Properties of Triangle Exercise # 3

EXERCISE # 3 ac Part-A Subjective Type Questions Q.1 Sol.  = a sin (B  C) b2  c2 = = k (sin B  sin C) 2 similarly 2 b sin (C  A) c a Hence proved. 2 2 sin = k (sin 2 B  sin 2 C) sin (B  C) sin( B  C) 2 = AC AC 2 = 2 cos sin 60 2 = k (sin B  sin C) 2 c sin (A  B) 1 1 and = k k a 2  b2 L.H.S. Hence proved. Q.4 Find the greatest angle of the triangle whose sides are x2 + x + 1, 2x + 1, x2 –1. Sol.  sides are positive.  x2 – 1 > 0  (x –1) (x + 1) > 0 x > 1 or x < –1 In a triangle ABC, the side c has two values. Prove that both the values satisfy the equation But x  – 1  a = x2 + x + 1 > x + x + 1 Taking L.H.S. we have  x2 + x + 1 > 2x + 1 (1  cos C)(a  b) 2  (1  cos C)(b  a ) 2 a>b and a – c = x2 + x + 1 – x2 + 1 > 0 1  cos 2 C = 2(a 2  b 2 )  4ab cos C sin 2 C =  greatest side is a 2c 2 sin 2 C so cos A = Then we get 2  a   c  2    =2   sin A   sin C  2 =  Hence both values of c satisfy the equation. Sol. (2x  1) 2  ( x 2  1) 2  ( x 2  x  1) 2 cos A = – In a triangle, the angles A, B, C are in AP. ac AC Show that 2 cos = 2 [a 2  ac  c 2 ] b2  c2  a 2 2bc 2(2x  1)(x 2  1) On solving we get Hence proved. Q.3  2x + 1 becomes – ve So x > 1 (a  b) 2 ( b  a ) 2 2a 2 + = 1  cos C 1  cos C sin 2 A Sol. AC AC  cos 2 2 sin B 2 sin 60 cos sin 2 B  sin 2 C 2 a  c sin A  sin C = sin B b using sine rule sin A sin (B  C) 1 = k Q.2 a 2  ac  c 2 In any ABC, prove that a sin (B  C) b sin (C  A ) c sin ( A  B) = = b2  c2 c2  a 2 a 2  b2 = Q.5 1  A = 120° 2 The two adjacent sides of cyclic quadrilateral are 2 and 5 and the angle between them is 60º. A, B, C are in A.P. If the area of the quadrilateral is 4 3 , find the so 2B = A + C remaining two sides. But A + B + C =  B = 60° and A + C = 120° Now b2 = c2 + a2 –2ca cos B b2 = c2 + a2 – ac From R.H.S. we have  B = 60° C Sol. D Sol. Taking R.H.S. we have k sin C cos (A – ) + k sin A cos (C + ) = k [sin C cos A cos  + sin C sin A sin  + sin A cos C cos  – sin A sin C sin ] = k cos  [sin C cos A + cos C sin A] = k cos  sin (A + C) = k cos  sin B = b cos  L.H.S. Hence proved. Q.8 In any ABC, prove that a(cos B + cos C – 1) + b(cos C + cos A – 1) + c(cos A + cos B – 1) = 0. Sol. Taking L.H.S. we have a cos B + a cos C – a + b cos C + b cos A – b + c cos A + c cos B – C = (a cos B + b cos A) + (a cos C + c cos A) + (b cos C + c cos B) – a – b – c =c+b+a–a–b–c by projection formula = 0 Q.9 In any ABC, prove that B 2 5 60° A Let BC = a, DC = b, BD = c then from ABD weight 25  4  c 2  c2 = 19 ….(1) 20 cos 60 = given that area of ABCD = 4 3 1 1 . 2.5 sin 60 + . a.b sin 120 = 4 3 2 2  ab = 6 …….(2) from BCD we have  a 2  b2  c2 2ab 2 2  a + b = 13 from (2) & (3) we get a = 2, b = 3 or a = 3, b = 2 cos 120 = Q.6 In any  ABC, prove that b sec B  c sec C c secC  a sec A = tan B  tan C tan C  tan A = Sol. …….(3) a sec A  b sec B . tan A  tan B Taking = b cos C  c cos B sin B cos C  cos C sin B = a a = sin (B  C) sin A a sec A  b sec B c = tan A  tan B sin C from sine rule, we have a b c = = sin A sin B sin C and In any  ABC, prove that, if  be any angle, then b cos = c cos(A – ) + a cos(C + ). 1 A   +  2  (b  c) (b  a ) 1 1  B C tan   + tan   = . 2 2  ( c  a ) ( c  b )     Sol. Taking L.H.S. tan b sec B  c sec C tan B  tan C (By projection formula) c sec C  a sec A b Similarly = tan C  tan A sin B Q.7 1 tan ( a  b ) ( a  c)  A = 2 (s  b) (s  c) etc. s (s  a ) (s  b) (s  c) s (s  a ) ( a  b ) ( a  c) + + (s  a ) (s  c) s (s  b) ( b  c) ( b  a ) (s  a ) (s  b) s (s  c) (c  a ) (c  b )  (s –b) (s –c) (c – b) + (a – c) (s – a) (s – c) + (b – a) (s – a) (s – b) _____________________________________ (a – b) (b –c) (c –a) s(s  a) (s  b) (s  c) On solving we get = 1 s(s  a ) (s  b) (s  c) R.H.S. Hence proved. = 1  Q.10 Sol. one triangle we can formed which is right angled C A In any ABC, if 3 tan   .tan   = 1, prove 2 2 that a, b, c are in A. P. A C 3 tan .tan = 1 2 2 3 (s  b) (s  c) s(s  a ) (s  a ) (s  b) =1 s(s  c) at B. Q.13 If a = b sin A and A is obtuse (A) 1 (B) 2 (C) 0 (D) Infinite Sol. [C] If a = b sin A and A obtuse.  3 (s – b) = s No any triangle are formed.  2s = 3b  2b = a + c a, b, c are in A.P. Hence proved. Part-B Passage based objective questions Passage-I (Q. 11 to 13) In a  ABC two sides a, b and A are given. Then the numbers of triangles that are formed in the following cases are On the basis of above passage, answer the following questions : Q.11 Sol. If a < b sin A (A) 1 (C) 0 Passage-II (Q. 14 to 16) The internal bisectors of the angle of ABC meet the sides BC, CA and AB at D, E and F respectively. Then On the basis of above passage, answer the following questions : Q.14 Sol. The length BE is given by (A) 2bc A cos   bc 2 (B) bc A cos   bc 2 (C) 2ac  B cos   ac 2 (D) ca  B cos   ca 2 [C] B (B) 2 (D) Infinite a [C] B/2 B/2 c From sine rule, we have b sin B = sin A a C …(i) A   = 1 + 2  sin B > 1 which is impossible.  No triangles are formed. Sol. c E Let BE = x If a < b sin A Q.12 a If a = b sin A and A is acute (A) 1 (B) 2 (C) 0 (D) Infinite 1 1 B 1 B ac sin B = ax sin + cx sin 2 2 2 2 2 x= 2ac cos ac sin B (a  c) sin B 2 = ac B 2 [A] If a = b sin A and A acute then from (i), we get sin B = 1 Q.15 A B C 1 1 1 cos + cos + cos = CF BE AD 2 2 2 1 1 1 + + a b c (C) a + b + c [A] (A) Sol. (B) abc (D) ab + bc + ca ac bc Similarly, BD = we know that BE = B A 2ac 2 bc cos , AD = cos , ac bc 2 2 CF = 2ab C cos ab 2  1 A 1 B C 1 cos + cos + cos AD 2 CF 2 2 BE = bc ac ab + + 2 bc 2ac 2ab  y= ....(3) c ab BF = ay = BF = ac ab ac ab (from (5) .....(6) 1 × BD × BF sin B 2 1 ac ac = × × sin B ab bc 2 ar (BDF) = 1 1 1 = + + a b c Area of BDF is given by ac sin B ac (A) (B) (a  b)(b  c) 2(a  b) ac sin B (a  b)(b  c) (C) (from (1)) AF BF = = y (let) b a AF = by ......(4) and BF = ay ......(5) AF + BF = y(a + b) = c {From (4) & (5)) 2(ab  bc  ca) = 2abc Q.16 ac bc BD = cx = Sol.[B] Area of BDF = (D) =  ac , ( = area of ABC) (a  b)(b  c) ab sin C (a  b ) 1 × BD × BF × sin B 2 A c F B E b D a C c BD = = x (let) CD b  BD CD = =x b c  BD = cx .....(1) and CD = bx ......(2)  BD + CD = cx + bx = a (from (1) & (2))  (x) (b + c) = a x= a bc EXERCISE # 3 4. 120º 5. 2 & 3 11. (C) 12. (A) 13. (C) 14. (C) 15. (A) 16. (B)