Index1. Theory2. Short Revision3. Exercise4. Assertion & Reason5. Que. from Compt. Exams

Subject : MathematicsTopic: Trigonometric Equation &Properties & Solution of Triangle

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1. Trigonometric Equation :An equation involving one or more trigonometric ratios of an unknown angle is called a trigonometricequation.

2. Solution of Trigonometric Equation :A solution of trigonometric equation is the value of the unknown angle that satisfies the equation.

e.g. if sinθ = 21

⇒ θ = 4π

, 4

3π,

49π

, 4

11π, ...........

Thus, the trigonometric equation may have infinite number of solutions (because of their periodic nature) andcan be classified as :(i) Principal solution (ii) General solution.

2.1 Principal solutions:The solutions of a trigonometric equation which lie in the interval[0, 2π) are called Principal solutions.

e.g Find the Principal solutions of the equation sinx = 21

.Solution.

∵ sinx = 21

∵ there exists two values

i.e. 6π

and 65π

which lie in [0, 2π) and whose sine is 21

∴ Principal solutions of the equation sinx = 21

are 6π

, 65π

Ans.2.2 General Solution :

The expression involving an integer 'n' which gives all solutions of a trigonometric equation is calledGeneral solution.General solution of some standard trigonometric equations are given below.

3. General Solution of Some Standard Trigonometric Equations :

(i) If sin θ = sin α ⇒ θ = n π + (−1)n α where α ∈

ππ−

2,

2, n ∈ Ι.

(ii) If cos θ = cos α ⇒ θ = 2 n π ± α where α ∈ [0, π], n ∈ Ι.

(iii) If tan θ = tan α ⇒ θ = n π + α where α ∈

ππ−

2,

2, n ∈ Ι.

(iv) If sin² θ = sin² α ⇒ θ = n π ± α, n ∈ Ι.

(v) If cos² θ = cos² α ⇒ θ = n π ± α, n ∈ Ι.

(vi) If tan² θ = tan² α ⇒ θ = n π ± α, n ∈ Ι. [ Note: α is called the principal angle ]Some Important deductions :

(i) sinθ = 0 ⇒ θ = nπ, n ∈ Ι

(ii) sinθ = 1 ⇒ θ = (4n + 1) 2π

, n ∈ Ι

(iii) sinθ = – 1 ⇒ θ = (4n – 1) 2π

, n ∈ Ι

(iv) cosθ = 0 ⇒ θ = (2n + 1) 2π

, n ∈ Ι(v) cosθ = 1 ⇒ θ = 2nπ, n ∈ Ι(vi) cosθ = – 1 ⇒ θ = (2n + 1)π, n ∈ Ι(vii) tanθ = 0 ⇒ θ = nπ, n ∈ Ι

Solved Example # 1

Solve sin θθθθθ = 23 .

Solution.

Trigonometric Equation

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∵ sin θ = 23

⇒ sinθ = sin3π

∴ θθθθθ = nπππππ + (– 1)n 3π

, n ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι Ans.

Solved Example # 2

Solve sec 2θθθθθ = – 32

Solution.

∵ sec 2θ = – 32

⇒ cos2θ = – 23

⇒ cos2θ = cos 65π

⇒ 2θ = 2nπ ± 6

5π , n ∈ Ι

⇒ θ θ θ θ θ = nπππππ ± 125π

, n ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι Ans.

Solved Example # 3Solve tanθθθθθ = 2

Solution.∵ tanθ = 2 ............(i)Let 2 = tanα⇒ tanθ = tanα⇒⇒⇒⇒⇒ θ θ θ θ θ = nπππππ + ααααα, where ααααα = tan–1(2), n ∈ Ι ∈ Ι ∈ Ι ∈ Ι ∈ Ι

Self Practice Problems:1. Solve cotθ = – 1

2. Solve cos3θ = – 21

Ans. (1) θ = nπ – 4π

, n ∈ Ι (2) 3n2 π

± 92π

, n ∈ Ι

Solved Example # 4

Solve cos2θθθθθ = 21

Solution.

∵ cos2θ = 21

⇒ cos2θ = 2

21

⇒ cos2θ = cos2

4π

⇒ θθθθθ = nπππππ ± 4π

, n ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι Ans.Solved Example # 5

Solve 4 tan2θθθθθ = 3sec2θθθθθ

Solution.∵ 4 tan2θ = 3sec2θ .............(i)

For equation (i) to be defined θ ≠ (2n + 1) 2π

, n ∈ Ι∵ equation (i) can be written as:

θθ

2

2

cossin4

= θ2cos

3∵ θ ≠ (2n + 1)

2π

, n ∈ Ι∴ cos2θ ≠ 0

⇒ 4 sin2θ = 3

⇒ sin2θ =

2

23

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9⇒ sin2θ = sin2 3π

⇒ θθθθθ = nπππππ ± 3π

, n ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι Ans.

Self Practice Problems :

1. Solve 7cos2θ + 3 sin2θ = 4.

2. Solve 2 sin2x + sin22x = 2

Ans. (1) nπ ± 3π

, n ∈ Ι (2) (2n + 1) 2π

, n ∈ Ι or nπ ± 4π

, n ∈ Ι

Types of Trigonometric Equations :

Type -1

Trigonometric equations which can be solved by use of factorization.

Solved Example # 6

Solve (2sinx – cosx) (1 + cosx) = sin2x.

Solution.∵ (2sinx – cosx) (1 + cosx) = sin2x⇒ (2sinx – cosx) (1 + cosx) – sin2x = 0⇒ (2sinx – cosx) (1 + cosx) – (1 – cosx) (1 + cosx) = 0⇒ (1 + cosx) (2sinx – 1) = 0⇒ 1 + cosx = 0 or 2sinx – 1 = 0

⇒ cosx = – 1 or sinx = 21

⇒ x = (2n + 1)π, n ∈ Ι or sin x = sin 6π

⇒ x = nπ + (– 1)n 6π

, n ∈ Ι∴ Solution of given equation is

(2n + 1)πππππ, n ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι or nπππππ + (–1)n 6π

, n ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι Ans.Self Practice Problems :

1. Solve cos3x + cos2x – 4cos2

2x

= 0

2. Solve cot2θ + 3cosecθ + 3 = 0

Ans. (1) (2n + 1)π, n ∈ Ι

(2) 2nπ –2π , n ∈ Ι or nπ + (–1)n + 1

6π , n ∈ Ι

Type - 2

Trigonometric equations which can be solved by reducing them in quadratic equations.

Solved Example # 7Solve 2 cos2x + 4cosx = 3sin2x

Solution.∵ 2cos2x + 4cosx – 3sin2x = 0⇒ 2cos2x + 4cosx – 3(1– cos2x) = 0⇒ 5cos2x + 4cosx – 3 = 0

⇒

+−−5

192xcos

−−−5

192xcos = 0 ........(ii)

∵ cosx ∈ [– 1, 1] ∀ x ∈ R

∴ cosx ≠ 5

192 −−

∴ equation (ii) will be true if

cosx = 5

192 +−

⇒ cosx = cosα, where cosα = 5

192 +−

⇒ x = 2nπππππ ± ααααα where α α α α α = cos–1

+−5

192, n ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι Ans.

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9Self Practice Problems : 1. Solve cos2θ – ( 2 + 1)

−θ

21cos = 0

2. Solve 4cosθ – 3secθ = tanθ

Ans. (1) 2nπ ± 3π , n ∈ Ι or 2nπ ±

4π

, n ∈ Ι

(2) nπ + (– 1)n α where α = sin–1

−−8

171, n ∈ Ι

or nπ + (–1)n β where β = sin–1

+−8

171, n ∈ Ι

Type - 3

Trigonometric equations which can be solved by transforming a sum or difference of trigonometricratios into their product.

Solved Example # 8Solve cos3x + sin2x – sin4x = 0

Solution.cos3x + sin2x – sin4x = 0 ⇒ cos3x + 2cos3x.sin(– x) = 0

⇒ cos3x – 2cos3x.sinx = 0 ⇒ cos3x (1 – 2sinx) = 0⇒ cos3x = 0 or 1 – 2sinx = 0

⇒ 3x = (2n + 1) 2π

, n ∈ Ι or sinx = 21

⇒ x = (2n + 1) 6π

, n ∈ Ι or x = nπ + (–1)n 6π

, n ∈ Ι∴ solution of given equation is

(2n + 1) 6π

, n ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι or nπππππ + (–1)n 6π

, n ∈ Ι ∈ Ι ∈ Ι ∈ Ι ∈ Ι Ans.

Self Practice Problems :

1. Solve sin7θ = sin3θ + sinθ

2. Solve 5sinx + 6sin2x +5sin3x + sin4x = 0

3. Solve cosθ – sin3θ = cos2θ

Ans. (1) 3nπ

, n ∈ Ι or2

nπ ±

12π

, n ∈ Ι

(2)2

nπ, n ∈ Ι or 2nπ ± 3

2π, n ∈ Ι

(3) 3n2 π

, n ∈ Ι or 2nπ – 2π

, n ∈ Ι or nπ + 4π

, n ∈ Ι

Type - 4Trigonometric equations which can be solved by transforming a product of trigonometric ratios into theirsum or difference.

Solved Example # 9Solve sin5x.cos3x = sin6x.cos2x

Solution.∵ sin5x.cos3x = sin6x.cos2x ⇒ 2sin5x.cos3x = 2sin6x.cos2x⇒ sin8x + sin2x = sin8x + sin4x ⇒ sin4x – sin2x = 0⇒ 2sin2x.cos2x – sin2x = 0 ⇒ sin2x (2cos2x – 1) = 0⇒ sin2x = 0 or 2cos2x – 1 = 0

⇒ 2x = nπ, n ∈ Ι or cos2x = 21

⇒ x = 2

nπ, n ∈ Ι or 2x = 2nπ ± 3

π, n ∈ Ι

⇒ x = nπ ± 6π

, n ∈ Ι∴ Solution of given equation is

2nπ , n ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι or nπππππ ± 6

π, n ∈ Ι ∈ Ι ∈ Ι ∈ Ι ∈ Ι Ans.

Type - 5

Trigonometric Equations of the form a sinx + b cosx = c, where a, b, c ∈ R, can be solved by dividing

both sides of the equation by 22 ba + .

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9Solved Example # 10Solve sinx + cosx = 2

Solution.∵ sinx + cosx = 2 ..........(i)Here a = 1, b = 1.∴ divide both sides of equation (i) by 2 , we get

sinx . 21

+ cosx.21

= 1

⇒ sinx.sin4π

+ cosx.cos4π

= 1

⇒ cos

π−

4x = 1

⇒ x – 4π

= 2nπ, n ∈ Ι

⇒ x = 2nπ + 4π

, n ∈ Ι∴ Solution of given equation is

2nπ π π π π + 4π

, n ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι Ans.

Note : Trigonometric equation of the form a sinx + b cosx = c can also be solved by changing sinx and cosxinto their corresponding tangent of half the angle.

Solved Example # 11

Solve 3cosx + 4sinx = 5Solution.

∵ 3cosx + 4sinx = 5 .........(i)

∵ cosx =

2xtan1

2xtan1

2

2

+

−& sinx =

2xtan1

2xtan2

2+

∴ equation (i) becomes

⇒ 3

+

−

2xtan12xtan1

2

2

+ 4

+2xtan1

2xtan2

2 = 5 ........(ii)

Let tan 2x

= t∴ equation (ii) becomes

3

+−

2

2

t1t1

+ 4

+ 2t1t2

= 5

⇒ 4t2 – 4t + 1 = 0⇒ (2t – 1)2 = 0

⇒ t = 21

∵ t = tan2x

⇒ tan 2x

= 21

⇒ tan 2x

= tanα, where tanα = 21

⇒2x

= nπ + α

⇒ x = 2nπππππ + 2ααααα where ααααα = tan–1

21

, n ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι Ans.

Self Practice Problems :

1. Solve 3 cosx + sinx = 2

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92. Solve sinx + tan 2x

= 0

Ans. (1) 2nπ + 6π

, n ∈ Ι (2) x = 2nπ, n ∈ Ι

Type - 6

Trigonometric equations of the form P(sinx ± cosx, sinx cosx) = 0, where p(y, z) is a polynomial, canbe solved by using the substitution sinx ± cosx = t.

Solved Example # 12

Solve sinx + cosx = 1 + sinx.cosx

Solution.∵ sinx + cosx = 1 + sinx.cosx ........(i)Let sinx + cosx = t⇒ sin2x + cos2x + 2 sinx.cosx = t2

⇒ sinx.cosx = 2

1t2 −

Now put sinx + cosx = t and sinx.cosx = 2

1t2 − in (i), we get

t = 1 + 2

1t2 −

⇒ t2 – 2t + 1 = 0⇒ t = 1 ∵ t = sinx + cosx⇒ sinx + cosx = 1 .........(ii)divide both sides of equation (ii) by 2 , we get

⇒ sinx.21

+ cosx.21

= 21

⇒ cos

π−

4x = cos

4π

⇒ x – 4π

= 2nπ ± 4π

(i) if we take positive sign, we get

x = 2nπππππ + 2π

, n ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι Ans.(ii) if we take negative sign, we get

x = 2nπππππ, n ∈ Ι ∈ Ι ∈ Ι ∈ Ι ∈ Ι Ans.

Self Practice Problems:

1. Solve sin2x + 5sinx + 1 + 5cosx = 0

2. Solve 3cosx + 3sinx + sin3x – cos3x = 0

3. Solve (1 – sin2x) (cosx – sinx) = 1 – 2sin2x.

Ans. (1) nπ – 4π

, n ∈ Ι (2) nπ – 4π

, n ∈ Ι

(3) 2nπ + 2π

, n ∈ Ι or 2nπ, n ∈ Ι or nπ + 4π

, n ∈ ΙType - 7

Trigonometric equations which can be solved by the use of boundness of the trigonometric ratiossinx and cosx.

Solved Example # 13

Solve sinx

− xsin2

4xcos + xcosxcos2

4xsin1

−+ = 0

Solution.

∵ sinx

− xsin2

4xcos + xcosxcos2

4xsin1

−+ = 0 .......(i)

⇒ sinx.cos4x

– 2sin2x + cosx + sin4x

.cosx – 2cos2x = 0

⇒

+ xcos.

4xsin

4xcos.xsin – 2 (sin2x + cos2x) + cosx = 0

⇒ sin4x5 + cosx = 2 ........(ii)

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9Now equation (ii) will be true if

sin4x5 = 1 and cosx = 1

⇒ 4x5 = 2nπ +

2π

, n ∈ Ι and x = 2mπ, m ∈ Ι

⇒ x = 5

)2n8( π+, n ∈ Ι ........(iii) and x = 2mπ, m ∈ Ι ........(iv)

Now to find general solution of equation (i)

5)2n8( π+

= 2mπ

⇒ 8n + 2 = 10m

⇒ n = 41m5 −

if m = 1 then n = 1if m = 5 then n = 6

......... ......... .........

......... ......... .........if m = 4p – 3, p ∈ Ι then n = 5p – 4, p ∈ Ι

∴ general solution of given equation can be obtained by substituting either m = 4p – 3 inequation (iv) or n = 5p – 4 in equation (iii)

∴ general solution of equation (i) is(8p – 6)πππππ, p ∈ Ι∈ Ι∈ Ι∈ Ι∈ Ι Ans.

Self Practice Problems :

1. Solve sin3x + cos2x = – 2

2. Solve 3xcosx5sin3 2 −− = 1 – sinx

Ans. (1) (4p – 3) 2π

, p ∈ Ι (2) 2mπ + 2π

, m ∈ Ι

Exercise -1Part : (A) Only one correct option

1. The solution set of the equation 4sinθ.cosθ – 2cosθ – 2 3 sinθ + 3 = 0 in the interval (0, 2π) is

(A)

ππ

47,

43

(B)

ππ

35,

3 (C)

ππππ

35,

3,,

43

(D)

πππ

611,

65,

6

2. All solutions of the equation, 2 sinθ + tanθ = 0 are obtained by taking all integral values of m and n in:

(A) 2nπ +23π

, n ∈ Ι (B) nπ or 2m π ±23π

where n, m ∈ Ι

(C) nπ or m π ± π3 where n, m ∈ Ι (D) nπ or 2m π ±

π3 where n, m ∈ Ι

3. If 20 sin2 θ + 21 cos θ − 24 = 0 &74π

< θ < 2π then the values of cotθ2 is:

(A) 3 (B) 153

(C) − 153

(D) − 3

4. The general solution of sinx + sin5x = sin2x + sin4x is:(A) 2 nπ ; n ∈ Ι (B) nπ ; n ∈ Ι (C) nπ/3 ; n ∈ Ι (D) 2 nπ/3 ; n ∈ Ι

5. A triangle ABC is such that sin(2A + B) = 21

. If A, B, C are in A.P. then the angle A, B, C are

respectively.

(A) 125π

, 4π

, 3π

(B) 4π

, 3π

, 125π

(C) 3π

, 4π

, 125π

(D) 3π

, 125π

, 4π

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96. The maximum value of 3sinx + 4cosx is(A) 3 (B) 4 (C) 5 (D) 7

7. If sin θ + 7 cos θ = 5, then tan (θ/2) is a root of the equation(A) x2 − 6x + 1 = 0 (B) 6x2 − x − 1 = 0 (C) 6x2 + x + 1 = 0 (D) x2 − x + 6 = 0

8.sin cossin cos

3 3θ θθ θ

−−

−cos

cot

θ

θ1 2+ − 2 tan θ cot θ = − 1 if:

(A) θ ∈ 02

, π

(B) θ ∈ ππ

2,

(C) θ ∈ π π, 32

(D) θ ∈32

2π π,

9. The number of integral values of a for which the equation cos 2x + a sin x = 2a − 7 possesses a solutionis(A) 2 (B) 3 (C) 4 (D) 5

10. The principal solution set of the equation, 2 cos x = 2 2 2+ sin x is

(A)

ππ

813,

8 (B)

ππ

813,

4 (C)

ππ

1013,

4 (D)

ππ

1013,

8

11. The number of all possible triplets (a1, a2, a3) such that : a1 + a2 cos 2x + a3 sin2x = 0 for all x is(A) 0 (B) 1 (C) 2 (D) infinite

12. If 2tan2x – 5 secx – 1 = 0 has 7 different roots in

π

2n,0 , n ∈ N, then greatest value of n is

(A) 8 (B) 10 (C) 13 (D) 15

13. The solution of |cosx| = cosx – 2sinx is

(A) x = nπ, n ∈ Ι (B) x = nπ + 4π

, n ∈ Ι

(C) x = nπ + (–1)n 4π

, n ∈ Ι (D) (2n + 1)π + 4π

, n ∈ Ι

14. The arithmetic mean of the roots of the equation 4cos3x – 4cos2x – cos(π + x) – 1 = 0 in the interval[0, 315] is equal to(A) 49π (B) 50π (C) 51π (D) 100π

15. Number of solutions of the equation cos 6x + tan2 x + cos 6x . tan2 x = 1 in the interval [0, 2π] is :(A) 4 (B) 5 (C) 6 (D) 7

Part : (B) May have more than one options correct

16. sinx − cos2x − 1 assumes the least value for the set of values of x given by:(A) x = nπ + (−1)n+1 (π/6) , n ∈ Ι (B) x = nπ + (−1)n (π/6) , n ∈ Ι(C) x = nπ + (−1)n (π/3), n ∈ Ι (D) x = nπ − (−1)n (π/6) , n ∈ Ι

17. cos4x cos8x − cos5x cos9x = 0 if(A) cos12x = cos 14 x (B) sin13 x = 0(C) sinx = 0 (D) cosx = 0

18. The equation 2sin2x

. cos2x + sin2x = 2 sin2x

. sin2x + cos2x has a root for which

(A) sin2x = 1 (B) sin2x = – 1 (C) cosx = 21

(D) cos2x = – 21

19. sin2x + 2 sin x cos x − 3cos2x = 0 if(A) tan x = 3 (B) tanx = − 1(C) x = nπ + π/4, n ∈ Ι (D) x = nπ + tan−1 (−3), n ∈ Ι

20. sin2x − cos 2x = 2 − sin 2x if(A) x = nπ/2, n ∈ Ι (B) x = nπ − π/2, n ∈ Ι(C) x = (2n + 1) π/2, n ∈ Ι (D) x = nπ + (−1)n sin−1 (2/3), n ∈ Ι

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29Exercise -21. Solve cotθ = tan8θ

2. Solve cot

2x

– cosec

2x

= cotx

3. Solve cot2θ +

+

313 cotθ + 1 = 0.

4. Solve cos2θ + 3 cosθ = 0.

5. Solve the equation: sin 6x = sin 4x − sin 2x .

6. Solve: cos θ + sin θ = cos 2 θ + sin 2 θ .

7. Solve 4 sin x . sin 2x . sin 4x = sin 3x .

8. Solve sin2nθ – sin2(n – 1)θ = sin2θ, where n is constant and n ≠ 0, 1

9. Solve tanθ + tan2θ + 3 tanθ tan2θ = 3 .

10. Solve: sin3 x cos 3 x + cos3 x sin 3 x + 0.375 = 0

11. Solve the equation, sin cos

sin

3 32 2

2

x x

x

−

+ =

cos x3 .

12. Solve the equation: sin 5x = 16 sin5 x .

13. If tan θ + sin φ = 32

& tan² θ + cos² φ =74

then find the general value of θ & φ .

14. Solve for x, the equation 13 18− tanx = 6 tan x − 3, where − 2 π < x < 2 π .

15. Find the general solution of sec 4 θ − sec 2 θ = 2 .

16. Solve the equation 3

2 sin x − cos x = cos² x .

17. Solve for x: 2 34

1 8 2 22sin sin . cosx x x+

= +π.

18. Solve the equation for 0 ≤ θ ≤ 2 π; ( )22cos32sin θ+θ − 5 = cos

θ−π 2

6 .

19. Solve: tan2 x . tan2 3 x . tan 4 x = tan2 x − tan2 3 x + tan 4 x .

20. Find the values of x, between 0 & 2 π, satisfying the equation; cos 3x + cos 2x = sin32x

+ sinx2

.

21. Solve: cos23x

cos 6 x = − 1 .

22. Solve the equation, sin2 4 x + cos2 x = 2 sin 4 x cos4 x .

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29

ANSWEREXERCISE 1

1. D 2. B 3. D 4. C 5. B 6. C 7. B 8. B 9. D 10. A

11. D 12. D 13. D 14. C 15. D 16. AD 17. ABC 18. ABCD 19. CD

20. BC

EXERCISE 2

1.

+

21n

9π

, n ∈ Ι 2. x = 4nπ ± 32π

, n ∈ Ι

3. θ = nπ – 3π

, n ∈ Ι or nπ – 6π

, n ∈ Ι 4. 2nπ ± α where α = cos–1

−4

317, n ∈ Ι

5.n π4

, n ∈ Ι or n π ±π6

, n ∈ Ι 6. 2 n π, n ∈ Ι or 2

3n π

+π6

, n ∈ Ι

7. x = n π, n ∈ Ι or n π3

±π9

, n ∈ Ι 8. mπ, m ∈ Ι or 1n

m−π

, m ∈ Ι or

+

21m

nπ

, m ∈ Ι

9.

+

31n 3

π, n ∈ Ι 10. x =

n π4

+ (− 1)n + 1 π24

, n ∈ Ι

11. x = (4 n + 1)π2

, n ∈ Ι 12. x = n π ; x = n π ±π6

, n ∈ Ι

13. θ = n π +π4

, φ = n π + (−1)nπ6

, n ∈ I 14. α − 2 π; α − π, α, α + π, where tan α = 23

15. 5n2 π

± 10π

or 2nπ ± 2π

, n ∈ Ι 16. x = (2 n + 1)π, , n ∈ Ι or 2 n π ±π3

, n ∈ Ι

17. (24" + 1) 12π

, " ∈ Ι or x = (24k – 7) 12π

, k ∈ Ι 18. θ =712

π, 19

12π

19. ( )2 14

n + π, k π, where n, k ∈ Ι 20. π π π π π7

57

97

137

, , , ,

21. φ 22. x = (2 n + 1)π2

, n∈ I

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29Properties & Solution of Triangle1. Sine Rule:

In any triangle ABC, the sines of the angles are proportional to the opposite sides i.e.

Csinc

Bsinb

Asina == .

Example : In any ∆ABC, prove that cba +

=

2Csin

2BAcos

−

.

Solution. ∵ We have to prove cba +

=

2Csin

2BAcos

−

.

∵ From sine rule, we know that

Asina

= Bsinb

= Csinc

= k (let)⇒ a = k sinA, b = k sinB and c = k sinC

∵ L.H.S. = cba +

= Csink)BsinA(sink +

=

2Ccos

2Csin

2BAcos

2BAsin

−

+

=

2Ccos

2Csin

2BAcos

2Ccos

−

=

2Csin

2BAcos

−

= R.H.S.Hence L.H.S. = R.H.S. Proved

Example : In any ∆ABC, prove that(b2 – c2) cot A + (c2 – a2) cot B + (a2 – b2) cot C = 0

Solution. ∵ We have to prove that(b2 – c2) cot A + (c2 – a2) cot B + (a2 – b2) cot C = 0

∵ from sine rule, we know thata = k sinA, b = k sinB and c = k sinC

∴ (b2 – c2) cot A = k2 (sin2B – sin2C) cot A∵ sin2B – sin2C = sin (B + C) sin (B – C)∴ (b2 – c2) cot A = k2 sin (B + C) sin (B – C) cotA ∵ B + C = π – A

∴ (b2 – c2) cot A = k2 sin A sin (B – C) AsinAcos

∵ cosA = – cos(B + C)= – k2 sin (B – C) cos (B + C)

= – 2

k2 [2sin (B – C) cos (B + C)]

⇒ (b2 – c2) cot A = – 2

k2 [sin 2B – sin 2C] ..........(i)

Similarly (c2 – a2) cot B = – 2

k2 [sin 2C – sin 2A] ..........(ii)

and (a2 – b2) cot C = – 2

k2 [sin 2A – sin 2B] ..........(iii)

adding equations (i), (ii) and (iii), we get(b2 – c2) cot A + (c2 – a2) cot B + (a2 – b2) cot C = 0 Hence Proved

Self Practice ProblemsIn any ∆ABC, prove that

1. a sin

+B

2A

= (b + c) sin

2A

.

2.CsinBsin

)CBsin(a2

+−

+ AsinCsin

)ACsin(b2

+−

+ BsinAsin)BAsin(c2

+−

= 0 3. bac− =

2Btan

2Atan

2Btan

2Atan

−

+.

2. Cosine Formula:(i) cos A =

b c ab c

2 2 2

2+ −

or a² = b² + c² − 2bc cos A = b2 + c2 + 2bc cos (B + C)

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29(ii) cos B = c a b

ca

2 2 2

2+ − (iii) cos C =

a b ca b

2 2 2

2+ −

Example : In a triangle ABC if a = 13, b = 8 and c = 7, then find sin A.

Solution. ∵ cosA = bc2

acb 222 −+ = 7.8.2

1694964 −+

⇒ cosA = – 21

⇒ A = 32π

∴ sinA = sin 32π

= 23 Ans.

*Example : In a ∆ABC, prove that a(b cos C – c cos B) = b2 – c2

Solution. ∵ We have to prove a (b cosC – c cosB) = b2 – c2.∵ from cosine rule we know that

cosC = ab2

cba 222 −+ & cos B =

ac2bca 222 −+

∴ L.H.S. = a

−+−

−+ac2

bcacab2

cbab222222

= 2

cba 222 −+ –

2)bca( 222 −+

= (b2 – c2) = R.H.S.Hence L.H.S. = R.H.S. Proved

Example : If in a ∆ABC, ∠ A = 60° then find the value of

++

cb

ca1

−+

ba

bc1 .

Solution. ∵ ∠ A = 60°

∵

++

cb

ca1

−+

ba

bc1 =

++

cbac

−+

bacb

= bc

a)cb( 22 −+

= bc

bc2)acb( 222 +−+

= bc

acb 222 −+ + 2

= 2

−+bc2

acb 222

+ 2

= 2cosA + 2 ∵ ∠ A = 60° ⇒ cos A = 21

∴

−+

++

ba

bc1

cb

ca1 = 3 Ans.

Self Practice Problems :1. The sides of a triangle ABC are a, b, 22 baba ++ , then prove that the greatest angle is 120°.

2. In a triangle ABC prove that a(cosB + cosC) = 2(b + c) sin2 2A

.

3. Projection Formula:(i) a = b cosC + c cosB (ii) b = c cosA + a cosC (iii) c = a cosB + b cosA

Example : In a triangle ABC prove that a(b cosC – c cosB) = b2 – c2

Solution. ∵ L.H.S. = a (b cosC – c cosB)= b (a cosC) – c (a cosB) ............(i)

∵ From projection rule, we know thatb = a cosC + c cosA ⇒ a cosC = b – c cosA

& c = a cosB + b cosA ⇒ a cosB = c – b cosAPut values of a cosC and a cosB in equation (i), we get

L.H.S. = b (b – ccos A) – c(c – b cos A)= b2 – bc cos A – c2 + bc cos A= b2 – c2 = R.H.S.

Hence L.H.S. = R.H.S. ProvedNote: We have also proved a (b cosC – ccosB) = b2 – c2 by using cosine – rule in solved *Example.Example : In a ∆ABC prove that (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c.

Solution. ∵ L.H.S. = (b + c) cos A (c + a) cos B + (a + B) cos C= b cos A + c cos A + c cos B + a cos B + a cos C + b cos C= (b cos A + a cos B) + (c cos A + a cos C) + (c cos B + b cos C)= a + b + c= R.H.S.

Hence L.H.S. = R.H.S. Proved

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Self Practice Problems

In a ∆ABC, prove that

1. 2

+

2Bcosc

2Ccosb 22 = a + b + c.

2. CcosBcos

= AcoscbAcosbc

−−

.

3. CcosbBcoscAcos

+ + AcoscCcosaBcos

+ + AcosbBcosaCcos

+ = abc2

cba 222 ++.

4. Napier’s Analogy - tangent rule:(i) tan

2CB −

= cbcb

+−

cot 2A

(ii) tan2

AC − = ac

ac+−

cotB2

(iii) tanA B−

2 =

a ba b

−+

cotC2

Example : Find the unknown elements of the ∆ABC in which a = 3 + 1, b = 3 – 1, C = 60°.

Solution. ∵ a = 3 + 1, b = 3 – 1, C = 60°∵ A + B + C = 180°∴ A + B = 120° .......(i)∵ From law of tangent, we know that

tan

−

2BA

= baba

+−

cot 2C

= )13()13()13()13(

−++−−+

cot 30°

= 32

2 cot 30°

⇒ tan

−

2BA

= 1

∴ 2

BA − = 4π

= 45°⇒ A – B = 90° .......(ii)From equation (i) and (ii), we get

A = 105° and B = 15°Now,

∵ From sine-rule, we know that Asina

= Bsinb

= Csinc

∴ c = AsinCsina

= °

°+105sin

60sin)13(

=

221323)13(

+

+∵ sin105° =

2213 +

⇒ c = 6∴ c = 6 , A = 105°, B = 15° Ans.

Self Practice Problem

1. In a ∆ABC if b = 3, c = 5 and cos (B – C) = 257

, then find the value of tan 2A

.

Ans. 31

2. If in a ∆ABC, we define x = tan

−

2CB

tan 2A

, y = tan

−

2AC

tan 2B

and z = tan

−

2BA

tan 2C

then show that x + y + z = – xyz.

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5. Trigonometric Functions of Half Angles:

(i) sinA2

=( ) ( )s b s c

b c− −

; sinB2

=( ) ( )s c s a

ca− −

; sinC2

= ( ) ( )s a s b

a b− −

(ii) cosA2

=s s a

b c( )−

; cosB2

=s s b

ca( )−

; cosC2

= s s c

a b( )−

(iii) tanA2

= ( ) ( )

( )s b s c

s s a− −

− =∆

s s a( )− where s =a b c+ +

2 is semi perimetre of triangle.

(iv) sin A = )cs)(bs)(as(sbc2 −−− =

bc2∆

6. Area of Triangle (∆)

∆ = 21

ab sin C = 21

bc sin A = 21

ca sin B = s s a s b s c( ) ( ) ( )− − −

Example : In a ∆ABC if a, b, c are in A.P. then find the value of tan 2A

. tan 2C

.

Solution. ∵ tan 2A

= )as(s −∆

and tan 2C

= )cs(s −∆

∴ tan 2A

. tan2C

= )cs)(as(s2

2

−−∆

∵ ∆2 = s (s – a) (s – b) (s – c)

∴ tan 2A

. tan 2C

= sbs −

= 1 – sb

........(i)∵ it is given that a, b, c are in A.P.⇒ 2b = a + c

∵ s = 2

cba ++ =

2b3

∴ sb

= 32

put in equation (i)

∴ tan 2A

. tan 2C

= 1 – 32

⇒ tan 2A

. tan 2C

= 31

Ans.

Example : In a ∆ABC if b sinC(b cosC + c cosB) = 42, then find the area of the ∆ABC.

Solution. ∵ b sinC (b cosC + c cosB) = 42 ........(i) given∵ From projection rule, we know that

a = b cosC + c cosB put in (i), we getab sinC = 42 ........(ii)

∵ ∆ = 21

ab sinC∴ ∆ = 21 sq. unit Ans.

Example : In any ∆ABC prove that (a + b + c)

+

2Btan

2Atan = 2c cot

2C

.

Solution. ∵ L.H.S. = (a + b + c)

+

2Btan

2Atan

∵ tan2A

= )as(s)cs)(bs(

−−−

and tan2B

= )bs(s)cs)(as(

−−−

∴ L.H.S. = (a + b + c)

−

−−+−

−−)bs(s

)cs)(as()as(s

)cs)(bs(

= 2s scs −

−−+

−−

bsas

asbs

= 2 )cs(s −

−−−+−

)bs)(as(asbs

∵ 2s= a + b + c

∴ 2s – b – a = c

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= 2 )cs(s −

−− )bs)(as(c

= 2c )bs)(as()cs(s−−

−∵ cot

2C

= )bs)(as()cs(s−−

−

= 2c cot2C

= R.H.S.Hence L.H.S. = R.H.S. Proved

7. m - n Rule:

(m + n) cot θ = −= −

m nn B m C

cot cotcot cot

α β

Example : If the median AD of a triangle ABC is perpendicular to AB, prove that tan A + 2tan B = 0.

Solution. From the figure, we see that θ = 90° + B (as θ is external angle of ∆ABD)

Now if we apply m-n rule in ∆ABC, we get(1 + 1) cot (90 + B) = 1. cot 90° – 1.cot (A – 90°)⇒ – 2 tan B = cot (90° – A)⇒ – 2 tan B = tan A⇒ tan A + 2 tan B = 0 Hence proved.

Example : The base of a triangle is divided into three equal parts. If t1, t2, t3 be the tangents of the anglessubtended by these parts at the opposite vertex, prove that

4

+ 2

2t11 =

+

21 t1

t1

+

32 t1

t1

.

Solution. Let point D and E divides the base BC into three equal parts i.e. BD = DE = DC = d (Let) andlet α, β and γ be the angles subtended by BD, DE and EC respectively at their opposite vertex.⇒ t1 = tanα, t2 = tanβ and t3 = tanγNow in ∆ABC∵ BE : EC = 2d : d = 2 : 1∴ from m-n rule, we get

(2 + 1) cotθ = 2 cot (α + β) – cotγ⇒ 3cotθ = 2 cot (α + β) – cotγ .........(i)again∵ in ∆ADC∵ DE : EC = x : x = 1 : 1∴ if we apply m-n rule in ∆ADC, we get

(1 + 1) cotθ = 1. cotβ – 1 cotγ2cotθ = cotβ – cotγ .........(ii)

from (i) and (ii), we get

θθ

cot2cot3

= γ−βγ−β+α

cotcotcot)cot(2

⇒ 3cotβ – 3cotγ = 4cot (α + β) – 2 cotγ⇒ 3cotβ – cotγ = 4 cot (α + β)

⇒ 3cotβ – cotγ = 4

α+β−βα

cotcot1cot.cot

⇒ 3cot2β + 3cotα cotβ – cotβ cotγ – cotα cotγ = 4 cotα cotβ – 4⇒ 4 + 3cot2β = cotα cotβ + cotβ cotγ + cotα cotγ⇒ 4 + 4cot2β = cotα cotβ + cotα cotγ + cotβ cotγ + cot2β⇒ 4(1 + cot2β) = (cotα + cotβ) (cotβ + cotγ)

⇒ 4

β+ 2tan

11 =

β

+α tan

1tan

1

γ

+β tan

1tan

1

⇒ 4

+ 2

2t11 =

+

21 t1

t1

+

32 t1

t1

Hence proved

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Self Practice Problems :

1. In a ∆ABC, the median to the side BC is of length 3611

1

− and it divides angle A into the angles of

30° and 45°. Prove that the side BC is of length 2 units.8. Radius of Circumcirlce :

R =Csin2

cBsin2

bAsin2

a == = ∆4cba

Example : In a ∆ABC prove that sinA + sinB + sinC = Rs

Solution. In a ∆ABC, we know that

Asina

= Bsinb

= Csinc

= 2R

∴ sin A =R2a

, sinB = R2b

and sinC = R2c

.

∴ sinA + sinB + sinC = R2

cba ++∵ a + b + c = 2s

= R2s2

⇒ sinA + sinB + sinC = Rs

.Example : In a ∆ABC if a = 13 cm, b = 14 cm and c = 15 cm, then find its circumradius.

Solution. ∵ R = ∆4

abc.......(i)

∵ ∆ = )cs)(bs)(as(s −−−

∵ s = 2

cba ++ = 21 cm

∴ ∆ = 6.7.8.21 = 222 3.4.7⇒ ∆ = 84 cm2

∴ R = 84.415.14.13

= 865

cm

∴ R = 865

cm.

Example : In a ∆ABC prove that s = 4R cos2A

. cos2B

. cos2C

.Solution. In a ∆ABC,

∵ cos2A

= bc)as(s −

, cos2B

= ca)bs(s −

and cos2C

= ab)cs(s −

and R = ∆4

abc

∵ R.H.S. = 4R cos2A

. cos2B

. cos2C

.

= ∆

abc. s 2)abc(

)cs)(bs)(as(s −−−∵ ∆ = )cs)(bs)(as(s −−−

= s= L.H.S.

Hence R.H.L = L.H.S. proved

Example : In a ∆ABC, prove that as1− + bs

1− + cs

1− – s

1 =

∆R4

.

Solution.as

1−

+ bs

1−

+ cs

1−

– s1

= ∆R4

∵ L.H.S. =

−+

− bs1

as1

+

−

− s1

cs1

= )bs)(as(bas2

−−−−

+ )cs(s)css(

−+−

∵ 2s = a + b + c

= )bs)(as(c

−− + )cs(sc−

= c

−−−−−+−

)cs)(bs)(as(s)bs)(as()cs(s

= c

∆+++−

2

2 ab)cba(ss2

∴ L.H.S. = c

∆+−

2

2 ab)s2(ss2 = 2

abc∆

= 2R4∆

∆ =

∆R4

∵ R = ∆4

abc

⇒ abc = 4R∆

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Self Practice Problems :In a ∆ABC, prove the followings :

1. a cot A + b cotB + cos C = 2(R + r).

2. 4

−1

as

−1bs

−1

cs

= Rr

.

3. If α, β, γ are the distances of the vertices of a triangle from the corresponding points of contact with the

incircle, then prove that y

y+β+α

αβ = r2

9. Radius of The Incircle :(i) r =

∆s

(ii) r = (s − a) tanA2

= (s − b) tanB2

= (s − c) tanC2

(iii) r =a B C

A

sin sincos

2 2

2

& so on (iv) r = 4R sinA2

sinB2

sinC2

10. Radius of The Ex- Circles :(i) r1 =

∆s a−

; r2 =∆

s b−; r3 = ∆

s c−(ii) r1 = s tan

A2

; r2 = s tanB2

; r3 = s tanC2

(iii) r1 =a B C

A

cos coscos

2 2

2

& so on (iv) r1 = 4 R sinA2

. cosB2

. cosC2

Example : In a ∆ABC, prove that r1 + r2 + r3 – r = 4R = 2a cosecA

Solution. ∵ L.H.S = r1 + r2 + r3 – r

= as −∆

+ bs −∆

+ cs −∆

– s∆

= ∆

−+

− bs1

as1

+ ∆

−

− s1

cs1

= ∆

−+−+

−−−+−

)cs(scss

)bs)(as(asbs

= ∆

−

+−− )cs(s

c)bs)(as(

c

= c∆

−−−−−+−

)cs)(bs)(as(s)bs)(as()cs(s

= c∆

∆+++−

2

2 ab)cba(ss2∵ a + b + c = 2s

= ∆

abc∵ R =

∆4abc

= 4R = 2acosecA ∵ Asina

= 2R = acosecA

= R.H.S.Hence L.H.S. = R.H.S. proved

Example : If the area of a ∆ABC is 96 sq. unit and the radius of the escribed circles are respectively8, 12 and 24. Find the perimeter of ∆ABC.

Solution. ∵ ∆ = 96 sq. unitr1 = 8, r2 = 12 and r3 = 24

∵ r1 = as −∆

⇒ s – a = 12 .........(i)

∵ r2 = bs −∆

⇒ s – b = 8 .........(ii)

∵ r3 = cs −∆

⇒ s – c = 4 .........(iii)∴ adding equations (i), (ii) & (iii), we get

3s – (a + b + c) = 24s = 24

∴ perimeter of ∆ABC = 2s = 48 unit.

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Self Practice Problems

In a ∆ABC prove that

1. r1r2 + r2r3 + r3r1 = s2

2. rr1 + rr2 + rr3 = ab + bc + ca – s2

3. If A, A1, A2 and A3 are the areas of the inscribed and escribed circles respectively of a ∆ABC, then prove

that A1

= 1A

1 +

2A1

+ 3A

1.

4. arr1 −

+ brr2 −

= 3rc

.

11. Length of Angle Bisectors, Medians & Altitudes :

(i) Length of an angle bisector from the angle A = βa =2 2bc

b c

Acos+

;

(ii) Length of median from the angle A = ma =12

2 22 2 2b c a+ −

& (iii) Length of altitude from the angle A = Aa = a2∆

NOTE : m m ma b c2 2 2+ + =

34 (a2 + b2 + c2)

Example : AD is a median of the ∆ABC. If AE and AF are medians of the triangles ABD and ADC

respectively, and AD = m1, AE = m2 , AF = m3 , then prove that m22 + m3

2 – 2m12 =

8a2

.Solution. ∵ In ∆ABC

AD2 = 41

(2b2 + 2c2 – a2) = m12 .........(i)

∵ In ∆ABD, AE2 = m22 =

41

(2c2 + 2AD2 – 4

a2) .........(ii)

Similarly in ∆ADC, AF2 = m32 =

41

−+

4ab2AD2

222

........(iii)

by adding equations (ii) and (iii), we get

∵ m22 + m3

2 = 41

−++

2ac2b2AD4

2222

= AD2 + 41

−+

2ac2b2

222

= AD2 + 41

+−+

2aac2b2

2222

= AD2 + 41

(2b2 + 2c2 – a2) + 8

a2

= AD2 + AD2 +8

a2

= 2AD2 + 8

a2

∵ AD2 = m12

= 2m12 +

8a2

∴ m22 + m3

2 – 2m12 =

8a2

Hence ProvedSelf Practice Problem :

3. In a ∆ABC a = 5, b = 4, c = 3. ‘G’ is the centroid of triangle, then find circumradius of ∆GAB.

Ans.125

13

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29

12. The Distances of The Special Points from Vertices and Sides of Triangle:

(i) Circumcentre (O) : OA = R & Oa = R cos A

(ii) Incentre (I) : IA = r cosec2A

& Ia = r

(iii) Excentre (I1) : I1 A = r1 cosec2A

& I1a = r1

(iv) Orthocentre (H) : HA = 2R cos A & Ha = 2R cos B cos C

(v) Centroid (G) : GA = 222 ac2b231 −+ & Ga = a3

2∆

Example : If x, y and z are respectively the distances of the vertices of the ∆ABC from its orthocentre,then prove that

(i)xa

+ yb

+ zc

= xyzabc

(ii) x y + z = 2(R + r)

Solution. ∵ x = 2R cosA, y = 2R cosB, z = 2R cosC andand a = 2R sinA, b = 2R sinB, c = 2R sinC

∴xa

+ yb

+ zc

= tanA + tan B + tan C .........(i)

& xyzabc

= tanA. tanB. tanC ........(ii)∵ We know that in a ∆ABC ΣΣΣΣΣtanA = ΠΠΠΠΠ tanA∴ From equations (i) and (ii), we get

xa

+ yb

+ zc

= xyzabc

∵ x + y + z = 2R (cosA + cosB + cosC)

∵ in a ∆ABC cosA + cosB + cosC = 1 + 4sin2A

sin2B

sin2C

∴ x + y + z = 2R

+

2Csin.

2Bsin.

2Asin41

= 2

+

2Csin.

2Bsin.

2AsinR4R ∵ r = 4R sin

2A sin

2B

sin2C

∴ x + y + z = 2(R + r)Self Practice Problems

1. If Ι be the incentre of ∆ABC, then prove that ΙA . ΙB . ΙC = abc tan2A tan

2B tan

2C .

2. If x, y, z are respectively be the perpendiculars from the circumcentre to the sides of ∆ABC, then prove

that xa

+ yb

+ zc

= xyz4abc

.

13. Orthocentre and Pedal Triangle:The triangle KLM which is formed by joining the feet of the altitudes is called the Pedal Triangle.(i) Its angles are π − 2A, π − 2B and π − 2C.

(ii) Its sides are a cosA = R sin 2A, b cosB = R sin 2B and c cosC = R sin 2C

(iii) Circumradii of the triangles PBC, PCA, PAB and ABC are equal.14. Excentral Triangle:

The triangle formed by joining the three excentres Ι1, Ι2 and Ι3 of ∆ ABC is calledthe excentral or excentric triangle.(i) ∆ ABC is the pedal triangle of the ∆ Ι1 Ι2 Ι3. (ii) Its angles are

π2 2

− A,π2 2

− B &

π2 2

− C.

(iii) Its sides are 4 R cosA2 ,

4 R cosB2 & 4 R cos

C2 .

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Y S

UH

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G S

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H:

G S

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H:

G S

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H:

G S

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29(iv) Ι Ι 1 = 4 R sinA2 ;

Ι Ι 2 = 4 R sinB2 ; Ι Ι 3 = 4 R sin

C2 .

(v) Incentre Ι of ∆ ABC is theorthocentre of the excentral ∆ Ι1 Ι2 Ι3.

15. Distance Between Special Points :(i) Distance between circumcentre and orthocentre

OH2 = R2 (1 – 8 cosA cos B cos C)(ii) Distance between circumcentre and incentre

OΙ2 = R2 (1 – 8 sin2A

sin 2B

sin 2C

) = R2 – 2Rr(iii) Distance between circumcentre and centroid

OG2 = R2 – 91

(a2 + b2 + c2)Example : In Ι is the incentre and Ι1, Ι2, Ι3 are the centres of escribed circles of the ∆ABC, prove that

(i) ΙΙ 1. ΙΙ 2 . ΙΙ 3 = 16R2r (ii) ΙΙ 12 + Ι2Ι3

2 = ΙΙ 22 + Ι3Ι1

2 = ΙΙ 32 + Ι1Ι2

2

Solution. (i) ∵ We know that

ΙΙ 1 = a sec2A

, ΙΙ 2 = b sec2B

and ΙΙ 3 = c sec2C

∵ Ι1Ι2 = c. cosec2C

, Ι2 Ι3 = a cosec2A

and Ι3Ι1 = b cosec2B

∵ ΙΙ 1 . ΙΙ 2 . ΙΙ 3 = abc sec2A

sec2B

.sec2C

........(i)∵ a = 2R sin A, b = 2R sinB and c = 2R sinC∴ equation (i) becomes

∵ ΙΙ 1. ΙΙ 2 . ΙΙ 3 = (2R sin A) (2R sin B) (2R sinC) sec2A

sec2B

sec2C

= 8R3 .

2Ccos.

2Bcos.

2Acos

2Ccos

2Csin2

2Bcos

2Bsin2

2Acos

2Asin2

= 64R3 sin2A

sin2B

sin2C

∵ r = 4R sin2A

sin2B

sin2C

∴ ΙΙ 1 . ΙΙ 2 . ΙΙ 3 = 16R2r Hence Proved

(ii) ΙΙ 12 + Ι2Ι3

2 = ΙΙ 22 + Ι3Ι1

2 = ΙΙ 32 + Ι1Ι2

2

∵ ΙΙ 12 + Ι2Ι3

2 = a2 sec2 2A

+ a2 cosec2 2A

=

2Acos

2Asin

a22

2

∵ a = 2 R sinA = 4R sin 2A

cos 2A

∴ ΙΙ 12 + Ι2Ι3

2 =

2Acos.

2Asin

2Acos.

2AsinR16

22

222

= 16R2

Similarly we can prove ΙΙ 22 + Ι3Ι1

2 = ΙΙ 32 + Ι1Ι2

2 = 16R2

Hence ΙΙ 12 + Ι2Ι3

2 = ΙΙ 22 + Ι3Ι1

2 = ΙΙ 32 + Ι1Ι2

2

Self Practice Problem :

1. In a ∆ABC, if b = 2 cm, c = 3 cm and ∠ A = 6π

, then find distance between its circumcentre andincentre.

Ans. 32 − cm

Exercise -1Part : (A) Only one correct option

1. In a triangle ABC, (a + b + c) (b + c − a) = k. b c, if :(A) k < 0 (B) k > 6 (C) 0 < k < 4 (D) k > 4

2. In a ∆ABC, A = 32π

, b – c = 3 3 cm and ar (∆ABC) =239

cm2. Then a is

(A) 6 3 cm (B) 9 cm (C) 18 cm (D) none of these

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of

293. If R denotes circumradius, then in ∆ABC, Ra2cb 22 −

is equal to

(A) cos (B – C) (B) sin (B – C) (C) cos B – cos C (D) none of these

4. If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ (= PR), then the angle P is

(A) 6π

(B) 3π

(C) 2π

(D) 32π

5. In a ∆ ABC, the value of cba

CcoscBcosbAcosa++

++ is equal to:

(A) rR

(B) Rr2

(C) Rr (D)

2rR

6. In a right angled triangle R is equal to

(A) 2

rs +(B)

2rs −

(C) s – r (D) ars +

7. In a ∆ABC, the inradius and three exradii are r, r1, r2 and r3 respectively. In usual notations the value ofr. r1. r2. r3 is equal to

(A) 2∆ (B) ∆2 (C) R4

abc(D) none of these

8. In a triangle if r1 > r2 > r3, then(A) a > b > c (B) a < b < c (C) a > b and b < c (D) a < b and b > c

9. With usual notation in a ∆ ABC

+

+

+

133221 r1

r1

r1

r1

r1

r1

= 222

3

cbaRK

,

where 'K' has the value equal to:(A) 1 (B) 16 (C) 64 (D) 128

10. The product of the arithmetic mean of the lengths of the sides of a triangle and harmonic mean of thelengths of the altitudes of the triangle is equal to:(A) ∆ (B) 2 ∆ (C) 3 ∆ (D) 4 ∆

11. In a triangle ABC, right angled at B, the inradius is:

(A) 2

ACBCAB −+ (B)

2BCACAB −+

(C)2

ACBCAB ++(D) None

12. The distance between the middle point of BC and the foot of the perpendicular from A is :

(A) a2

cba 222 ++−(B) a2

cb 22 −(C)

bccb 22 +

(D) none of these

13. In a triangle ABC, B = 60° and C = 45°. Let D divides BC internally in the ratio 1 : 3, then, CADsinBADsin

∠∠

=

(A) 32

(B) 31

(C) 61

(D) 31

14. Let f, g, h be the lengths of the perpendiculars from the circumcentre of the ∆ ABC on the sides a, b and

c respectively. If hc

gb

fa ++ = λ

hgfcba

then the value of λ is:

(A) 1/4 (B) 1/2 (C) 1 (D) 2

15. A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length3, 4 and 5 units. Then area of the triangle is equal to:

(A) 2)31(39

π+

(B) 2)13(39

π−

(C) 22)31(39

π+

(D) 22)13(39

π−

16. If in a triangle ABC, the line joining the circumcentre and incentre is parallel to BC, thencos B + cos C is equal to:(A) 0 (B) 1 (C) 2 (D) none of these

17. If the incircle of the ∆ ABC touches its sides respectively at L, M and Nand if x, y, z be the circumradii of the triangles MIN, NIL and LIM whereI is the incentre then the product xyz is equal to:

(A) R r2 (B) r R2 (C) 21

R r2 (D) 21

r R2

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2918. If in a ∆ABC,1rr

=21

, then the value of tan2A

+

2Ctan

2Btan is equal to :

(A) 2 (B) 21

(C) 1 (D) None of these

19. In any ∆ABC, then minimum value of 3321

rrrr

is equal to(A) 3 (B) 9 (C) 27 (D) None of these

20. In a acute angled triangle ABC, AP is the altitude. Circle drawn with AP as its diameter cuts the sidesAB and AC at D and E respectively, then length DE is equal to

(A) R2∆

(B) R3∆

(C) R4∆

(D) R∆

21. AA1, BB1 and CC1 are the medians of triangle ABC whose centroid is G. If the concyclic, then pointsA, C1, G and B1 are(A) 2b2 = a2 + c2 (B) 2c2 = a2 + b2 (C) 2a2 = b2 + c2 (D) None of these

22. In a ∆ABC, a, b, A are given and c1, c2 are two values of the third side c. The sum of the areas of twotriangles with sides a, b, c1 and a, b, c2 is

(A)21

b2 sin 2A (B)21

a2 sin 2A (C) b2 sin 2A (D) none of these

23. In a triangle ABC, let ∠ C = 2π

. If r is the inradius and R is the circumradius of the triangle, then 2(r + R)is equal to [IIT - 2000](A) a + b – c (B) b + c (C) c + a (D) a + b + c

24. Which of the following pieces of data does NOT uniquely determine an acute - angled triangleABC (R being the radius of the circumcircle )? [IIT - 2002](A) a , sin A, sin B (B) a, b, c (C) a, sin B, R (D) a, sin A, R

25. If the angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the longest side to the perimeter is(A) 3 : (2 + 3 ) (B) 1 : 6 (C) 1 : 2 + 3 (D) 2 : 3 [IIT - 2003]

26. The sides of a triangle are in the ratio 1 : 3 : 2, then the angle of the triangle are in the ratio[IIT - 2004]

(A) 1 : 3 : 5 (B) 2 : 3 : 4 (C) 3 : 2 : 1 (D) 1 : 2 : 3

27. In an equilateral triangle, 3 coincs of radii 1 unit each are kept so that they touche each other and alsothe sides of the triangle. Area of the triangle is [IIT - 2005]

(A) 4 + 2 3 (B) 6 + 4 3 (C) 12 + 437 (D) 3 +

437

28. If P is a point on C1 and Q is a point on C2, then 2222

2222

QDQCQBQAPDPCPBPA

++++++

equals(A) 1/2 (B) 3/4 (C) 5/6 (D) 7/8

29. A circle C touches a line L and circle C1 externally. If C and C1 are on the same side of the line L, thenlocus of the centre of circle C is(A) an ellipse (B) a circle (C) a parabola (D) a hyperbola

30. Let " be a line through A and parallel to BD. A point S moves such that its distance from the line BD andthe vertex A are equal. If the locus of S meets AC in A1, and " in A2 and A3, then area of ∆A1 A2A3 is(A) 0.5 (unit)2 (B) 0.75 (unit)2 (C) 1 (unit)2 (D) (2/3) (unit)2

Part : (B) May have more than one options correct

31. In a ∆ABC, following relations hold good. In which case(s) the triangle is a right angled triangle?(A) r2 + r3 = r1 − r (B) a2 + b2 + c2 = 8 R2 (C) r1 = s (D) 2 R = r1 − r

32. In a triangle ABC, with usual notations the length of the bisector of angle A is :

(A) cb2Acosbc2

+ (B) cb2Asinbc2

+ (C) )cb(R22Aeccosabc

+ (D) 2Aeccos.

cb2+∆

33. AD, BE and CF are the perpendiculars from the angular points of a ∆ ABC upon the opposite sides,then :

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of

29(A) Rr

ABCofPerimeterDEFofPerimeter =

∆∆ (B) Area of ∆DEF = 2 ∆ cosA cosB cosC

(C) Area of ∆AEF = ∆ cos2A (D) Circum radius of ∆DEF = 2R

34. The product of the distances of the incentre from the angular points of a ∆ ABC is:

(A) 4 R2 r (B) 4 Rr2 (C) ( )

sRcba

(D) ( )

srcba

35. In a triangle ABC, points D and E are taken on side BC such that BD = DE = EC. If angleADE = angle AED = θ, then:(A) tanθ = 3 tan B (B) 3 tanθ = tanC

(C) 9tantan62 −θ

θ = tan A (D) angle B = angle C

36. With usual notation, in a ∆ ABC the value of Π (r1 − r) can be simplified as:

(A) abc Π tanA2 (B) 4 r R2 (C)

( )( )

a bc

R a b c

2

2+ +(D) 4 R r2

Exercise -21. If in a triangle ABC, cos cos

cos cosA CA B

++

22

= sinsin

BC

, prove that the triangle ABC is either isosceles or right

angled.

2. In a triangle ABC, if a tan A + b tan B = (a + b) tanA B+

2

, prove that triangle is isosceles.

3. If

−

−

3

1

2

1

rr1

rr1 = 2 then prove that the triangle is the right triangle.

4. In a ∆ ABC, ∠ C = 60° & ∠ A = 75°. If D is a point on AC such that the area of the ∆ BAD is 3 timesthe area of the ∆ BCD, find the ∠ ABD.

5. The radii r1, r2, r3 of escribed circles of a triangle ABC are in harmonic progression. If its area is 24 sq.cm and its perimeter is 24 cm, find the lengths of its sides.

6. ABC is a triangle. D is the middle point of BC. If AD is perpendicular to AC, then prove that

cos A. cos C = ( )

ca3ac2 22 −

.

7. Two circles, of radii a and b, cut each other at an angle θ. Prove that the length of the common chord is

θ++

θ

cosab2ba

sinab222

.

8. In the triangle ABC, lines OA, OB and OC are drawn so that the angles OAB, OBC and OCA are eachequal to ω, prove that(i) cot ω = cot A + cot B + cot C(ii) cosec2 ω = cosec2 A + cosec2 B + cosec2 C

9. In a plane of the given triangle ABC with sides a, b, c the points A′ , B′ , C′ are taken so that the∆ A′ BC, ∆ AB′C and ∆ ABC′ are equilateral triangles with their circum radii Ra, Rb, Rc; in−radii ra, rb, rc& ex−radii ra′ , rb′ & rc′ respectively. Prove that;

(i) Π ra: Π Ra: Π ra′ = 1: 8: 27 (ii) r1 r2 r3 = ( )[ ]

2Atan

3648

r2r6R33

aaaΠ

′++∑10. The triangle ABC is a right angled triangle, right angle at A. The ratio of the radius of the circle

circumscribed to the radius of the circle escribed to the hypotenuse is, ( )23:2 + . Find the acuteangles B & C. Also find the ratio of the two sides of the triangle other than the hypotenuse.

11. The triangle ABC is a right angled triangle, right angle at A. The ratio of the radius of the circlecircumscribed to the radius of the circle escribed to the hypotenuse is, ( )23:2 + . Find the acuteangles B & C. Also find the ratio of the two sides of the triangle other than the hypotenuse.

12. If the circumcentre of the ∆ ABC lies on its incircle then prove that,cosA + cosB + cosC = 2

13. Three circles, whose radii area a, b and c, touch one another externally and the tangents at their pointsof contact meet in a point; prove that the distance of this point from either of their points of contacts

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: 25

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Exercise -1

1. C 2. B 3. B 4. D 5. A 6. B 7. B 8. A 9. C 10. B

11. A 12. B 13. C 14. A 15. A 16. B 17. C 18. B 19. C 20. D

21. C 22. A 23. A 24. D 25. A 26. A 27. B 28. B 29. C 30. C

31. ABCD 32. ACD 33. ABCD 34. BD 35. ACD 36. ACD

Exercise -2

4. ∠ ABD = 30° 5. 6, 8, 10 cms

10. B =125π

, C =12π

, cb

= 32 + 11. B =125π

, C =12π

, cb

= 32 +