Properties of Triangle Exercise # 1

Properties of triangle EXERCISE # 1 Questions based on Q.1 Sine & Cosine Rule Q.4 In a triangle ABC, (a + b + c) (b + c – a) = bc if (A) < 0 (B) > 0 (C) 0 < < 4 (D) > 4 Sol.[C]  (b + c)2 – a2 = bc  b2 + c2 – a2 + 2bc = bc  2bc cos A + 2bc = bc (from cosine rule)  2(cos A + 1) =    cos A = –1 2 But – 1 < cos A < 1   –1< –1<1 0< < 2  0 <  < 4 2 2 Q.2 a 2  b2 sin (A  B) = , 2 2 sin (A  B) a b then the triangle is (A) Right angled or isosceles (B) Right angled and isosceles (C) Equilateral (D) None of these Sol.[A] sin (A  B) sin 2 A  sin 2 B = sin (A  B) sin 2 A  sin 2 B  sin( A  B) sin( A  B) sin( A  B)  sin C sin 2 A  sin 2 B sin C  1    sin (A – B)  =0 2 2   sin C sin A  sin B  Let ABC be a triangle such that A = 45º, either sin (A – B) = 0  A = B B = 75º then a + c 2 is equal to (A) 0 (B) b (C) 2b (D) –b or sin2A + sin2B = sin2C Sol.[C]   a2 + b 2 = c 2 a b c   sin A sin B sin C   Triangle is right angled or isosceles a b c   sin 45 sin 75 sin 60  a 2 a= b. 2 2  3 1 2b 3 1  a +c 2  Q.3 If in a triangle ABC, Q.5 a 2 sin( B  C) b 2 sin( C  A) c 2 sin( A  B)   sin B  sin C sin C  sin A sin A  sin B is equal to (A) a + b + c (B) a + b – c (C) a – b + c (D) 0 2c 3 6b , c= 2b 3 1 3 1  2b 3 3 1 In any triangle ABC, Sol.[D] From sine rule we take 1st term = 2b  2 sin 2 A sin( B  C) sin B  sin C  2 sin A sin( B  C) sin( B  C) sin B  sin C  2 sin A(sin 2 B  sin 2 C) sin B  sin C Angles A, B and C of a triangle ABC are in A.P. If (A)  6 3 b = , then angle A is equal to c 2 (B)  4 (C) Sol.[C] Given that 2B = A + C and 5 12 b  c (D)  2  2 sin A (sin B – sin C) similarly we solve remain term and add we get 2 [sin A (sin B – sin C) + sin B (sin C – sin A) + sin C (sin A – sin B)] = 0 3 2 b c b c    sin 60 sin 45 3 / 2 1/ 2 B = 60°, C = 45° 5 = 120° = A + 45°  A = 75° = 12  Q.6 The expression (a  b  c) ( b  c  a ) (c  a  b ) (a  b  c) 4b 2 c 2 is equal to (A) cos2 A (B) sin2 A (C) cosA cosB cosC (D) None of these Sol.[B] E = = = Sol.[A] (s  a ) (s  b) =1 s(s  c) C C  tan2   = 1  tan =1 2 2 [(b  c) 2  a 2 ][a 2  (b  c) 2 ]  2 2 4b c [b 2  c 2  a 2  2bc][a 2  b 2  c 2  2bc] 4b 2 c 2 Questions based on Area of Triangle, Solutions of Triangle, m–n theorem & Ambiguous case Q.10 In a  ABC, if  [(b 2  c 2  a 2 ) 2  4b 2 c 2 ] 4b 2 c 2 2  2  b  c 2  a 2   1 =      2bc    2 2 = – [cos A – 1] = sin A cos A cos B cos C , and the   a b c side a = 2, then area of the triangle is - (A) 1 Sol.[D]  Questions based on Q.7 Projection formula, Napier's Analogy, & Half-Angled Formulae (D) 15º  BC bc  A  cot  Sol.[A]  tan    2  bc  2   tan 45° =  cot  3 1 A cot   2 3 1 A 3 1 = cot 15°  2 3 1 (B) 2 3 2 (D) 3 cos A cos B  a b b2  c2  a 2 a 2  c2  b2  a2 = b2 a = b  2abc 2abc cos B cos C b = c  b c so a = b = c = 2 given triangle is equilateral so Area = Q.11 3 2 a = 4 3 We are given b, c and sin B such that B is acute and b < c sin B. Then (A) No triangle is possible (B) One triangle is possible (C) Two triangles are possible (D) A right angled triangle is possible Sol.[A] We know that c sin B b c  sin C =  sin B sin C b given that b < c sin B  sin C > 1 which is impossible  No triangle is possible Q.12 In the adjacent figure 'P' is any interior point of the equilateral triangle ABC of side length 2 unit A A = 15°  A = 30° 2 If in a triangle ABC, (s – a)(s – b) = s(s – c), then angle C is equal to (A) 90º (B) 45º (C) 30º (D) 60º (C) similarly If in a triangle ABC, b = 3 , c = 1 and B – C = 90º then A is (A) 30º (B) 45º (C) 75º Q.9  In a triangle ABC, a2 cos 2B + b2 cos 2A + 2ab cos(A – B) = (A) a2 (B) c2 (C) b2 (D) a2 + b2 Sol.[B] a2(2 cos2 B – 1) + b2 (2 cos2A – 1) + 2ab cos (A – B) = 2(a2 cos2 B + b2 cos2A) – a2 – b2 + 2ab cos (A – B) = 2(a cos B + b cos A)2 – a2 – b2 + 4ab cos A cos B + 2ab cos (A – B) = 2c2 – a2 – b2 – 2ab cos A cos B + 2ab sin A sin B = 2c2 – a2 – b2 – 2ab cos (A + B) = 2c2 – a2 – b2 + 2ab cos C = 2c2 – c2 = c2 Q.8 C = 45°  C = 90° 2 •P B C If xa, xb and xc represent the distance of P from the sides BC, CA and AB respectively then xa + xb + xc is equal to (A) 6 Sol. (B) 3 (C) 2 3 C 60° (D) 2 3 [B] A A (A) 5 • P (C) B C Given that AB = BC = CA = 2 xa, xb and xc are the distances of P from the sides BC, CA and AB 1 1  area of BPC = BC. xa = .2xa = xa 2 2 similarly Area of CPA = xb and Area of APB = xc then the area of ABC or (ABC) = ar(BPC)+ar(CPA) + ar (APB)  ABC is equilateral ar (ABC) = xa + xb + xc = Q.13 Sol. 3 7 3 cm 7 (B) cm (D) none of these [A] 60° 30° 2 cm A 3 cm D 5 cm 5 3  sin (CDA) = 5 2 7 Apply sine rule in ACD, we get AD AC =  5  sin 60  sin  sin 1 2 7   cot (CDA) = If two sides a, b and the angle A be such that two triangles are formed, then the sum of the two values of the third side is (A) b2 – a2 (B) 2b cosA bc (C) 2b sinA (D) bc AC = 3. 2 3 [B] In the figure, ABC is triangle in which C = 90º and AB = 5 cm. D is a point on AB such that AD = 3 cm and ACD = 60º. Then the length of AC is B Using (m + n) cot  = n cot  – m cot  we get (3 + 2) cot (CDA) = 2 cot 30 – 3 cot 60 3 b2  c2  a 2 2b c 2  c – 2bc cos A + b2 – a2 = 0 Let c1 and c2 be two values of c then c1 + c2 = 2b cos A 3 cm 7 C  AB = 2 cos A = Q.14 Sol. 3 B 3 cm D 5 cm . 5 =5 2 7 3 7  Fill in The Blanks type Questions Q.15 If one angle of a triangle be 60º, the area 10 3 sq. cm and the perimeter 20 cm, then the lengths of the sides are .......... Sol. Let A = 60° given = 10 3 sq. cm. and a + b + c = 20 cm We know that  =  1 bc sin A 2 1 bc sin 60° = 10 3  bc = 40 2 b 2  c 2  a 2 (b  c) 2  2bc  a 2 = 2bc 2bc  b + c = 20 – a, A = 60°, bc = 40°  cos A = 1 (20  a ) 2  80  a 2 = 80 2  3 1 cos B  sin B 2 2 =2 sin B  (b + c)2 = 169  (b –c)2 = 169 – 4bc  b – c = 3 on solving we get b = 8, c = 5 side are 7, 8, 5  3 3 cot B = 2 2 If the sines of the angles of a triangle are in the ratio 3 : 5 : 7, their cotangents are in the ratio .............. and A = 90°   40 = 400 – 80 – 40a  a = 7  a + b + c = 20  b + c = 13 Q.16 Sol.  cot B = Q.18 C a : b : c = 3 : 5 : 7 or a = 3k, b = 5k, c = 7k = b2  c2  a 2 2bc Sol. 4 5 2 1 bc sin A  sin A = bc 2 cot A = (True) If the sides of a triangle are 4m, 5m, 6m then the smallest angle is half of the greatest angle. By sine rule we have cos A = 3  B = 30° A 6 cos A b 2  c 2  a 2 65k 2 = = 4 4 sin A cos A = 36  25  16 45 3   60 60 4 c 2  a 2  b 2 33k 2 = 4 4 cos C = 25  16  36 1  40 8 similarly cot B = C = 2 a 2  b2  c2  15k 2 and cot C = = 4 4 cos  cot A : cot B : cot C = 65 : 33 : – 15 A= B 1  cos C 3  2 4 C True 2  True or False type Questions Q.17 If one side of a triangle is double of another side and the angles opposite to these differ by 60°, then the triangle is right angled. Sol. Let a = 2b and A – B = 60° Then by sine rule 2b b = sin (60  B) sin B  sin (60  B) =2 sin B EXERCISE # 1 Q.No. Ans. 15. 5, 7, 8 1 C 2 C 3 C 16. 65 : 33 : –15 4 A 5 D 6 B 17. True 7 B 8 A 9 A 10 D 11 A 18. True 12 B 13 B 14 A