Properties of Triangle Exercise # 2

EXERCISE # 2 Part-A Only Single Correct Answer type Questions Q.1 Sol.  2 If in a triangle ABC, sinA = sin B and 2 cos2A = 3cos2B, then the ABC is (A) right angled (B) obtuse angled (C) isosceles (D) equilateral .. (i) …(ii) sin B AE sin C AE = and = … (ii) EC sin z BE sin ( x  y) From (i) DC sin B sin ( y  z) = . sin x BD sin C From (ii) sin C sin ( x  y) BE = . EC sin B sin z  [B] sin A = sin2 B 2 cos2A = 3 cos2B …(iii) …(iv) DC BE = 2 and =2 EC BD Then from (iii) and (iv), we get sin ( x  y) sin ( y  z) =4 sin x sin z from (i) and (ii) 2 (1 – sin2A) = 3 (1 – sin A)  2 sin2A – 3 sin A + 1 = 0 Q.3 If the base angles of a triangle are 22½º and 112½º, the height of the triangle is equal to (A) half the base (B) the base (C) twice the base (D) four times the base Sol. [A]  (2 sin A –1) (sin A –1) = 0 sin A = and 1 or sin A = 1 2 But sin A  1  A = 30° A from (i) sin2B = 1 2 45°  B = 45° and C = 180° – 75° = 105° ABC is obtuse angled. Q.2 Points D, E are taken on the side BC of a triangle ABC such that BD = DE = EC. If BAD = x, DAE = y, EAC = z, then the sin( x  y) sin( y  z) value of = sin x sin z (A) 1 Sol. 112½° 67½° 22½° B C D (B) 2 (C) 4 From sine rule, we have BC AC = sin 45 sin 22½ Also (D) none [C] z x y B  P P D AD = sin 67½° AC  AD = AC cos 22½° A P …(i) E  AD = BC sin 22½° cos 22½° from (i) sin 45  AD = BC sin 45 . 2 sin 45  AD = 1 1 BC = of base 2 2 C sin C AD sin B AD = and = …(i) sin x DC sin( y  z) BD Q.4 In a ABC if median AD is perpendicular to AB, then tan A + 2 tan B is equal to - (A) 1 Sol. (B) 3 (C) 0 (D)1/2 then b2 + c2 = [C] B triangle is right triangled, A = 90° x cos B = z D y 90° x A C Q.6 z 2y 90° N DAB = 90° Given that BD = DC ; Sol. C 1 = ; B = 60° and C = 30° 1 2 If in a ABC, a2 cos2 A = b2 + c2, then   (A) A < (B) < A< 2 4 2 (C) A > Draw a perpendicular CN at extend BA In BCN ; DA =  2 (D) A = Given a2 cos2 A = b2 + c2 1 CN = y and AB = AN = z 2 we know that cos A = 2AD  tan A = – AB or cos A = –  tan A = – 2 tan B A >  tan A + 2 tan B = 0 Q.5 2 Sol. a 2 sin 2 A = – ve 2bc  2 Q.7 In a ABC, if a4 + b4 + c4 = 2c2 (a2 + b2), then C is equal to(A) 60º (B) 135º (C) 90º (D) 75º Sol. [B] 2 In a triangle, a + b + c = ca + ab 3 , then the triangle is (A) equilateral (B) right-angled and isosceles (C) right-angled with A = 90º, B = 60º, C = 30º (D) None of these b2  c2  a 2 2bc a 2 cos2 A  a 2 2bc  cos A = CN 2y =– AN z 2  2 [C]  tan A = tan ( – CAN) = – tan (CAN) =– a2 3a 2 + = a2 4 4 a4 + b4 + c4 –2a2c2 –2b2c2 = 0 a4 + b4 + c4 –2a2c2 –2b2c2 + 2a2b2 = 2a2b2  (a2 + b2 – c2)2 = 2a2b2 [C]  a 2  b2  c2    2ab  a2 3a 2 Write a = + 4 4 2 2   = 1  2  then  a2    c 2  ca  +  4     3a  a   b = 0    c +    2   2  a and b = 2 1 1  cos C = ± 2 2  C = 45°, 135° 2 2 c=  cos2C =  3a 2    ab 3  b 2  = 0  4    Q.8 3a 2 In ABC, a = 5, b = 4 and tan side c is (A) 6 (B) 3 Sol. [A] (C) 2 C 7 = . The 2 9 (D) None C 2 cos C = 2 C 1  tan 2 1  tan 2 C  tan = 2 B=c then from (i) 7 9 7 9  cos C = 1  cos C = 7 8 1 9  cos C = a b c 2ab  =1 sin 2 B  sin A = 1  A = 90° B = C = 45° triangle is right angled isosceles.  a = 5, b = 4 In a ABC, A = If A is the area and 2s the sum of the sides of a triangle, thens2 s2 (A) A  (B) A  3 3 3 3 (C) A > s2 (D) none of these 3 c=6 ar(ABC) = Sol. 2 , b – c = 3 3 cm and 3 9 3 cm2. Then a is 2 (A) 6 3 cm (C) 18 cm Sol. Q.11 2 1 25  16  c 2 = 8 40  c2 = 36 Q.9 2 1  cos2 B sin2A = 1 2  cos   1 [A]  A2 = s(s – a) (s – b) (s – c) and 2s = a + b + c Now A.M.  G.M. (s  a )  (s  b)  (s  c)  3  [(s – a) (s – b) (s – c)]1/3 3 A2 s2 s     A s 3 3 3 (B) 9 cm (D) None of these [B] Given A = = 2 3 2 , b – c = 3 3 & ar (ABC) = 2 3 1 1 2 9 3 bc sin A  bc sin = 2 3 2 2 Part-B One or More Than one correct Answer Type Questions Q.12 In a ABC, tan A and tan B satisfy the in-equation 3x2 – 4x + 3 < 0. Then(A) a2 + b2 + ab > c2 (B) a2 + b2 – ab < c2 (C) a2 + b2 > c2 (D) none of these Sol. [A, B]  bc = 18 and b – c = 3 3  b2 + c2 = 27 + 2bc = 27 + 36  b2 + c2 = 63  cos A = b2  c2  a 2 2bc 1 2 63  a 2 63  a 2  cos = – = 3 2  18 36 2  a2 = 63 + 18 = 81  a = 9 Q.10 In a ABC, cos B.cos C + sinB. sinC sin2 A = 1. Then the triangle is(A) right-angle isosceles (B) isosceles whose equal angles are greater than /4 (C) equilateral (D) none of these Sol. [A] 1 cos B cos C 1 sin B sin C  1  cos B cos C + sin B sin C  1  cos (B – C)  1 = cos (B – C)  B – C = 0 sin2A = … (i) 3 x2 – 4x + 3 < 0  (x – 3 ) ( 3 x –1) < 0 1  c2 Q.13 Sol. In a ABC, A =  and b : c = 2 : 3. 3 (D) If Assertion is false but Reason is true 3  If tan  = ,0<< , then5 2 (A) B = 60º +  (B) C = 60º +  (C) B = 60º –  (D) C = 60º –  [B, C] CB c b A tan = cot cb 2 2 Q.15 Assertion (A) : If y2 + y + 1, 2y + 1 and y2 – 1 represents lengths of the sides of , then y > 1. Reason (R) : If a, b, c be the length of sides of , then a + b > c, b + c < a, c + a > b. Sol. [C] We know that in any triangle a + b > c, b + c > a, c + a > b (R) is false. and y2 + y + 1 + 2y + 1 > y2 – 1  3y + 3 > 0  y > – 1 … (i) and y2 + y + 1 + y2 – 1 > 2y + 1  2y2 – y – 1 > 0 1  y    ,   (1, ) … (ii) 2  and 2y + 1 + y2 – 1 > y2 + y + 1  y > 1 ….(iii) from (i), (ii) and (iii) we get y > 1 (A) is true. Q.16 Assertion (A) : If , ,  are the lengths of internal bisector of the angles A, B and C respectively of a triangle, then 1 A 1  B 1 C 1 1 1 cos   cos   cos      2  2  2 a b c Reason (R) : Since, the length of angle bisectors bc C  B  ab  A  ca are cos   , cos   , cos   2 2 a  b bc   ca   2 CB 1 3 = cot 30° = = tan  given. 5 5 2  C – B = 2 and C + B = 180° – 60° = 120° Solving we get C = 60° + , B = 60° –   tan Q.14 If in a ABC, a = 6, b = 3 and cos (A – B) = 4 5 then 2 (B) A = sin–1 4 5 (C) ar (ABC) = 9 (D) none of these [B, C] AB ab C 1 C tan = cot = cot ab 2 2 3 2 2 AB 1  tan 2  cos (A – B) = A B 1  tan 2 2 1 2 C 1  cot 4 9 2  tan2 C = 1  C =   = 5 1  1 cot 2 C 2 2 9 2 1 1 ar (ABC) = ab = . 6. 3 = 9 2 2 6 6 Also sin A = = 3 5 32  6 2 (A) C = Sol. A = sin–1 2 5 Part-C Assertion Reason type Questions The following questions 15 to 16 consists of two statements each, printed as Assertion and Reason. While answering these questions you are to choose any one of the following four responses. (A) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (B) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion. (C) If Assertion is true but the Reason is false. A Sol. [C] B c AA  b 2 2  D a C 1 1 A 1 A bc sin A = c sin + b sin 2 2 2 2 2 A A A A  2bc sin cos = c sin + b sin 2 2 2 2 A  2bc cos = (b + c) 2 A cos 2  1  1   2 b 2c B cos 2  1  1 similarly  2a 2c  C 2  1  1 and  2a 2b 1 B 1 A C 1 1 1 1 So cos + cos  cos     2  2 a b c  2  length of bisector is 2 bc A = cos bc 2  A is true but R is false (BE)2 = (BD)2 – (ED)2 (BE)2 = c2 cos2B – c2 sin2B cos2B = c2 cos4B BE = c cos2B 1 Area = c2 cos2B sin2B 4 cos Q.18 Match the following : Column-I Column-II (A) In a triangle ABC, if a is the (P) 9 arithmetic mean and b, c (b  c) are the two geometric means between any two positive real Part-D Column Matching type Questions Q.17 In a triangle ABC, AD is perpendicular to BC and DE is perpendicular to AB. Column-I Column-II (A) Area of ADB (P) (b2/4) sin 2C (B) Area of ADC (Q) (c2/4) cos2B sin 2B (C) Area of ADE (R) (c2/4) sin 2B (D) Area of BDE (S) (c2/4) sin2 B sin 2B Sol. A  R, B  P, C  S, D  Q numbers then is equal to (B) In a triangle ABC; (Q) 2 (a 2  b 2  c 2 ) tan A  (a 2  c 2  b 2 ) tan B 2 is equal to (C) In a triangle ABC, if B = 30º (R) 1 A and C = 3 b, then can 45º be equal to (D) In a ABC, (S) 0 a b c if b c a = 0, then c a b  4 sin A sin B = A E B D C 1 (AD) (BD) 2  c cos B = BD, c sin B = AD (A) Area of ADB = Sol. A  Q, B  S, C  Q, D  P (A) Let number are P, Q 1/ 3 then c2 1 2 c sin B cos B = sin 2B 4 2 1 (B) Area of ADC = (AD) (DC) 2  AD = b sin C, DC = b cos C b2 1 2 b sin C cos C = sin 2C 4 2 1 (C) Area of ADE = (AE) (ED) 2  ED = BD sin B = C sin B cos B  (AE)2 = (AD)2 – (ED)2 = c2 sin4B  AE = c sin2B 1 Area = c2sin2B sin2B 4 1 (D) Area of BDE = (ED) (BE) 2 ED = c sin B cos B PQ Q = a, b = P   2 P Q , c = P  P sin 3 B  sin 3 C b 3  c 3  sin A sin B sin C abc put the values of a,b,c then we get 2(P  Q)PQ =2 (P  Q)PQ Area = Area = sin 3 B  sin 3 C sin A sin B sin C  (a 2  b 2  c 2 ) tan A  (a 2  c 2  b 2 ) tan B 2 2bc cos A tan A  2ac cos B tan B = 2 = – bc sin A + ac sin B = – abc + abc = 0 (C) b sin C = c sin B (B) c= 3 b, B = 30° 3  C = 60° 2  A = 180° – (90°) = 90° sin C = 2/3 A =2 45 (D) solving we get (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = 0 but a + b + c  0 (side of )  a2 + b2 + c2 – ab – bc – ca = 0 1  [(a – b)2 + (b – c)2 + (c – a)2] = 0 2 a=b=c  is equilateral A = B = C = 60° 3 9  sin A sin B = 3 sin2A = 3.  4 4  EXERCISE # 2 PART-A Q.No. Ans. 1 B 2 C 3 A 4 C 5 C 6 C 7 B 8 A 9 B PART-B Q.No. 12 13 14 Ans. A,B B,C B,C PART-C Q.No. Ans. 15 C 16 C PART-D 17. A  R, B  P, C  S, D  Q 18. A  Q, B  S, C  Q, D  P 10 A 11 A