.

Solution 25 + 81 = 106 units
212 units
Here, →PQ = 5→i – 9→j , |→PQ| = 52 + (–9)2 = Q
–5, 9)
or, →QP = –5→i + 9→j

and Q→R = 4→i + 14→j , |Q→R| = 42 + 142 =

Let, Q be origin O(0, 0) P
→QP = →OP = (–5, 9)
Q→R = O→R = (4, 14)
Now, →PR = (4 + 5, 14 – 9) = (9, 5)

|→PR | = 92 + 52 = 106 units R 4, 14)

→QP . Q→R = (–5, 9) . (9, 5) = –45 + 45 = 0

Vedanta Optional Mathematics Teacher's Guide ~ 10 337

 PQR = 90° and |→PQ| = |→PR |
Hence ∆PQR is an right angled triangle. proved

7. a. Find the angle between →a = 4→i – 3→j and x-axis.

Solution

We know that a unit vector along x-axis is →i .

Let, O→B = →i

and →a = O→A = 4→i – 3→j
Let, θ be the angle between O→B and O→A.

cosθ = O→A . O→B = (4→i – 3→j ) . →i = 4
|O→A| |O→B| 42 + (–3)2 . 12 5

 θ = cos–1 4 .
5

b. Find the angle between 2→i + →j and y-axis.

Solution
Let, unit vector along y-axis be O→B = →j

and given vector O→A = 2→i + →j

Let, θ be the angle between O→A and O→B, then

cosθ = O→A . O→B = (2→i + →j ) . →j = 1
|O→A| |O→B| 22 + 1 . 1 5

 θ = 63.43°.

8. Show that the angle between two vector →a and →c is 90°, if →a + →b + →c = (0, 0).

Solution
Let, |→a| = 3, |→b| = 5, |→c| = 4
Here, →a + →b + →c = (0, 0)
or, →a + →c = – →b
Squaring on both sides, we get,
(→a + →c)2 = b2
or, a2 + 2→a . →c + c2 = b2
or, 9 + 2 |→a| |→c| cosθ + 16 = 25
(Where θ is the angle between →a and →c.)

338 Vedanta Optional Mathematics Teacher's Guide ~ 10

2 × 3 × 4 cosθ = 0
or, cosθ = 0
 cosθ = cos90°
θ = 90°. proved

9. If →a + →b + →c = O (0, 0) and |→a| = 3, |→b| = 5 and |→c| = 7, show that the angle between
→a and →b is 60°.

Solution
Here, |→a| = 3, |→b| = 5, |→c| = 7

→a + →b + →c = O(0, 0)

or, →a + →b = –→c
Squaring on both sides, we get,

(→a + →b)2 = (–→c)2

or, a2 + 2→a . →b + b2 = c2

or, 9 + 2 |→a| |→b| cosθ + 25 = 49
Where θ is the angle between →a and →b.

2 × 3 × 5 cosθ = 49 – 34

or, cosθ = 15 = 1 = cos60°
30 2
 θ = 60°. proved

10. Find the angle between →a and →b, if →a + →b + →c = O (0, 0), |→a| = 6, |→b| = 7 and →c = 127.

Solution 106
Here, |→a| = 6, |→b| = 7 and →c =
Now, →a + →b + →c = O (0, 0)

(→a + →b) = –→c

Squaring on both sides, we get,

a2 + b2 + 2→a . →b = c2

or, 36 + 49 + 2|→a| |→b| cosθ = 127

or, 2 × 6 × 7 cosθ = 42

or, cosθ = 1 = cos60°
2

 θ = 60°.

Vedanta Optional Mathematics Teacher's Guide ~ 10 339

11. If →a + →b and 2→a – →b are orthogonal vectors, →a and →b are unit vectors, find the angle
between →a and →b.

Solution
Here, →a + →b and 2→a – →b are orthogonal vectors, (→a + →b) . (2→a – →b) = O(0, 0) and
|→a| = 1, |→b| = 1
Now, (→a + →b) . (2→a – →b) = 0
or, 2a2 + 2→a . →b – →a . →b – b2 = 0
or, 2 + |→a| |→b| cosθ – 1 = 0
or, cosθ = –1
or, cosθ = cos180°
 θ = 180°.

12. If |→a – 3→b| = |→a + 3→b|, prove that →a and →b are orthogonal vectors.
Solution
Here, |→a – 3→b| = |→a + 3→b|

Squaring on both sides, we get,
(→a – 3→b)2 = (→a + 3→b)2
or, a2 + 9b2 – 2→a . 3→b = a2 + 9b2 + 2→a . 3→b
or, –6→a . →b = 6→a . →b
or, 12→a . →b = 0
or, →a . →b = 0
Hence, →a and →b are orthogonal vectors.

Questions for practice

1. If |→a| = 4, |→b| = 5 and →a . →b = 10, find the angle between →a and →b.
2. If →p + →q + →r = 0, |→p| = 6, |→q| = 7 and |→r | = 127, find the angle between →p and →q.
3. If →x = 3→i + m→j and →x = 4→i – 2→j are perpendicular to each other, find the value of m.
4. In ∆ABC if A→B = 3→i + 4→j and B→C = →i – 7→j , show that the triangle ABC is a right

angled triangle.
5. Prove that →p = 3→i + 5→j , →q = 5→i + 3→j and →r = 8→i – 2→j form an isosceles right

angled triangle.

340 Vedanta Optional Mathematics Teacher's Guide ~ 10

Vector Geometry Estimated Periods : 13

1. Objectives

Knowledge (K) To state triangle law of vector addition.
Understanding To state mid point theorem, section formula, centroid formula.
To state vector geometry theorems : mid point theorem, section
(U) formula, centroid formula, theorems related to triangles, theorems on
Skill/Application quadrilateral, semi-circle.
To prove vector geometry theorems and problems based on the
(S/A) theorems.
To prove the following vector geometry problems.
Higher Ability – the diagonals of a rectangle are equal.
(HA) – the diagonals of a parm bisect each other.
– the diagonals of a rhombus bisec each other at right angle.
– the mid point of the hypotenuse of a right angled triangle is at

equidistance from its vertices.

2. Required Teaching Materials
Chart paper with statements of vector geometry to prescribed course by CDC.

3. Teaching Learning Strategies

→ Review the triangle law, parallelogram law of vector geometry.

→ State and prove each of theorems stated as in above objectives.

→ Show relations of each of above stated theorems with same theorems on plane geometry.

→ After proving each of above theorems let the students do the some theorems with figure
labelled differnty for further practice for them.

List of Formula B

1. Mid point formula

m→ = O→M = 1 (→a + →b) →b m→ M
2

→a A B
n
2. Section formula →b P
→a m
→OP = m→b + n→a, (internal division) PO A
m+n B
→b A 341

O

3. Section formula,

→OP = m→b – n→a
m–n

Vedanta Optional Mathematics Teacher's Guide ~ 10

4. Centroid of triangle
→g = →a + →b + →c

3
(Study all theorem of vector geometry from text book Vedanta Excel In opt. Mathematics-10)

Some solved problems

1. If the position vector of the mid point of the line segment AB is (3→i – 2→j ), where the
position vector of B is (5→i + 2→j ). Find the position vector of A.

Solution

Let, O→M be the position vector of the mid point of AB. O→M = 3→i – 2→j B
M
then O→B = 5→i + 2→j 5→i + 2→j A

Now, O→M = 1 ( O→A + O→B )
or, 2 O→A + O→B

2OM =

or, O→A = 2O→M – O→B O

= 2(3→i – 2→j ) – (5→i + 2→j )

= 6→i – 4→j – 5→i – 2→j
 O→A = →i – 6→j

2. The position vectors of the points A and B of the line segment AB are respectively
→a = 3→i + 4→j and →b =→i – 2→j . If C divides AB in the ratio of 3:2 internally, find the
position vector of C.

Solution O B
Here, O→A = →a = 3→i + 4→j
O→B = →i – 2→j 2
m:n = 3:2 P
i.e. m = 3, n = 2 3
A
C divides AB in ratio of 3:2 internally.
Now, O→C = m→b + n→a

m+n

= 3(→i – 2→j ) + 2(3→i + 4→j )
3+2

= 9→i + 2→j
5

= 95→i + 25→j

342 Vedanta Optional Mathematics Teacher's Guide ~ 10

3. The position vectors of A and B are respectively (→i + →j ) and (3→i + 5→j ). Find the position
vector of P which divides AB in ratio of 2:1 externally.

Solution C
Here, O→A = →i + →j

O→B = 3→i + 5→j

C divides AB in ratio of 2:1 externally. →b 3→i + 5→j
O i→+ j→
i.e. m = 2, n = 1 B
then, O→C = m→b – n→a A

m–n

= 2(3→i + 5→j ) – 1(→i + →j )
2–1

= 6→i + 10→j – →i – →j
1

= 5→i + 9→j

4. Find the position vector of centroid of ∆PQR whose position vector of vertices are
respectively (3→i + 4→j ), (4→i + 5→j ) and (5→i + 6→j ).

Solution
Let, →OP = →p = 3→i + 4→j
O→Q = →q = 4→i + 5→j
O→R = →r = 5→i + 6→j
Let, O→G = →g be the position vector of the centroid of ∆PQR, then

O→G = →g = →p + →q + →r
3

= 3→i + 4→j + 4→i + 5→j + 5→i + 6→j
3

= 12→i + 15→j
3

= 4→i + 5→j
5. In ∆LMN, OL = 4→i – 5→j , O→M = 6→i + 4→j and the position vector of centroid G,

O→G = 2→i + →j . Find the O→N.

Solution
Let, OL = 4→i – 5→j

Vedanta Optional Mathematics Teacher's Guide ~ 10 343

O→M = 6→i + 4→j

O→G = 2→i + →j

Using formula, O→G = →a + →b + →c
3
O→G = O→M + O→N + →OL
3
or, 3 O→G = 6→i + 4→j + O→N + 4→i – 5→j

or, 3(2→i + →j ) = 10→i – →j + O→N

 O→N = –4→i + 4→j

5. If the position vector of A and B are respectively →a and →b. Find the position vector of C
in AB produced such that →AC = 3 →BC .

Solution C
Here, O→A = →a, O→B = →b B
A
C is a point on AB produced and →AC = 3 →BC

i.e. →AC = 1
→BC 3

It means that C divides AB in ratio of 3:1 externally. O

Hence, O→C = m→b – n→a
m–n

Where, m = 3, n = 1

O→C = 3→b – →a = 3→b – →a
Alernatively, 3 – 1 2

Here, →AC = 3 →BC

or, O→C – O→A = 3 ( O→C – O→B )

or, O→C – →a = 3 O→C – 3→b

or, 3→b – →a = 2 O→C

 O→C = 3→b – →a.
2

= →c, find the vector →PQ and show that →PQ // O→B .

Solution
Here, OABC is a parallelogram, CP:PO = CQ:QB = 1:3

344 Vedanta Optional Mathematics Teacher's Guide ~ 10

O→A = →a, O→C = →c 3 B
A
or, C→Q = 1 →CB, →PC = 1 O→C C1 Q
4 4 O

Now, →PQ = →PC + C→Q 1

= 1 O→C + 1 →CB P
4 4 3

= 1 ( O→C + →CB)
4

= 1 O→B ( O→C + →CB = O→B )
4

 →PQ = 1 O→B
4

and →PQ // O→B . proved

7. In the given figure, | →CB | = 32, then show that : 2 →AB + →AC = 3 A→D .
| →CD |
Solution

Here, CB:CD = 3:2

It shows that BC = 3 parts, CD = 2 parts and BD = 3 – 2 = 1 part A

It means that D divides BC in ratio of 1:2 internally,

m = 1, n = 2

Now, AD = m→b + n→a
2+1

where, →b = →AC , →a = →AB

or, A→D = 1 . →AC + 2 . →AB B DC
3

 A→D = 2 →AB + →AC . Proved

8. In ∆ABC, the medians AD, BE and CF are drawn from the vertices A, B and C respectively.
G is centroid, then prove that:

i) A→D + →BE + →CF = O

ii) G→A + →GB + →GC = O

Solution
i) To prove: A→D + →BE + →CF = O

Vedanta Optional Mathematics Teacher's Guide ~ 10 345

By using mid point theorem, we get

A→D = +21 (→→ABEB + →AC ), →BE = 1 ( →BA + →BC ), →CF = 1 ( →CB + →CA )
LHS = A→D + →CF 2 2 A

= 1 ( →AB + →AC ) + 1 ( →BA + →BC ) + 1 ( →CB + →CA )
2 2 2

= 1 ( →AB + →AC + →BA + →BC + →CB + →CA )
2

= 1 ( →AB + →BA + →AC + →CA + →BC + →CB ) F E
2
G
( →AB = – →BA )

= 1 . 0
2

=0 B DC

ii) To prove: G→A + →GB + →GC = O(0, 0)

Since, AD, BE and CF are the medians of triangle ABC and G is the point of intersection

of medians. i.e. centroid. G divides the medians in ratio of 2:1.
LHS = G→A + →GB + →GC

=  2 A→D + 2 →BE + 2 →CF 
3 3 3

= 2 ( →AB + →AC + →BA + →BC + →CB + →CA )
3

= 2 . 0 (as in (i))
3

= 0 Proved

9. In the figure, ∆ABC is an isoceles triangle AD is median, then show that : A→D . →BC = 0.

or

In an isoceles triangle, the median drawn from the vertex to the base is perpendicular
to the base.

Solution A
Here, AD is the median from vertex A to the base BC of ∆ABC.

∆ABC is an isoceles triangle i.e. AB = AC

Now, by mid point theorem,

A→D = 1 (→AB + →AC )
2
B DC

346 Vedanta Optional Mathematics Teacher's Guide ~ 10

Also, →BC = →BA + →AC = – →AB + →AC

Taking the dot product of A→D and →BC , we get

A→D . →BC = 1 (→AB + →AC ) . (– →AB + →AC )
2

= 1 (AC2 – AB2)
2

= 1 . 0  | →AC | = |→AB|
2

 A→D . →BC = 0

As the dot product of A→D and →BC is zero, AD is perpendicular to BC. Proved

10. If the diagonals of a quadrilateral bisect each other, prove by vector method that it is a
parallelogram.

Solution
Let, ABCD is a quadrilateral in which the diagonals AC and BD bisect at O.

Then, we can write, DC
A→O = O→C ... ... ... (i)

O→D = B→O ... ... ... (ii) O

Adding (i) and (ii), we get
A→O + O→D = O→C + B→O

or, A→D = →BC , By using triangle law of vector addition ( A→O +A O→D = A→D , B→O + O→C = →BCB)

⸫ A→D // →BC .
Also, we can show that →DC = →AB and →DC // →AB as above.

Hence, ABCD is a parallelogarm. Proved

11. If a line is drawn from the centre of a circle to the mid point of a chord, prove by vector
method that the line is perpendicular to the chord.

Solution
Let, O be the centre of the circle and PQ be a chord.
M is the mid point of PQ.
Join OP and OQ.
1. Since M is the mid point of PQ. We have by mid point theorem.

O→M = 1 ( →OP + O→Q )
2

Vedanta Optional Mathematics Teacher's Guide ~ 10 347

2. By using triangle law of vector addition

→PQ = →PO + O→Q

= (– →OP ) + O→Q O

3. Taking the dot product of O→M and →PQ , we get,

O→M . →PQ = 1 ( →OP + O→Q ) . (– →OP + O→Q )
2
1
= 2 (OQ2 – OP2) PMQ

 | →OP | = |O→Q|, radii of same circle

= 1 . 0
2
 O→M . →PQ = 0

Hence, OM is perpendicular to PQ. Proved

12. In the given figure, PQRS is a trapezium where PS//QR. M and N are the mid points of

→PQ and →SR respectively. Prove vectorically that:

i) M→N 1 ( →PS Q→R ) P S
2
= +

ii) M→N // Q→R M N
Solution

Here, In the figure PQRS is a trapezium.

M and N are the mid points of PQ and SR.

1. M→N = M→P + →PS + →SN , by polygon lawQof vector addition. R

2. M→N = M→Q + Q→R + R→N

3. 2M→N = M→P + →PS + →SN + M→Q + Q→R + R→N (adding (1) and (2)

= (M→P + M→Q) + ( →SN + R→N ) + ( →PS + Q→R )

= 0 + 0 + →PS + Q→R ( M→P = –M→Q, →SN = – R→N )

= →PS + Q→R

 M→N = 1 ( →PS + Q→R )
2

4. Let, →PS = k Q→R , where k is a scalar, →PS // Q→R

 M→N = 1 (k Q→R + Q→R )
2

348 Vedanta Optional Mathematics Teacher's Guide ~ 10

= 1 (k + 1) . Q→R
2
 M→N// Q→R

Also, M→N// Q→R . Proved

13. In the adjoining figure, PQRS is a parallelogram. G is the point of intersection of the

diagonals. If O is any point, prove that : O→G = 1 ( O→A + O→B + O→C + O→D ).
4
Solution

Here, In the given figure, PQRS is a parallelogram

and G is the point of intersection of diagonals AC and BD.

1. by using mid point theorem, we get D C

i) O→G = 122(O→O→GD + O→B ), G being mid point of BD.
or, = O→D + O→D

ii) O→G = 1 ( O→A + O→C ), G being mid point AC. O G
2 B
2 O→G = O→A + O→C
or,

2. Adding (i) and (ii) of (1), we get

4 O→G = O→A + O→C + O→D + O→B A

 O→G = 1 ( O→A + O→B + O→C + O→D ). Proved
4
14. ABCD is a parallelogram and O is the origin. If O→A = →a, O→B = →b and O→C = →c, find O→D

is terms of →a, →b and →c.

Solution D C
Here, In the given figure, ABCD is a parallelogram

O→A = →a, O→B = →b and O→C = →c →c
1. A→D = →BC , opposite sides of a parm.

2. By using triangle law of vector addition in ∆OAD.

O→D = O→A + A→D A B
→a
= O→A + →BC ( using (1)) O →b

= →a + ( O→C – O→B )
= →a + →c – →b
= →a – →b + →c.

Vedanta Optional Mathematics Teacher's Guide ~ 10 349

Questions for practice

1. If (3→i + 6→j ) and (5→i + 2→j ) are the position of vectors of the points P and Q respectively,
find the position vector of M which divides PQ internally in the ratio of 2:3.

2. In the given figure, PQR = 90°, prove vectorically that PR2 = PQ2 + QR2

3. If M is the mid point of AB with O→A = 3 and O→B = –1 B
5 –1 M

. Find the position vector of M.

O

4. In the given figure ABCDEF is a regular hexagon prove that E DA
→AB + →AC + A→D + →EA + →FA = 4 →AB C

F

5. In the given figure PQ = PR and QS = RS, then prove by AP B
vector method that PS is perpendicular to QR.

Q SR

6. In the given figure PQ = QR = RS = SP, prove by vector method PR is perpendicular to
QS. P S

Q R

350 Vedanta Optional Mathematics Teacher's Guide ~ 10