Properties of Triangle Exercise # 5

EXERCISE # 5 Q.1 A ABCD is a trapezium such that AB, DC are parallel and BC is perpendicular to them. If angle ADB = , BC = p & CD = q. Show that AB =   (p 2  q 2 ) sin  p cos   q sin  6k Sol.   A Given BD : CD : AD = 2 : 3 : 6  BD = 2k, CD = 3k, AD = 6k from right triangle ADB, we have B Let BDC =  = DBA so DAB =  – ( + ) and tan  = tan  = p q tan  = BD AB AD = = sin  sin(   (  )) sin  BD sin  sin (  )  AB = BD 2 sin  …(i) BD sin  cos   BD cos  sin  A = 45° Hence proved. Q.3 Sol. ABC is a triangle. D is the middle point of BC. If AD is perpendicular to AC, then prove that cos A .cos C = 2(c 2  a 2 ) . 3ac E (p 2  q 2 ) sin  p cos   q sin  Hence proved. The perpendicular AD to the base of a triangle ABC divides it into segments such that BD, CD and AD are in the ratio of 2, 3, and 6; prove that the vertical angle of the triangle is 45º. tan   tan  1  tan  tan   tan A = 1 from right triangle DCB, we have BD2 = p2 + q2, p = BD sin , q = BD cos  put these values in (i), we get Q.2 3k 1 = 6k 2  tan (+ ) = tan A = BD AB = sin  sin (  )  AB = 2k 1 = 6k 3 and from ADC, we have Now from ABD AB = C Let BAD =  and DAC =  A =  + p  3k D C  – (+  2k B q D 2y Sol. 90° b A  –A 90° c b y B C a/2 D a/2  Median AD perpendicular to AC Draw a perpendicular on extended AC such that AE = AC = b From property of triangle If AD = y then BE = 2y C  cos C = b 2b = a/2 a cos ( – A) = it gives a = a=2 3± 2 b b  cos A = – c c cos A . cos C = – area = 2 2b ac 4 3  48  40 2 … (i) 1 ab sin 30 2 =2 3± 2 in BEA and DAC, we have 4y2 + b2 = c2 and y2 + b2 = Q.5 a2 4 = 2 3 sin C, then the angle A is ….. [REE 96] On subtracting, we get 3b2 = a2 – c2  b2 = Sol. a 2  c2 3 so from (i) cos A. cos C = If in a triangle ABC, 3 sin A = 6 sin B [D] Given 3 sin A = 6 sin B = 2 3 sin C  2(c 2  a 2 ) 3ac 1 sin A 2 sin B = and = sin B 1 sin C 3  sin A : sin B : sin C = a : b : c =2:1: Hence proved. 3  a = 2k, b = k, c = Q.4 Two sides of a triangle are of lengths 6 and 4 and the angle opposite to smaller sides is 30º. How many such triangles are possible? Find the lengths of their third side and area. [REE 98] cos A = Q.6 If the angles of a triangle are 30º and 45º and the included side is ( 3 + 1) cms, then the area of the triangle is ................ [IIT-1988] 4 6 b 2  c 2  a 2 k 2  3k 2  4k 2 = 2bc 2 3k 2 cos A = 0 A = 90° Sol. A 3k Sol. 30° B  a C 4 6 = sin B sin 30  sin B = 2 6 = 2 3 B have two values which is supplementary so that two triangle are possible. Now cos C =  Q.7 a 2  b2  c2 2ab 3 a 2  16  6 = 2 2.4.a  a2 – 4 3 a + 10 = 0 Sol. If in a triangle ABC, [IIT-1993] 2 cos A cos B 2 cos C a b =    a b c bc ca then the value of the angle A is ............. 2 (b 2  c 2  a 2 ) a 2  c 2  b 2 + 2abc 2abc 2(a 2  b 2  c 2 ) a 2  b 2 = 2abc abc 2 2 2 2 2 2  2(b + c –a + a + b – c ) + a2 + c2 – b2 = 2a2 + 2b2 2 2 2  4b – b + a + c2 = 2a2 + 2b2  b2 + c2 = a2 ; A = 90° + Q.8 Sol. In a triangle ABC, AD is the altitude from A. abc Given b > c, C = 23º and AD = 2 , b  c2 then B = ........... [IIT-1994] A  We have AD = b sin C abc given AD = 2 b  c2 sin C abc a  2 = b sin C  = 2 2 c b c b  c2 sin A sin C sin A a  = 2  = 2 a c a b c  sin A = a2 b2  c2  sin A = B sin 2 B  sin 2 C Using A.M. > G.M. ssa sbsc > 4 4s  (a  b  c) >  4 4s  2s   4 s   2 2 4 Q.10  1 1 1    = 4 1  2     t2 t3   t2  Let BP = PQ = QC = x Also, t1 = tan , t2 = tan , t3 = tan  1 1     t1 t 2  Sol. 4(cot  cot   1) – 2 cot  cot   cot  3 cot– cot  cot  + 3 cot  cot  – cot  cot  = 4 cot  cot  – 4 4 + 4 cot2 = cot2 + cot  cot  + cot  cot  + cot  cot  4(1 + cot2) = (cot  + cot ) (cot  + cot ) 3 cot  – 3 cot  =     1 1 1 4 1  2  =     t   t1 t 2  2   Q.11 Sol. 1 1    t   2 t3  If in a triangle ABC,  is the angle determined by cos = (a – b)/c, prove that (a  b) sin  1 cos (A – B) = and 2 2 ab cos 2 The base of a triangle is divided into three equal parts. If t1, t2, t3 be the tangents of the angles subtended by these parts at the opposite vertex, prove that .....(2) 2 cot(  )  cot  3 = cot   cot  2 (s)(s  a)(s  b)(s  c) s s >   < 4 4  C Also Apply m : n Theorem in APC (x + x)cot  = x cot  – x cot  Divide (1) by (2) sin 2 A sin( B  C) sin( B  C)  sin (B – C) = 1  B – C = 90°  B = 90° + C = 90° + 23° = 113° Sol. Q P (2x + x) cot  = 2 x cot ( + ) – x cot  .....(1) sin 2 A If  denotes the area of any triangle and s its s2 semi perimeter, prove that  < . 4  Apply m : n theorem in ABC we get  sin A = Q.9  c sin  1 (A + B) = 2 2 ab ab (given) c On squaring  cos  = cos2  = a 2  b 2  2ab c2  a 2  b 2  2ab   1 – cos2  = 1 –    c2   sin2  = sin2 = 2ab  (a 2  b 2  c 2 ) c2 2ab (1  cos C) c2  sin  = ab C  2  2 sin  2  = c sin C / 2 sin   = c 2 ab  If C = 60º then A = 90º  triangle is right angled and if, C = 120º then A = 30º = B  a = b  triangle is isosceles when, A = 90º, C = 60º using sine rule, 2 1  cos C c (a  b) sin  .....(1) C a b =   sin 2  c  2 ab c a  sin C sin A C  sin A  sin B  =  [use sine rule]  sin sin C 2   150 C  A  B  A  B 2 sin   cos  sin 2  2   2  = C C 2 sin cos 2 2 AB = cos    2   a = 100 3 If B = 30º, C = 150 and b = 75 then, using sine rule, b c  sin B sin C Proved. 75 150  1 / 2 sin C sin C = 1  C = 90º  only one triangle is possible.  This would not be an ambiguous solution. Also from (1) c sin  = sin 2 ab Q.12 C AB = cos   2  2  =a 3/2 Proved. Given B = 30º, c = 150 and b = 50 3 , prove that of the two triangles which satisfy the data, one will be isosceles and the other right angled. Find the greater value of the third side. Would the solution have been ambiguous had the values been B = 30º, c = 150 and b = 75 ? Sol. Given :  B = 30º, b = 50 3 , c = 150 using sine rule, b c  sin B sin C 50 3 150  1/ 2 sin C 3 2 C = 60º or 120º sin C = EXERCISE # 5 4. Two triangles are possible, 2 3  2 5. 90° 12. 100 3 , No 6. 3 1 Sq. units 2 7. 90º 8. 113º