STUDY PACKAGE. Subject : Mathematics Topic: Trigonometric Equation & Properties & Solution of Triangle ENJOY MATHEMA WITH

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1 fo/u fopkjr Hkh# tu] ugha vkjehks dke] foifr ns[k NksMs rqjar e/;e eu dj ';kea iq#"k flag ladyi dj] lgrs foifr vusd] ^cuk^ u NksMs /;s; dks] j?kqcj jk[ks VsdAA jfpr% ekuo /kez iz.ksrk ln~xq# Jh j.knksmnklth egkjkt STUDY PACKAGE Subject : Mathematics Topic: Trigonometric Equation & Properties & Solution of Triangle Index. Theory. Short Revision. Exercise 4. Assertion & Reason 5. Que. from Compt. Exams ENJOY MATHEMA THEMATICS TICS WITH SUHAAG SIR Student s Name : Class Roll No. : : Head Office: 4-B, III- Floor, Near Hotel Arch Manor, Zone-I M.P. NAGAR, Main Road, Bhopal!:(0755) , Free Study Package download from website :

2 Trigonometric Equation. Trigonometric Equation : An equation involving one or more trigonometric ratios of an unknown angle is called a trigonometric equation.. Solution of Trigonometric Equation : A solution of trigonometric equation is the value of the unknown angle that satisfies the equation. 9 e.g. if sinθ θ,,,, Thus, the trigonometric equation may have infinite number of solutions (because of their periodic nature) and can be classified as : (i) Principal solution (ii) General solution.. Principal solutions: The solutions of a trigonometric equation which lie in the interval [0, ) are called Principal solutions. e.g Find the Principal solutions of the equation sinx. sinx i.e. there exists two values 5 and which lie in [0, ) and whose sine is Principal solutions of the equation sinx are, 6 6. General Solution : The expression involving an integer 'n' which gives all solutions of a trigonometric equation is called General solution. General solution of some standard trigonometric equations are given below.. General Solution of Some Standard Trigonometric Equations : (i) If sin θ sin α θ n ( ) n α where α,, n Ι. (ii) If cos θ cos α θ n ± α where α [0, ], n Ι. (iii) If tan θ tan α θ n α where α,, n Ι. (iv) If sin² θ sin² α θ n ± α, n Ι. (v) If cos² θ cos² α θ n ± α, n Ι. (vi) If tan² θ tan² α θ n ± α, n Ι. [ Note: α is called the principal angle ] Some Important deductions : (i) sinθ 0 θ n, n Ι Page : of 9 (0755) , G SIR PH: (0755) , (ii) sinθ θ (4n ),n Ι (iii) sinθ θ (4n ),n Ι (iv) cosθ 0 θ (n ),n Ι (v) (vi) cosθ cosθ θ n, θ (n ), n Ι n Ι (vii) tanθ 0 θ n, n Ι Solved Example # Solve sin θ.

3 sin θ sinθ sin Solved Example # θ n ( ) n, n Ι Solve sec θ sec θ cosθ 5 cosθ cos 6 5 θ n ±, n Ι 6 θ n ± Ι Solved Example # Solve tanθ Let tanθ tanα...(i) tanθ tanα θ n α, where α tan (), n Ι Self Practice Problems:. Solve cotθ. Solve cosθ n () θ n, n Ι () ±, n Ι 4 9 Solved Example # 4 Solve cos θ cos θ cos θ cos θ cos 4 θ n ±, n Ι 4 Solved Example # Page : of 9 (0755) , G SIR PH: (0755) , Solve 4 tan θ sec θ 4 tan θ sec θ...(i) For equation (i) to be defined θ (n ), n Ι equation (i) can be written as: 4sin θ cos θ cos θ θ (n ), n Ι cos θ 0 4 sin θ sin θ

4 sin θ sin θ n ±, n Ι Self Practice Problems :. Solve 7cos θ sin θ 4.. Solve sin x sin x () n ±, n Ι () (n ), n Ι or n ± 4, n Ι Types of Trigonometric Equations : Type - Trigonometric equations which can be solved by use of factorization. Solved Example # 6 Solve (sinx cosx) ( cosx) sin x. (sinx cosx) ( cosx) sin x (sinx cosx) ( cosx) sin x 0 (sinx cosx) ( cosx) ( cosx) ( cosx) 0 ( cosx) (sinx ) 0 cosx 0 or sinx 0 cosx or sinx x (n ), n Ι or sin x sin 6 Solution of given equation is x n ( ) n 6, n Ι (n ), n Ι or n ( ) n, n Ι 6 Self Practice Problems :. Solve x cos x cos x 4cos 0. Solve cot θ cosecθ 0 () (n ), n Ι () n, n Ι or n ( ) n, n Ι 6 Type - Trigonometric equations which can be solved by reducing them in quadratic equations. Solved Example # 7 Solve cos x 4cosx sin x cos x 4cosx sin x 0 cos x 4cosx ( cos x) 0 5cos x 4cosx 0 9 cos x 9 cos x cosx [, ] x R...(ii) 9 cosx 5 equation (ii) will be true if 9 cosx 5 9 cosx cosα, where cosα 5 x n ± α where α cos 9, n Ι Page : 4 of 9 (0755) , G SIR PH: (0755) ,

5 Self Practice Problems :. Solve cosθ ( ) cos θ 0. Solve 4cosθ secθ tanθ () n ±, n Ι or n ± 4, n Ι () n ( ) n α where α sin 7, n Ι 8 or n ( ) n β where β sin 7, n Ι 8 Type - Trigonometric equations which can be solved by transforming a sum or difference of trigonometric ratios into their product. Solved Example # 8 Solve cosx sinx sin4x 0 cosx sinx sin4x 0 cosx cosx.sinx 0 cosx cosx.sin( x) 0 cosx ( sinx) 0 cosx 0 or sinx 0 x (n ), n Ι or sinx x (n ), n Ι or x n ( ) n, n Ι 6 6 solution of given equation is (n ), n Ι or n ( ) n, n Ι 6 6 Self Practice Problems :. Solve sin7θ sinθ sinθ. Solve 5sinx 6sinx 5sinx sin4x 0. Solve cosθ sinθ cosθ () n n, n Ι or ±, n Ι () n, n Ι or n ±, n Ι () n, n Ι or n, n Ι or n, n Ι 4 Type - 4 Trigonometric equations which can be solved by transforming a product of trigonometric ratios into their sum or difference. Solved Example # 9 Solve sin5x.cosx sin6x.cosx sin5x.cosx sin6x.cosx sin5x.cosx sin6x.cosx sin8x sinx sin8x sin4x sin4x sinx 0 sinx.cosx sinx 0 sinx (cosx ) 0 sinx 0 or cosx 0 x n, n Ι or cosx Page : 5 of 9 (0755) , G SIR PH: (0755) , n x, n Ι or x n ±, n Ι Type - 5 x n ±, n Ι 6 Solution of given equation is n, n Ι or n ±, n Ι 6 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 a b.

6 Solved Example # 0 Solve sinx cosx sinx cosx Here a, b....(i) divide both sides of equation (i) by, we get sinx. cosx. sinx.sin 4 cosx.cos 4 cos x 4 x 4 n, n Ι x n 4, n Ι Solution of given equation is n, n Ι 4 Note : Trigonometric equation of the form a sinx b cosx c can also be solved by changing sinx and cosx into their corresponding tangent of half the angle. Solved Example # Solve cosx 4sinx 5 cosx 4sinx 5...(i) x x tan tan cosx & sinx x x tan tan equation (i) becomes x x tan tan 4 x x tan tan 5...(ii) Let x tan t equation (ii) becomes t t 4 t t 5 4t 4t 0 (t ) 0 t tan x t tan x Page : 6 of 9 (0755) , G SIR PH: (0755) , tan x tanα, where tanα x n α x n α where α tan Ι Self Practice Problems :. Solve cosx sinx

7 . Solve sinx tan x 0 () n 6, n Ι () x n, n Ι Type - 6 Trigonometric equations of the form P(sinx ± cosx, sinx cosx) 0, where p(y, z) is a polynomial, can be solved by using the substitution sinx ± cosx t. Solved Example # Solve sinx cosx sinx.cosx sinx cosx sinx.cosx...(i) Let sinx cosx t sin x cos x sinx.cosx t t sinx.cosx Now put sinx cosx t t and sinx.cosx in (i), we get t t t t 0 t t sinx cosx sinx cosx...(ii) divide both sides of equation (ii) by, we get sinx. cosx. cos x cos 4 4 x n ± 4 4 (i) (ii) Self Practice Problems: if we take positive sign, we get x n, n Ι if we take negative sign, we get x n, n Ι. Solve sinx 5sinx 5cosx 0. Solve cosx sinx sinx cosx 0. Solve ( sinx) (cosx sinx) sin x. Type - 7 () n 4, n Ι () n 4, n Ι () n, n Ι or n, n Ι or n 4, n Ι Page : 7 of 9 (0755) , G SIR PH: (0755) , Trigonometric equations which can be solved by the use of boundness of the trigonometric ratios sinx and cosx. Solved Example # Solve x x sinx cos sinx sin cos x cos x x x sinx cos sin x sin cos x cos x (i) sinx.cos 4 x sin x cosx sin 4 x.cosx cos x 0 x x sin x.cos sin.cos x (sin 4 4 x cos x) cosx 0 5x sin cosx...(ii) 4

8 Now equation (ii) will be true if 5x sin and cosx 4 5x 4 n, n Ι and x m, m Ι ( 8n ) x, n Ι...(iii) 5 Now to find general solution of equation (i) and x m, m Ι...(iv) ( 8n ) m 5 8n 0m 5m n 4 if m then n if m 5 then n if... m 4p, p Ι... then... n 5p 4, p Ι Self Practice Problems : general solution of given equation can be obtained by substituting either m 4p in equation (iv) or n 5p 4 in equation (iii) general solution of equation (i) is (8p 6), p Ι. Solve sinx cosx. Solve sin5x cos x sinx () (4p ), p Ι () m, m Ι Part : (A) Only one correct option Exercise -. The solution set of the equation 4sinθ.cosθ cosθ sinθ 0 in the interval (0, ) is (A) 7, 4 4 (B) 5, 5 (C),,, 4 5 (D),, All solutions of the equation, sinθ tanθ 0 are obtained by taking all integral values of m and n in: (A) n, n Ι (B) n or m ± where n, m Ι (C) n or m ± where n, m Ι (D) n or m ±. If 0 sin θ cos θ 4 0 & 7 θ < θ < then the values of cot 4 is: where n, m Ι Page : 8 of 9 (0755) , G SIR PH: (0755) , (A) (B) 5 (C) 5 (D) 4. The general solution of sinx sin5x sinx sin4x is: (A) n ; n Ι (B) n ; n Ι (C) n/ ; n Ι (D) n/ ; n Ι 5. A triangle ABC is such that sin(a B). If A, B, C are in A.P. then the angle A, B, C are respectively. 5 (A),, 4 5 (B),, 4 5 (C),, 4 5 (D),, 4

9 6. The maximum value of sinx 4cosx is (A) (B) 4 (C) 5 (D) 7 7. If sin θ 7 cos θ 5, then tan (θ/) is a root of the equation (A) x 6x 0 (B) 6x x 0 (C) 6x x 0 (D) x x sin θ sinθ cos θ cosθ cosθ cot θ (A) θ 0, (B) θ, tan θ cot θ if: (C) θ, (D) θ, 9. The number of integral values of a for which the equation cos x a sin x a 7 possesses a solution is (A) (B) (C) 4 (D) 5 0. The principal solution set of the equation, cos x sinx is (A), 8 8 (B), 4 8 (C), 4 0 (D), 8 0. The number of all possible triplets (a, a, a ) such that : a a cos x a sin x 0 for all x is (A) 0 (B) (C) (D) infinite n. If tan x 5 secx 0 has 7 different roots in 0,, n N, then greatest value of n is (A) 8 (B) 0 (C) (D) 5. The solution of cosx cosx sinx is (A) x n, n Ι (C) x n ( ) n 4, n Ι (B) x n 4, n Ι (D) (n ) 4, n Ι 4. The arithmetic mean of the roots of the equation 4cos x 4cos x cos( x) 0 in the interval [0, 5] is equal to (A) 49 (B) 50 (C) 5 (D) Number of solutions of the equation cos 6x tan x cos 6x. tan x in the interval [0, ] is : (A) 4 (B) 5 (C) 6 (D) 7 Part : (B) May have more than one options correct 6. sinx cos x assumes the least value for the set of values of x given by: (A) x n ( ) n (/6), n Ι (B) x n ( ) n (/6), n Ι (C) x n ( ) n (/), n Ι (D) x n ( ) n (/6), n Ι 7. cos4x cos8x cos5x cos9x 0 if (A) cosx cos 4 x (B) sin x 0 (C) sinx 0 (D) cosx Page : 9 of 9 (0755) , G SIR PH: (0755) , 8. The equation sin x. cos x sin x sin x. sin x cos x has a root for which (A) sinx (B) sinx (C) cosx (D) cosx 9. sin x sin x cos x cos x 0 if (A) tan x (B) tanx (C) x n /4, n Ι (D) x n tan ( ), n Ι 0. sin x cos x sin x if (A) x n/, n Ι (C) x (n ) /, n Ι (B) x n /, n Ι (D) x n ( ) n sin (/), n Ι

10 . Solve cotθ tan8θ. Solve x x cot cosec cotx. Solve cot θ cotθ Solve cosθ cosθ Solve the equation: sin 6x sin 4x sin x. 6. Solve: cos θ sin θ cos θ sin θ. 7. Solve 4 sin x. sin x. sin 4x sin x. Exercise - 8. Solve sin nθ sin (n )θ sin θ, where n is constant and n 0, 9. Solve tanθ tanθ tanθ tanθ. 0. Solve: sin x cos x cos x sin x Solve the equation, sin x cos x sin x. Solve the equation: sin 5x 6 sin 5 x.. If tan θ sin φ & tan² θ cos² φ 7 4 cos x. then find the general value of θ & φ. 4. Solve for x, the equation 8tanx 6 tan x, where < x <. 5. Find the general solution of sec 4 θ sec θ. 6. Solve the equation sin x cos x cos² x. 7. Solve for x: sin x 8sin x. cos x 4 8. Solve the equation for 0 θ ; ( ). sinθ cosθ 5 cos θ Solve: tan x. tan x. tan 4 x tan x tan x tan 4 x. 0. Find the values of x, between 0 &, satisfying the equation; cos x cos x sin x x sin.. Solve: cos x cos 6 x Page : 0 of 9 (0755) , G SIR PH: (0755) ,. Solve the equation, sin 4 x cos x sin 4 x cos4 x.

11 ANSWER EXERCISE. D. B. D 4. C 5. B 6. C 7. B 8. B 9. D 0. A. D. D. D 4. C 5. D 6. AD 7. ABC 8. ABCD 9. CD 0. BC. n EXERCISE, n Ι. x 4n ±, n Ι 9. θ n, n Ι or n 6, n Ι 4. n ± α where α cos 5. n 4, n Ι or n ± 6, n Ι 6. n, n Ι or n 6, n Ι 7. x n, n Ι or n ± 9, n Ι 8. m, m Ι or m, m Ι or n 9. n n, n Ι 0. x 4 ( )n 4, n Ι. x (4 n ), n Ι. x n ; x n ± 6, n Ι 7 4, n Ι. θ n 4, φ n ( ) n 6, n I 4. α ; α, α, α, where tan α 5. n ± 5 0 m, m Ι n or n ±, n Ι 6. x ( n ),, n Ι or n ±, n Ι (4" ), " Ι or x (4k 7), k Ι 8. θ, 9. ( n ), k, where n, k Ι ,,,, Page : of 9 (0755) , G SIR PH: (0755) ,. φ. x ( n ), n I

12 Properties & Solution of Triangle. Sine Rule: In any triangle ABC, the sines of the angles are proportional to the opposite sides i.e. a b c. sina sinb sinc Example : A B cos a b In any ABC, prove that c C sin. A B cos a b We have to prove c C sin From sine rule, we know that. a b c k (let) sin A sinb sinc a k sina, b k sinb and c k sinc a b L.H.S. c A B A B sin cos k (sin A sinb) k sinc C C sin cos C A B A B cos cos cos C C C sin cos sin R.H.S. Hence L.H.S. R.H.S. Proved Example : In any ABC, prove that (b c ) cot A (c a ) cot B (a b ) cot C 0 We have to prove that (b c ) cot A (c a ) cot B (a b ) cot C 0 from sine rule, we know that a k sina, b k sinb and c k sinc (b c ) cot A k (sin B sin C) cot A sin B sin C sin (B C) sin (B C) (b c ) cot A k sin (B C) sin (B C) cota B C A (b c ) cot A cos A k sin A sin (B C) sin A cosa cos(b C) k sin (B C) cos (B C) k [sin (B C) cos (B C)] k (b c ) cot A [sin B sin C]...(i) k Similarly (c a ) cot B [sin C sin A]...(ii) Page : of 9 (0755) , G SIR PH: (0755) , k and (a b ) cot C [sin A sin B]...(iii) adding equations (i), (ii) and (iii), we get (b c ) cot A (c a ) cot B (a b ) cot C 0 Hence Proved Self Practice Problems In any ABC, prove that A A. a sin B (b c) sin.. a sin(b C) sinb sinc b sin(c sinc. Cosine Formula: (i) cos A b c a bc A) sin A c sin(a sin A B) sinb 0. c a b A B tan tan. A B tan tan or a² b² c² bc cos A b c bc cos (B C)

13 (ii) cos B c a b (iii) cos C a b c ca ab Example : In a triangle ABC if a, b 8 and c 7, then find sin A. b c a cosa bc cosa A sina sin *Example : In a ABC, prove that a(b cos C c cos B) b c We have to prove a (b cosc c cosb) b c. from cosine rule we know that a b c a c b cosc & cos B ab ac L.H.S. a a b c a c b b c ab ac a b c (a c b ) (b c ) R.H.S. Hence L.H.S. R.H.S. Proved Example : a b c a If in a ABC, A 60 then find the value of. c c b b A 60 a b c a c a b b c a c c b b c b (b c) bc (b b c a a bc c a bc ) bc b c a bc cosa A 60 cos A a b c a c c b b Self Practice Problems :. The sides of a triangle ABC are a, b, a ab b, then prove that the greatest angle is 0.. In a triangle ABC prove that a(cosb cosc) (b c) sin A.. 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) b c L.H.S. a (b cosc c cosb) b (a cosc) c (a cosb)...(i) From projection rule, we know that b a cosc c cosa a cosc b c cosa & c a cosb b cosa a cosb c b cosa Put values of a cosc and a cosb in equation (i), we get L.H.S. b (b ccos A) c(c b cos A) b bc cos A c bc cos A b c R.H.S. Hence L.H.S. R.H.S. Proved Note: We have also proved a (b cosc ccosb) b c 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 Page : of 9 (0755) , G SIR PH: (0755) , 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

14 Self Practice Problems In a ABC, prove that. C B bcos c cos a b c... cosb cos C c bcos A. b c cos A cos A c cosb bcosc cosb acosc c cos A cosc acosb bcos A 4. Napier s Analogy - tangent rule: a b c abc B C b c A C A c a (i) tan cot (ii) tan cot B b c c a (iii) tan A B a b cot C a b Example : Find the unknown elements of the ABC in which a, b, C 60. a, b, C 60 A B C 80 A B 0...(i) From law of tangent, we know that A B a b C tan cot a b ( ( ) ( ) ( cot 0 ) cot 0 ) A B tan A B 45 4 A B 90...(ii) From equation (i) and (ii), we get A 05 and B 5 Now, c From sine-rule, we know that c 6 asinc sin A ( ( ) )sin60 sin05 a sin A sin05 c 6, A 05, B 5 Self Practice Problem b sinb c sinc Page : 4 of 9 (0755) , G SIR PH: (0755) , 7 A. In a ABC if b, c 5 and cos (B C), then find the value of tan. 5 B C. If in a ABC, we define x tan then show that x y z xyz. tan A, y tan C A tan B and z tan A B tan C

15 5. Trigonometric Functions of Half Angles: (i) sin A ( s b) ( s c) bc (ii) cos A s ( s a ) bc ; sin B ( s c) ( s a) ca ; cos B s ( s b ) ca (iii) tan A ( s b) ( s c) ss ( a) ss ( a) (iv) sin A s(s a)(s b)(s c) bc bc 6. Area of Triangle () ; sin C ( s a) ( s b) ab ; cos C s ( s c) ab where s a b c ab sin C bc sin A ca sin B ss ( a) ( s b) ( s c) is semi perimetre of triangle. A C Example : In a ABC if a, b, c are in A.P. then find the value of tan. tan. A C tan and tan s(s a) s(s c) Example : A C tan. tan s (s a)(s c) A C s b b tan. tan s s it is given that a, b, c are in A.P. b a c a b c b s b put in equation (i) s A C tan. tan tan A. tan C s (s a) (s b) (s c)...(i) In a ABC if b sinc(b cosc c cosb) 4, then find the area of the ABC. b sinc (b cosc c cosb) 4...(i) given From projection rule, we know that a b cosc c cosb put in (i), we get ab sinc 4...(ii) ab sinc sq. unit Example : A B C In any ABC prove that (a b c) tan tan c cot. A B L.H.S. (a b c) tan tan A tan (s b)(s c) B (s a)(s c) and tan s(s a) s(s b) (s b)(s c) (s a)(s c) L.H.S. (a b c) s(s a) s(s b) s s c s b s a s s a s b s(s c) s b s a (s a)(s b) s a b c s b a c Page : 5 of 9 (0755) , G SIR PH: (0755) ,

16 7. m - n Rule: s(s c) c c cot C R.H.S. Hence L.H.S. R.H.S. s(s c) (s a)(s b) (m n) cot θ m cot α n cot β n cot B m cot C c (s a)(s b) Proved C s(s c) cot (s a)(s b) Example : If the median AD of a triangle ABC is perpendicular to AB, prove that tan A tan B 0. Example : 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 ( ) cot (90 B). cot 90.cot (A 90 ) tan B cot (90 A) tan B tan A tan A tan B 0 Hence proved. The base of a triangle is divided into three equal parts. If t, t, t be the tangents of the angles subtended by these parts at the opposite vertex, prove that 4 t t t. t t Let point D and E divides the base BC into three equal parts i.e. BD DE DC d (Let) and let α, β and γ be the angles subtended by BD, DE and EC respectively at their opposite vertex. t tanα, t tanβ and t tanγ Now in ABC BE : EC d : d : from m-n rule, we get ( ) cotθ cot (α β) cotγ cotθ cot (α β) cotγ...(i) again in ADC DE : EC x : x : if we apply m-n rule in ADC, we get ( ) cotθ. cotβ cotγ cotθ cotβ cotγ...(ii) from (i) and (ii), we get cot θ cot( α β) cot γ cot θ cotβ cot γ cotβ cotγ 4cot (α β) cotγ cotβ cotγ 4 cot (α β) cot α.cotβ cotβ cotγ 4 cotβ cot α cot β cotα cotβ cotβ cotγ cotα cotγ 4 cotα cotβ 4 4 cot β cotα cotβ cotβ cotγ cotα cotγ 4 4cot β cotα cotβ cotα cotγ cotβ cotγ cot β 4( cot β) (cotα cotβ) (cotβ cotγ) 4 tan β tanα tanβ tanβ tan γ 4 t t t t t Hence proved Page : 6 of 9 (0755) , G SIR PH: (0755) ,

17 Self Practice Problems :. In a ABC, the median to the side BC is of length 6 0 and 45. Prove that the side BC is of length units. 8. Radius of Circumcirlce : Example : Example : a b c R sina sinb sinc abc 4 In a ABC prove that sina sinb sinc R s In a ABC, we know that a sin A b c R sinb sinc a, sinb R R a b R s R c. sin A and sinc b R c sina sinb sinc and it divides angle A into the angles of a b c s sina sinb sinc R s. In a ABC if a cm, b 4 cm and c 5 cm, then find its circumradius. abc R 4 s(s a)(s b)(s c) a b c s cm cm R R cm cm...(i) Example : A B C In a ABC prove that s 4R cos. cos. cos. In a ABC, A cos s(s a) B s(s b), cos bc ca C and cos Example : R.H.S. 4R cos A. cos B. cos C. s(s c) ab and R abc s(s a)(s b)(s c). s (abc) s(s a)(s b)(s c) s L.H.S. Hence R.H.L L.H.S. proved 4 R In a ABC, prove that. s a s b s c s 4R s a s b s c s L.H.S. s a s b s c s s a b (s s c) (s a)(s b) s(s c) s a b c c c (s a)(s b) s(s c) s(s c) (s a)(s b) s s(a b c) ab c c s(s a)(s b)(s c) s s(s) ab abc 4R 4R abc L.H.S. c R 4 abc 4R abc Page : 7 of 9 (0755) , G SIR PH: (0755) ,

18 4R L.H.S. Self Practice Problems : In a ABC, prove the followings :. a cot A b cotb cos C (R r). s s s. 4 r. a b c R. If α, β, γ are the distances of the vertices of a triangle from the corresponding points of contact with the αβy incircle, then prove that r α β y 9. Radius of The Incircle : (i) r s (iii) r a B C sin sin A cos 0. Radius of The Ex- Circles : (ii) r (s a) tan A (s b) tan B (s c) tan C & so on (iv) r 4R sin A sin B sin C (i) r ; r s a ; r s b (ii) r s c s tan A ; r s tan B ; r s tan C (iii) r a B C cos cos & so on (iv) r A cos 4 R sin A. cos B. cos C Example : In a ABC, prove that r r r r 4R a coseca L.H.S r r r r s a s b Example : s c s s a s b s c s s b s a s s c (s a)(s b) s(s c) c c (s a)(s b) s(s c) s(s c) (s a)(s b) c s(s a)(s b)(s c) s s(a b c) ab c abc 4R acoseca a b c s R a sin A abc 4 R acoseca R.H.S. Hence L.H.S. R.H.S. proved If the area of a ABC is 96 sq. unit and the radius of the escribed circles are respectively 8, and 4. Find the perimeter of ABC Page : 8 of 9 (0755) , G SIR PH: (0755) , 96 sq. unit r 8, r and r 4 r s a s a...(i) r s b s b 8...(ii) r s c 4 s c adding equations (i), (ii) & (iii), we get...(iii) s (a b c) 4 s 4 perimeter of ABC s 48 unit.

19 Self Practice Problems In a ABC prove that. r r r r r r s. rr rr rr ab bc ca s. If A, A, A and A are the areas of the inscribed and escribed circles respectively of a ABC, then prove that. A 4. r r a r r b A c r. A A. Length of Angle Bisectors, Medians & Altitudes : & Example : bc cos A (i) Length of an angle bisector from the angle A β a ; b c (ii) Length of median from the angle A m a b c a (iii) Length of altitude from the angle A A a a a b c NOTE : m m m 4 (a b c ) AD is a median of the ABC. If AE and AF are medians of the triangles ABD and ADC respectively, and AD m, AE m, AF m, then prove that m m m In ABC AD 4 (b c a ) m a In ABD, AE m (c AD ) 4 4 Similarly in ADC, AF a m AD b 4 4 by adding equations (ii) and (iii), we get m m a 4AD b c 4 a AD b c 4 a AD b c a 4 AD 4 (b c a ) a AD AD 8 a 8 a. 8...(i)...(ii)...(iii) Page : 9 of 9 (0755) , G SIR PH: (0755) , a AD AD m 8 m m m m Self Practice Problem : a 8 a 8 Hence Proved. In a ABC a 5, b 4, c. G is the centroid of triangle, then find circumradius of GAB. 5

20 . The Distances of The Special Points from Vertices and Sides of Triangle: (i) Circumcentre (O) : OA R & O a R cos A (ii) Incentre (I) : IA r cosec A & Ia r (iii) Excentre (I ) : I A r cosec A & Ia r (iv) Orthocentre (H) : HA R cos A & H a R cos B cos C (v) Centroid (G) : GA b c a & G a a Example : If x, y and z are respectively the distances of the vertices of the ABC from its orthocentre, then prove that (i) a b c abc x y z xyz (ii) x y z (R r) and x R cosa, y R cosb, z R cosc a R sina, b R sinb, c R sinc and a b c tana tan B tan C x y z...(i) abc & xyz tana. tanb. tanc...(ii) We know that in a ABC ΣtanA Π tana From equations (i) and (ii), we get a b c abc x y z xyz x y z R (cosa cosb cosc) A B C in a ABC cosa cosb cosc 4sin sin sin A B C x y z R 4sin. sin. sin A B C A B C R 4R sin. sin. sin r 4R sin sin sin x y z (R r) Self Practice Problems. If Ι be the incentre of ABC, then prove that ΙA. ΙB. ΙC abc tan A tan B tan C.. If x, y, z are respectively be the perpendiculars from the circumcentre to the sides of ABC, then prove that a b c x y z abc. 4xyz. 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 A, B and C. (ii) Its sides are a cosa R sin A, b cosb R sin B and c cosc R sin C (iii) Circumradii of the triangles PBC, PCA, PAB and ABC are equal. 4. Excentral Triangle: The triangle formed by joining the three excentres Ι, Ι and Ι of ABC is called the excentral or excentric triangle. (i) ABC is the pedal triangle of the Ι Ι Ι. (ii) Its angles are A, B & C Page : 0 of 9 (0755) , G SIR PH: (0755) , (iii) Its sides are 4 R cos A, 4 R cos B & 4 R cos C.

21 (iv) Ι Ι 4 R sin A ; (v) Ι Ι 4 R sin B ; Ι Ι 4 R sin C. Incentre Ι of ABC is the orthocentre of the excentral Ι Ι Ι. 5. Distance Between Special Points : (i) Distance between circumcentre and orthocentre OH R ( 8 cosa cos B cos C) (ii) Distance between circumcentre and incentre A B C OΙ R ( 8 sin sin sin ) R Rr (iii) Distance between circumcentre and centroid OG R 9 (a b c ) Example : In Ι is the incentre and Ι, Ι, Ι are the centres of escribed circles of the ABC, prove that (i) ΙΙ. ΙΙ. ΙΙ 6R r (ii) ΙΙ Ι Ι ΙΙ Ι Ι ΙΙ Ι Ι (i) We know that A B C ΙΙ a sec, ΙΙ b sec and ΙΙ c sec Ι Ι c. cosec C, Ι Ι a cosec A and Ι Ι b cosec B ΙΙ. ΙΙ. ΙΙ abc sec A sec B.sec C a R sin A, b R sinb and c R sinc equation (i) becomes...(i) ΙΙ. ΙΙ. ΙΙ (R sin A) (R sin B) (R sinc) sec A sec B sec C A A B B C C sin cos sin cos sin cos 8R. A B C cos. cos.cos A B C A B C 64R sin sin sin r 4R sin sin sin ΙΙ. ΙΙ. ΙΙ 6R r Hence Proved (ii) ΙΙ Ι Ι ΙΙ Ι Ι ΙΙ Ι Ι A A a ΙΙ Ι Ι a sec a cosec A A sin cos A A 6 R sin.cos A A a R sina 4R sin cos ΙΙ Ι Ι A A sin.cos Similarly we can prove ΙΙ Ι Ι ΙΙ Ι Ι 6R Hence ΙΙ Ι Ι ΙΙ Ι Ι ΙΙ Ι Ι Self Practice Problem : 6R. In a ABC, if b cm, c cm and A, then find distance between its circumcentre and 6 incentre Page : of 9 (0755) , G SIR PH: (0755) , cm Part : (A) Only one correct option Exercise -. 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 9. In a ABC, A, b c cm and ar (ABC) cm. Then a is (A) 6 cm (B) 9 cm (C) 8 cm (D) none of these

22 b c. If R denotes circumradius, then in ABC, is equal to a R (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) (B) (C) (D) 6 acosa bcosb ccosc 5. In a ABC, the value of is equal to: a b c (A) r R (B) R r 6. In a right angled triangle R is equal to (A) s r (B) s r (C) R r (C) s r (D) r R 7. In a ABC, the inradius and three exradii are r, r, r and r respectively. In usual notations the value of r. r. r. r is equal to abc (A) (B) (C) (D) none of these 4R 8. In a triangle if r > r > r, 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 KR r r r r r r, a b c where 'K' has the value equal to: (A) (B) 6 (C) 64 (D) 8 0. The product of the arithmetic mean of the lengths of the sides of a triangle and harmonic mean of the lengths of the altitudes of the triangle is equal to: (A) (B) (C) (D) 4. In a triangle ABC, right angled at B, the inradius is: AB BC AC AB AC BC AB BC AC (A) (B) (C) (D) None. The distance between the middle point of BC and the foot of the perpendicular from A is : (A) a b a c b c b (B) (C) a c bc (D) s r a (D) none of these. In a triangle ABC, B 60 and C 45. Let D divides BC internally in the ratio :, then, (A) (B) (C) 6 (D) sin BAD sin CAD 4. Let f, g, h be the lengths of the perpendiculars from the circumcentre of the ABC on the sides a, b and a b c abc c respectively. If λ then the value of λ is: f g h fgh (A) /4 (B) / (C) (D) Page : of 9 (0755) , G SIR PH: (0755) , 5. A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length, 4 and 5 units. Then area of the triangle is equal to: 9 ( ) 9 ( ) 9 ( ) 9 ( ) (A) (B) (C) (D) 6. If in a triangle ABC, the line joining the circumcentre and incentre is parallel to BC, then cos B cos C is equal to: (A) 0 (B) (C) (D) none of these 7. If the incircle of the ABC touches its sides respectively at L, M and N and if x, y, z be the circumradii of the triangles MIN, NIL and LIM where I is the incentre then the product xyz is equal to: (A) R r (B) r R (C) R r (D) r R

23 8. r If in a ABC, r A, then the value of tan (A) (B) B C tan tan is equal to : (C) (D) None of these r r r 9. In any ABC, then minimum value of is equal to r (A) (B) 9 (C) 7 (D) None of these 0. In a acute angled triangle ABC, AP is the altitude. Circle drawn with AP as its diameter cuts the sides AB and AC at D and E respectively, then length DE is equal to (A) (B) (C) (D) R R 4R R. AA, BB and CC are the medians of triangle ABC whose centroid is G. If the concyclic, then points A, C, G and B are (A) b a c (B) c a b (C) a b c (D) None of these. In a ABC, a, b, A are given and c, c are two values of the third side c. The sum of the areas of two triangles with sides a, b, c and a, b, c is (A) b sin A (B) a sin A (C) b sin A (D) none of these. In a triangle ABC, let C. If r is the inradius and R is the circumradius of the triangle, then (r R) is equal to [IIT - 000] (A) a b c (B) b c (C) c a (D) a b c 4. Which of the following pieces of data does NOT uniquely determine an acute - angled triangle ABC (R being the radius of the circumcircle )? [IIT - 00] (A) a, sin A, sin B (B) a, b, c (C) a, sin B, R (D) a, sin A, R 5. If the angles of a triangle are in the ratio 4 : :, then the ratio of the longest side to the perimeter is (A) : ( ) (B) : 6 (C) : (D) : [IIT - 00] 6. The sides of a triangle are in the ratio : :, then the angle of the triangle are in the ratio [IIT - 004] (A) : : 5 (B) : : 4 (C) : : (D) : : 7. In an equilateral triangle, coincs of radii unit each are kept so that they touche each other and also the sides of the triangle. Area of the triangle is [IIT - 005] (A) 4 (B) 6 4 (C) 7 4 (D) PA PB PC PD 8. If P is a point on C and Q is a point on C, then equals QA QB QC QD (A) / (B) /4 (C) 5/6 (D) 7/8 9. A circle C touches a line L and circle C externally. If C and C are on the same side of the line L, then locus of the centre of circle C is (A) an ellipse (B) a circle (C) a parabola (D) a hyperbola Page : of 9 (0755) , G SIR PH: (0755) , 0. Let " be a line through A and parallel to BD. A point S moves such that its distance from the line BD and the vertex A are equal. If the locus of S meets AC in A, and " in A and A, then area of A A A is (A) 0.5 (unit) (B) 0.75 (unit) (C) (unit) (D) (/) (unit) Part : (B) May have more than one options correct. In a ABC, following relations hold good. In which case(s) the triangle is a right angled triangle? (A) r r r r (B) a b c 8 R (C) r s (D) R r r. In a triangle ABC, with usual notations the length of the bisector of angle A is : bc cos A sin A A bc abc cosec (A) (B) (C) (D).cos ec A b c b c R(b c) b c. AD, BE and CF are the perpendiculars from the angular points of a ABC upon the opposite sides, then :

24 (A) Perimeter of DEF Perimeter of ABC r R (C) Area of AEF cos A (B) Area of DEF cosa cosb cosc (D) Circum radius of DEF R 4. The product of the distances of the incentre from the ( angular points of a ABC is: abc ) R ( abc ) r (A) 4 R r (B) 4 Rr (C) (D) s s 5. In a triangle ABC, points D and E are taken on side BC such that BD DE EC. If angle ADE angle AED θ, then: (A) tanθ tan B (B) tanθ tanc 6 tanθ (C) tan A (D) angle B angle C tan θ 9 6. With usual notation, in a ABC the value of Π (r r) can be simplified as: (A) abc Π tan A (B) 4 r R ( abc) (C) R( a b c) Exercise - (D) 4 R r cosa cosc. If in a triangle ABC, sin B, prove that the triangle ABC is either isosceles or right cosa cosb sinc angled.. In a triangle ABC, if a tan A b tan B (a b) tan A B, prove that triangle is isosceles.. r If r then prove that the triangle is the right triangle. r r 4. In a ABC, C 60 & A 75. If D is a point on AC such that the area of the BAD is times the area of the BCD, find the ABD. 5. The radii r, r, r of escribed circles of a triangle ABC are in harmonic progression. If its area is 4 sq. cm and its perimeter is 4 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 ( c ) a cos A. cos C. ac 7. Two circles, of radii a and b, cut each other at an angle θ. Prove that the length of the common chord is ab sinθ. a b abcosθ 8. In the triangle ABC, lines OA, OB and OC are drawn so that the angles OAB, OBC and OCA are each equal to ω, prove that (i) (ii) cot ω cot A cot B cot C cosec ω cosec A cosec B cosec 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 R a, R b, R c ; in radii r a, r b, r c & ex radii r a, r b & r c respectively. Prove that; [ ( R 6r r )] a a a A (i) Π r a : Π R a : Π r a : 8: 7 (ii) r r r Πtan 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, : ( ). Find the acute angles B & C. Also find the ratio of the two sides of the triangle other than the hypotenuse.. 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, : ( ). Find the acute angles B & C. Also find the ratio of the two sides of the triangle other than the hypotenuse.. If the circumcentre of the ABC lies on its incircle then prove that, cosa cosb cosc. Three circles, whose radii area a, b and c, touch one another externally and the tangents at their points of contact meet in a point; prove that the distance of this point from either of their points of contacts abc is. a b c Page : 4 of 9 (0755) , G SIR PH: (0755) ,

25 Exercise -. C. B. B 4. D 5. A 6. B 7. B 8. A 9. C 0. B. A. B. C 4. A 5. A 6. B 7. C 8. B 9. C 0. D. C. A. A 4. D 5. A 6. A 7. B 8. B 9. C 0. C. ABCD. ACD. ABCD 4. BD 5. ACD 6. ACD Exercise - 4. ABD , 8, 0 cms 5 b 5 b 0. B, C, c. B, C, c Page : 5 of 9 (0755) , G SIR PH: (0755) ,