Mathematics-26.Unit-24 Properties of Triangle

INVERSE CIRCULAR FUNCTIONS (INVERSE TRIGONOMETRIC FUNCTIONS) If sin = x, then  may be any angle whose sine is x, and we write  = sin–1x. It means that  is an angle which can be determined from its sine. Thus tan–1 is an angle whose tangent is , i.e.  = tan–1 = n + where is the least positive value of . The functions sin–1 x, cos–1 x, tan–1 x, cot–1 x, sec–1 x and cosec–1 x are called inverse circular or inverse trigonometric functions. If x is positive, the principal values of all the inverse circular functions lie between 0 and . If x is negative, the principal values of sin–1 x, cosec–1 x and tan–1 x lie between and 0, and those of cos–1 x, sec–1 x and cot–1 x lie between From now onwards we take only principal values. Hence sin  = x   = sin–1 x where   and x  [–1, 1] Similarly cos q = x Þ q = cos–1 x where q Î [0, p] and x Î [–1, 1] tan q = x Þ q = tan–1 x where q Î and x Î (–¥, ¥) cot q = x Þ q = cot–1 x where q Î (0, p) and x Î (–¥, ¥) sec q = x Þ q = sec–1 x where q Î È and x Î (–¥, –1] È [1, ¥) cosec q = x Þ q = cosec–1 x where q Î and x Î (–¥, –1] È [1, ¥) Note: • sin–1(sin ) =  if and only if –    and sin(sin–1 x) = x where –1  x  1 • cos–1(cos ) =  if and only if 0     and cos(cos–1 x) = x where –1  x  1 • tan–1(tan ) =  if and only if – <  < and tan(tan–1 x) = x where –  < x <  • cot–1(cot ) =  if and only if 0 <  <  and cot(cot–1 x) = x where – < x <  • sec–1(sec ) =  if and only if 0   < or <   . and sec(sec–1 x) = x where – < x  –1 or 1  x <  • cosec–1(cosec ) =  if and only if –   < 0 or 0 <   and cosec(cosec–1 x) = x where – < x  –1 or 1  x <  Illustration 1: Evaluate (i) sin–1 (ii) cot–1(cot4) (iii) cos–1 (cos10) (iv) tan–1 (tan5) (v) cos(tan–12) (vi) sin(cos–1 ) Solution: (i) sin–1 = sin–1 = sin–1 = (ii) cot–1(cot4) = cot–1 (cot( + (4 – )) = cot–1(cot(4 – )) = (4 – ) (iii) cos–1 (cos10) = cos–1( cos(4 –10)) = 4 –10 (iv) tan–1(tan 5) = tan–1 [tan(2 + 5–2)] = tan–1 ( tan(5–2)) = 5 – 2 (v) Let  = tan–12  tan = 2  cos = (vi) Let cos–1 =   cos =  sin = = = . SOME IMPORTANT RESULTS • sin–1 x + cos–1 x = |x|  1 • tan–1 x + cot–1 x = |x|  R • sec–1 x + cosec–1 x = |x|  1 • sin–1(–x) = –sin–1(x) • cosec–1(–x) = –cosec–1(x) • tan–1(–x) = –tan–1(x) • cos–1(–x) =  – cos–1(x) • sec–1(–x) =  – sec–1(x) • cot–1(–x) =  – cot–1(x) • sin–1x = = cosec–1 , x > 0 • sin–1x = cos–1 = , x > 0 • = cosec–1 , x > 0 • • cot–1 x = cos–1 = sec–1 • , x  0, y  0 • cos–1x – cos–1y = • sin–1x + sin–1y = • sin–1x – sin–1y = sin–1 , x  0, y  0 • tan–1x + tan–1y = • x  0, y  0 • 2cos–1x = cos–1 (2x2 – 1), x  0 • 2sin–1x = • 2tan–1x = x  0 • 2tan–1 x = sin–1 = tan–1 , |x| < 1 • 2tan–1 x =  – sin–1 =  + tan–1 , x > 1 Illustration 2: Solve the equation . Solution: Given i.e, For x  2, L.H.S.  tan–1 (3)  tan–1 while R.H.S.  tan–1 (–7) is negative.  no value of x satisfies the given equation. Illustration 3: Prove that Solution: Case (i) : If , then cot  > 1, cot3  > 1 since and Case (ii) : If Illustration 4: Sum to n terms the series Solution:  tan–1 (r  1) – tan–1 r Putting r  1, 2, 3, ….. n and adding Illustration 5: If tan–1 y = 4tan–1 x (|x| < tan ), find y as an algebraic function of x and hence prove that tan is a root of the equation x4 – 6x2 + 1 = 0. Solution: We have tan–1y = 4tan–1x = 2 (as |x| < 1) =  y = If x = tan  tan–1y = 4tan–1x =  y =   x4 – 6x2 + 1 = 0. Illustration 6: If , prove that Solution: Let so that and and so that and the given equation becomes      Now cos   cos (  ) cos  cos  - sin  sin  Hence 9x2  4y2 – 12xy cos   36 sin2  Illustration 7: Consider the equation . Find the values of parameter ‘a’ so that the given equation has a solution. Solution: ( Now, 0 (32a – 1) 27 Thus, the required values of ‘a’ are . Illustration 8: Solve for x; , given that x  (0, 1) Solution: tan–1 x + tan–1 (1 –x) = We have 0  x (1 –x) < 1  x  (0, 1) tan–1 x + tan–1 (1 –x) =  =  7 = 9 – 9(x) (1 –x)  9x2 –9x + 2 = 0  x = . PROPERTIES OF TRIANGLES In a triangle ABC, the angles are denoted by capital letters A, B, and C and the lengths of the sides opposite to these angles are denoted by small letters a, b, and c respectively. Semi-perimeter of the triangle is written as and its area denoted by S or . • Sine rule: , where R is the radius of the circumcircle of the ABC. • Cosine rule: • Projection rule: a = b cosC + c cosB, b = c cosA + a cosC, c = a cosB + b cosA. • Napier's analogy: Auxiliary Formulae • Trigonometric ratios of half - angles • Area of a triangle where R and r are the radii of the circumcircle and the incircle of the  ABC Circles Connected With Triangle Circumcircle The circle passing through the vertices of the triangle ABC is called the circum-circle. Its radius R is called the circum-radius. In the triangle ABC, . In-Circle The circle touching the three sides of the triangle internally is called the inscribed or the in-circle of the triangle. Its radius r is called the in-radius of the triangle. In the triangle ABC, Remark: • From r = 4R sin sin sin , we find that r  4R. because sin sin sin   2r  R. Here equality holds for the equilateral triangle. Escribed Circles The circle touching BC and the two sides AB and AC produced of  ABC externally is called the escribed circle opposite A. Its radius is denoted by r1. Similarly r2, and r3 denote the radii of the escribed circles opposite to angles B and C respectively. r1, r2, r3 are called the ex-radii of  ABC. Here , r1 + r2 + r3 = 4R + r, . Regular Polygon A regular polygon is a polygon which has all its sides as well as all its angles equal. If the polygon has ‘n’ sides, sum of its internal angles is (n – 2) and each angle is . Remarks: • Sum of the exterior angles of a polygon taken in one direction (clockwise or anticlockwise) remains constant and it is equal to 360. • In the regular polygon the centroid, the circumcentre, and the incentre are same. Area of Regular Polygon Area = cot = sin = nr2 tan (Where a is length of side, n is number of sides of polygon, R is radius of circumscribing circle and r is radius of incircle of the polygon). Area of Sector Area included between two radius and circumference. Area = , where  is in radians. Area of Segment Area between a circumference and a chord Area = ( – sin ) SOME BASIC DEFINITIONS The Orthocentre and the Pedal Triangle Orthocentre is a point of concurrence of altitude from the vertices to the opposite side of the triangle. The triangle LMN which is formed by joining the feet of these three altitudes is called the pedal triangle. Remarks: • Orthocentre of the triangle is the incentre of the pedal triangle. • If I1, I2 and I3 be the centres of escribed circles which are opposite to A, B and C respectively and I the centre of incircle then triangle ABC is the pedal triangle of the triangle I1 I2 I3 and I is the orthocentre of the triangle I1 I2I3 . • The centroid of the triangle lies on the line joining the circumcentre to the orthocentre and divides it in the ratio 1 : 2 • Circle circumscribing the pedal triangle of a given triangle bisects the sides of the given triangle and also the lines joining the vertices of the given triangle to the orthocentre of the given triangle. This circle is known as nine point circle. • Circum-centre of the pedal triangle of a given triangle bisects the line joining the circum-centre of the triangle to the orthocentre. Cyclic Quadrilateral: A cyclic Quadrilateral is a quadrilateral whose all the vertices lies on a circle. Note: • Sum of the opposite angles of a cyclic quadrilateral is 180 • In a cyclic quadrilateral sum of the products of the opposite sides is equal to the product of the diagonals. This is known as Ptolemy’s theorem. • If sum of the opposite sides of a quadrilateral is equal, then and only then a circle can be inscribed in the quadrilateral . SOLUTION OF TRIANGLES The three sides a, b, c and the three angles A, B, C are called the elements of the triangle ABC. When any three of these six elements (except all the three angles) of a triangle are given, the triangle is known completely; that is the other three elements can be expressed in terms of the given elements and can be evaluated. This process is called the solution of triangles. In this chapter we will discuss the solution of oblique triangles only. Type I: Problems based on finding the angles when three sides are given. If the data given in sine we use the following formula which ever is applicable. If the given data are in cosine first of all try the following formula which ever is needed. and see whether of logarithm of the number on R.H.S can be determined from the given data. If s proceed further, if not then try the following formula which ever is needed. If the given data are in tangent use the following formula which ever is applicable. Type II: Problem based on finding the angles when any two sides and the angles between them are given or any two sides and the difference of the angles opposite to them are given: Working Rule: Use the following formula whichever is needed. (a) tan = (b) tan = (c) tan = Type III: Problems based on finding the sides and angles when any two angles and side opposite to one of them are given: Working Rule: Use the following formula whichever is needed. (a) (b) A + B + C = 180. Type IV: When all the three angles are given In this case unique solution of triangle is not possible. In this case only the ratio of the sides can be determined. For this the formula. can be used. Type V: • If two sides b and c and the angle B (opposite to side b) are given, then , A = 180° - (B + C) and give the remaining elements. If b < c sin B, there is no triangle possible (fig1). If b = c sin B and B is an acute angle, then there is only one triangle possible (fig 2).If c sin B < b < c and B is an acute angle, then there are two value of angle C (fig 3). If c < b and B is an acute angle, then there is only one triangle (fig 4). This is, sometimes, called an ambiguous case. Alternative Method By applying cosine rule , we have cosB =  a2 – ( 2c cosB) a + ( c2 –b2) = 0  a = c cosB   a = c cosB  This equation leads to following cases: Case -I If b < c sinB, no such triangle is possible. Case -II Let b = c sin B. There are further following case: (a) B is an obtuse angle  cos B is negative. There exists no such triangle. (b) B is an acute angle  cos B is positive. There exists only one such triangle. Case -III Let b > c sin B. There are further following cases: (a) B is an acute angle  cos B is positive. In this case two values of a will exists if and only if c cos B > or c > b  Two such triangle is possible. If c < b, only one such triangle is possible. (b) B is an obtuse angle  cos B is negative . In this case triangle will exist if and only if > |c cos B|  b > c . So in this case only one such triangle is possible. If b < c there exists no such triangle . • If one side a and angles B and C are given, then A = 180° – (B + C), and . • If the three angles A, B, C are given, we can only find the ratios of the sides a, b, c by using sine rule (since there are infinite similar triangles possible). Illustration 9: In a cyclic quadrilateral ABCD, prove that tan2 (B/2) = , a, b, c and d being the lengths of sides AB, BC, CD and DA respectively and ‘s’ is semi-perimeter of quadrilateral. Solution: In DABC AC2 = a2 + b2 – 2ab cos B …(i) In DADC AC2 = c2 + d2 – 2cd cos D = c2 + d2 – 2cd cos (180 - B) = c2 + d2 + 2cd cos B …(ii) from (i) and (ii) cos B = Since tan2(B/2) = = = = = where s = a + b + c+ d Illustration 10: In a D ABC prove that cosA + cosB + cosC £ . Solution: cosA + cosB + cosC = 2 cos = 2 sin = 4 sin + 1 £ 1 + . 4 £ Illustration 11: For a triangle it is given that cos A + cos B + cos C = . Prove that the triangle is equilateral. Solution: If a, b, c are the sides of D ABC then given that cos A + cos B + cos C = Þ Þ ab2 + ac2 – a3 + bc2 + ba2 – b3 + ca2 + cb2 – c3 = 3abc Þ ab2 + ac2 + bc2 + ba2 + ca2 + cb2 – 6abc = a3 + b3 + c3 - 3abc Þ a(b –c)2 + b(c – a)2 + c(a – b)2 = [(a – b)2 + (b – c)2 + (c – a)2] Þ (a + b – c ) (a – b)2 + [b + c – a)(b – c)2 + (c + a – b)(c – a2) = 0 Now a + b > c, b + c > a, c + a > b Since each term on the left side has positive coefficient multiplied by perfect square, each must be separately zero. Þ a = b = c. Hence, the triangle is equilateral. Illustration 12: In a triangle ABC, prove that . Solution: LHS = = = = = = = Illustration 13: If the median AD of a triangle ABC divides the angle Ð BAC in ratio 1: 2 then show that . Solution: Ð BAD = , Ð DAC = \ From D ADC but BD = CD \ = Illustration 14: If I is the in centre of DABC and R1, R2, R3 are the radii of the circumcircles of the triangles IBC, ICA and IAB respectively, then show that R1R2R3 £ R3 Solution: If D IBC apply sine rule Þ R1 = 2R sin A/2 similarly R2 = 2R sin B/2 & R3 = 2R sin C/2 Now R1R2R3 = R3 8 sin A/2 sin B/2 sin C/2 £ R3 Illustration 15: If H is the orthocenre of a ABC, then find AH, BH & CH in terms of angle A, B, C and circumradius R. Solution: Applying sine rule in AHB,  AH = =  AH = 2R cos A. Similarly BH = 2R cos B and CH = 2R cos C. HEIGHTS AND DISTANCE DEFINITIONS Angle of elevation: If ‘O’ be the observer’s eye and OX be the horizontal line through O. If the object P is at a higher level than eye, then angle POX =  is called the angle of elevation Angle of Depression: If the object P is at a lower level than O, then angle POX is called the angle of depression. Illustration 16: A boat is being rowed away from a cliff 150 meter height. At the top of the cliff the angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat. Solution Let PQ be the cliff  PQ = 150 meters. Let the two position of the boat be at A and B so that AP = h cot60 = 150/3 BP = 150 cot45° = 150  AB = 150 - (3 - 1) The speed of the boat is . (3 - 1)  60 metres per hour. = (3 - 1) meters per hour. Illustration 17: The altitude of a rock is observed to be 470. After walking 1000 m towards it, up a slope inclined at 320 to the horizon, the altitude is 770. Find the vertical height of the rock above the first point of observation, given that sin 470 = 0.73. Solution: In  ABD  l = 2000.  l = 1000  AC = l sin 470  AC = 1000 . 0.73 = 730 m OBJECTIVE ASSIGNMENT 1. Length of the shadow of a person is x when the angle of elevation of the sun is 450. If the length of the shadow increases by ( – 1)x, then the angle of elevation of sun should become (A) 150 (B) 180 (C) 250 (D) 300 Solution: Let h be the height of the person  = tan 450 = 1  h = x Now, tan  = =   = 300 Hence (D) is the correct answer. 2. At a certain point the angle of elevation of a tower is found to be such that its cotangent is 3/5; on walking 32 m directly towards the tower its angle of elevation is an angle whose cotangent is 2/3. Then the height of tower is (A) 120 m (B) 80 m (C) 160 m (D) 210 m Solution: AB = tower cot ADB =   AB = (32 + BC) ……(1) Again, cot ACB =  BC = AB ……(2) From (1) and (2) BC = (32 + BC)  BC = 64 m  AB =  64 = 160 m. Hence (C) is the correct answer. 3. The angle of elevation of the top of a T.V. tower from three points A, B, C in a straight line in the horizontal plane through the foot of the tower are , 2, 3 respectively. If AB = a, the height of the tower is (A) a tan  (B) a sin  (C) a sin 2 (D) a sin 3 Solution: AB = a  BP = a from isosceles ABP  h = BP sin 2 = a sin 2 Hence (C) is the correct answer. 4. If minimum value of is , then value of k is (A) 4 (B) 6 (C) 8 (D) None of these. Solution:  Its minimum value is  k = 8. Hence (C) is the correct answer. 5. A man on the top of a cliff 100 meter high, observes the angles of depression of two points on the opposite sides of the cliff as 30° and 60° respectively. Find the distance between the two points. (A) meters. (B) 100 meter (C) 400 meter (D) none of these Solution Let PQ be the cliff and A and B be the points under observation.  PQ = 100 meters AP = 100 cot30° = 1003, BP = 100 cot60° = 100/3. Hence AB = AP+BP = 100 = m Hence (A) is the correct answer. 6. The top of a hill observed from the top and bottom of a building of height h is at angles of elevation p and q respectively. The height of the hill is (A) (B) (C) (D) none of these Solution: Let AD be the building of height h and BP be the hill. Then Tan q = and tan p =  tan q =  xcot p = (h + x) cotq   h + x = + h = Hence (B) is the correct answer. 7. A tree is broken by wind, its upper part touches the ground at a point 10 metres from the foot of the tree and makes an angle of 45° with the ground. The entire length of the tree is (A) 15 m (B) 20 m (C) m (D) m Solution: Let AQ (= PQ) be the broken part of the tree OP. It is given that OA = 10 mt and OAQ = 45° In OAQ, tan 45° =  OQ = 10 m Also, AQ2 = OA2 + OQ2  AQ = Hence, OP = OQ +PQ = OQ +AQ = 10 + 102 = 10 (2 + 1) mts Hence (C) is the correct answer. 8. A flag-staff of 6 meters high placed on the top of a tower throws a shadow of metres along the ground. Then the angle (in degrees) that the sun makes with the ground is (A) 60° (B) 30° (C) 45° (D) None of these Solution: Let OA and AB be the shadows of the tower OP and the flag staff PQ respectively on the ground. Suppose the sun makes an angle  with the ground. Let OA = x. In triangles OAP and OBQ, we have tan = and tan =  =  = 6x  x =  tan =   = 60° Hence (A) is the correct answer. 9. If the length of arc AB, BC and CA of a circle is 3, 4, 5 respectively. Then the area of triangle ABC is (A) (B) (C) (D) none of these Solution: Angle subtended by the chord AB, BC and CA at centre of circle is in ratio 3 : 4 : 5 i.e 900 , 1200 , 1500. ÐB= 750 , ÐC = 450, ÐA = 600 radius of circle 2pr = 12 r = Area of triangle = 2r2 sinA sinB sinC = = = Hence (A) is the correct answer. 10. The area of right angled triangle in terms of r and r1 if ÐA = 90°.( where r, r1 have their usual meanings), is: (A) r + r1 (B) rr1 (C) r – r1 (D) r1 – r Solution: r1 = s tan = s r = (s – a) tan = s – a Also, a2 = b2 +c2 Þ r1 + r = b + c Þ r1 - r = a (r1 + r)2 – (r1 – r)2 = 4r1r = 2bc or, D = = rr1 Hence (B) is the correct answer. 11. The number of points inside or on the circle x2 + y2 = 4 satisfying tan4x + cot4x + 1 = 3sin2y is: (A) one (B) two (C) four (D) infinite Solution: tan4x + cot4x + 1 = (tan2x - cot2x)2 + 3 ³ 3 3 sin2y £ 3 Þ tan2 x = cot2 x , sin2y =1 Þ tanx = ± 1, siny = ±1 Þ x = ± p/4, ± 3p/4 , . . . But x2 £ 4 Þ -2 £ x £ 2 Þ x = ± p/4 only Siny = ± 1 Þ y = ± p/2, ± 3p/2, . . .. But y2 £ 4 Þ y = ± p/2 only. So four solutions are possible. Hence (C) is the correct answer. 12. If in a D ABC (whose circumcentre is origin), a £ sinA, then for any point (x, y) inside the circumcircle of DABC: (A) |xy| < 1/8 (B) |xy| > 1/8 (C) 1/8 < xy < 1/2 (D) none of these Solution: a £ sinA Þ Þ R £ 1/2 So for any point (x, y) inside the circumcircle, x2 +y2 < 1/4 Þ |xy| < 1/8. Hence (A) is the correct answer. 13. If the sine of the angles of a triangle ABC satisfy the equation c3x3 –c2 (a +b+ c) x2 + lx + m =0 (where a, b, c are the sides of DABC), then triangle ABC is: (A) always right angled for any l, m (B) right angled only when l = c(ab + bc + ca), m = -abc (C) right angled only when l = , m = (D) never right angled Solution: sin A + sin B + sin C = but sin A + sin B + sin C = , comparing both we get c = 2R Þ triangle is a right angled triangle Put same value of c, we get l = ab + bc + ca, m = - abc Hence (B) is the correct answer. 14. If sinA and sinB of a triangle ABC satisfy c2x2 – c(a+b)x + ab = 0, then the triangle is: (A) Equilateral (B) Isosceles (C) Right angled (D) Acute angled Solution: sinA + sinB = = Thus C = p/2 Þ D ABC is right angled. Hence (C) is the correct answer. 15. ABCD is a quadrilateral circumscribed about a circle of unit radius, then: (A) ABsin . sin = CDsin .sin (B) ABsin . sin= =CD.sin .sin (C) ABsin . sin = CDsin .sin (D) ABsin . cos= =CDsin .cos Solution: Let ‘O’ be the centre of circle and ‘D’ be it’s point of contact with side AB, Thus, AD = OD. cot = cot and, DB = OC. cot = cot AD + DB = cot + cot = Similarly, CD = Since A + B + C = 2  sin AB. sin .sin = CD sin .sin Hence (B) is the correct answer. 16. If in triangle ABC, line joining the circumcentre and orthocenter is parallel to side AC, then value of tanA. tanC is equal to : (A) (B) 3 (C) 3 (D) None of these Solution: Distance of circumcenter from side AC = R cos B and distance of orthocentre from side AC = 2R cos A  cos C  R cos B = 2 R cos A  cos C  – cos (A + C) = 2 cos A  cos C  sin A  sin C = 3 cos A  cos C  tan A  tan C = 3 Hence (B) is the correct answer. 17. A , B C are the medians of triangle ABC whose centroid is G. If the points A, , and are concyclic, then : (A) 2 (B) 2 (C) 2 (D) None of these Solution: Since A, C1, G and B1 are concyclic, therefore BG. BB1 = BC1 . BA  (BB1)2 = c    Hence (C) is the correct answer. 18. If tanx = n. tany, n , then maximum value of (x – y) is equal to: (A) (B) (C) (D) Solution: tanx = n tany, cos(x – y) = cosx. cosy + sinx.siny. cos(x – y) = cosx.cosy(1 + tanx.tany) = cosx. cosy (1 + n tan2y) Now, Hence (D) is the correct answer. 19. From the top of a tower of height h, the angle of depression of two objects lying on a horizontal plane and in a line passing through the foot of the tower are 450 - and 450 + . The distance between the objects is (A) tan  (B) 2h tan  (C) 2h tan 2 (D) none of these Solution: tan (45  ) = … (1) tan (45 + ) = solving (1) = =  y =  = x (1 + tan ) + h (1  tan ) = x (1 + tan ) = x (1 + tan ) x = = 2h tan 2 Hence (C) is the correct answer. 20. If , then the value of is : (A) 1 (B) 2 (C) 3 (D) 4 Solution: We have, as we know that Hence above result is possible only when. , , x = 1, y = 1, z = 1 + {expanding summation} {using (i)} 1 + 1 + 1 3 Hence (C) is the correct answer. 21. If cos–1 , then (A) sin2 (B) cos2 (C) sin  (D) sec2 Solution: cos–1 =   cos–1 =   cos  = = squaring both sides, we get = sin2 Hence (A) is the correct answer. 22. The positive integral solution of the equation is : (A) x = 2, y = 1 (B) x = 1, y = 2 (C) x = 1, y = 7 (D) x = 2, y = 2 Solution: We have tan–1x+ tan–1 = tan–13 tan–1x – tan–13 = – tan–1 tan–13 – tan–1x = tan–1  tan–1 since x and y are positive integers.  x = 1, 2  x = 1, y = 2 x = 2, y = 7. Hence (B) is the correct answer. 23. The value of is : (A) (B) (C) (D) None of these Solution: tan–1 + tan–1 + tan–1 = L. H. S. = Since x2 +y2 +z2 = r2 = = = R.H.S. Hence (A) is the correct answer. 24. The angle of elevation of the top of a tower standing on a horizontal plane from a point A is . After walking a distance d towards the foot of the tower, the angle of elevation is found to be . The height of the tower is (A) (B) (C) (D) Solution: Apply sine rule in  ABD BD = In  BDC sin  =  h = Hence (B) is the correct answer. 25. A flagstaff of length l is fixed on the top of a tower of height h. The angles of elevation of the top and bottom of the flagstaff at a point on the ground are 60° and 30° respectively. Then (A) l = 2h (B) 2l = h (C) l = 3h (D) 3l = h Solution: Let OP be the tower of height h and PQ is the flagstaff of height l. ÐPRO = 30° ÐQRO = 60° Let OR = x In DPRO = =tan30° DQRO, …(1) = tan60° Þ …(2) From (1) and (2) Þ 3h = h + l Þ 2h = l Hence (A) is the correct answer. 26. The area of a triangle ABC, where a = 2(Ö3 + 1), B = 450, C = 600 is : (A) Ö3(Ö3 + 1) sq. unit (B) 2(Ö3 + 1) sq. unit (C) 2Ö3(Ö3 + 1) sq. unit (D) Ö3(2Ö3 + 1) sq. unit Solution: D = angle A = 750 hence D = = 2Ö3(Ö3 + 1) sq. unit. Hence (C) is the correct answer. 27. In a triangle ABC, the value of is : (A) 0 (B) 1 (C) 2 (D) 3 Solution: Hence (A) is the correct answer. 28. is equal to (A) (B) (C) (D) Solution: . Hence (D) is the correct answer. 29. In a triangle ABC if cos A + 2cos B + cos C = 2. The sides of the triangle are in : (A) H.P. (B) G.P. (C) A.P. (D) None of these Solution: cos A + 2 cos B + cos C =2 or, cos A + cos B = 2(1 – cos B)  2 cos  cos  cos   cot = 3    s = 3s –3b  2s = 3b  a + c =2b  a, b, c are in A.P. Hence (C) is the correct answer. 30. A pole of 50 meter high stands on a building 250 m high. To an observer at a height of 300 m, the building and the pole subtend equal angles. The distance of the observer from the top of the pole is (A) m (B) 50 m (C) m (D) 25 m Solution: Let BC be the building and AB = 50 be the pole. Let O be the observer so that OA = x. If ÐAOB = a and ÐAOC = 2a, then tan 2a = Þ or 3x2 –7500 = x2 or x2 = 3750 Þ x = 25 Hence (A) is the correct answer. 31. A tower subtends an angle of 30° at a point on the same level as its root, and at a second point h m above the first, the depression of the foot of the tower is 60°. Then the height of the tower is (A) h m (B) 3h m (C) h m (D) (h/3) m Solution: Let AB = k be the tower. If x is the distance between the base of two structures, then tan 300 = Þ x = k and tan 600 = Þ x = Þ k = Hence (D) is the correct answer. 32. ABC is a vertical tower such that AB = BC = x. From the foot of A of the tower, a horizontal lien AQP is drawn. If the angle of elevation of the point B from P is a and that of C from Q is 2a, then AP, x and PQ are in (A) A.P (B) G.P (C) H.P (D) none of these Solution: tan 2 = tan  = AP = PQ = AP  AQ = = (1  tan2 ) (1  1 + tan2) = x tan  AP = , x, PQ = x tan   AP, x and PQ are in G.P. Hence (B) is the correct answer. 33. If sin-1 ( sinx) = p - x, then x belongs to (A) ( -¥, ¥) (B) [ 0, p] (C) (D) [p, 2p] Solution: (C) p -x = sin-1( sinx) = sin-1( sin( p - x)) Þ Þ . Hence (C) is the correct answer. 34. Range of f(x) = x + x + x is: (A) [0, ] (B) (C) (D) Solution: Range of f(x) is Hence (A) is the correct answer. 35. If [ ( ( ( (x))))] = 1, where [.] denotes the greatest integer function, is: (A) [tan (sin (cos 1))tan (sin (sin 1))] (B) [tan (sin (cos 1)), tan (sin (cos (sin 1)))] (C) [tan (cos (sin 1)), 1] (D) None of these Solution: Hence (B) is the correct answer. 36. If , then value of is (A) (B) (C) (D) None of these. Solution: . Hence (C) is the correct answer. 37. A man on a cliff observes a boat sailing towards the shore at a uniform speed to the point immediately beneath him, the angle of depression of the boat is 300. 3 minutes later the depression is found to be 600. The time taken by the boat to reach the shore is (A) 2 minutes (B) 1.5 minutes (C) 3 minutes (D) 4 minutes Solution: Let C be the cliff with its base at B. Let BC = h mt. Also, the first position of the boat be at A and the second at D. CAB = 300 and CDB = 600  AB = h cot 300 = h and DB = h cot 600 =  AD = h Hence the speed of the boat = mt/ minute Hence time taken to cover distance DB = minutes. Hence (B) is the correct answer. 38. A man from the top of a 100 meters high tower sees a car moving towards the tower at angle of depression of 300. After some time, the angle of depression becomes 600. The distance (in meters) traveled by the car during this time is (A) 100 (B) (C) (D) 200 Solution: tan 300 =  x + y = 100 tan 600 =  y = x = 100  = . Hence (B) is the correct answer. 39. In a D ABC, r1 + r2 + r3 – r = (A) R (B) 2R (C) 3R (D) 4R Solution: r1 + r2 +r3 – r = = D.c = D.c = 4R Hence (D) is the correct answer. 40. In a triangle, (A) (B) (C) (D) None of these Solution: 1 - tan = 1 - = 1 - . Hence (B) is the correct answer. 41. In a triangle ABC a2b2c2(sin 2A + sin 2B + sin 2C) = (A) D3 (B) 8 D3 (C) 16 D3 (D) 32 D3 Solution: a2b2c2(sin 2A + sin 2B + sin 2C) = a2b2c2 (2sin(A + B) cos (A – B) + 2 sin C cos C) = a2b2c2 2 sin C[cos (A – B) – cos (A + B)] = a2b2c2 [4sin A sin B sin C] = . Hence (D) is the correct answer. 42. If ex-radii r1, r2, r3 of a triangle are in H.P. then its sides a, b, c are in : (A) A.P. (B) G.P. (C) H.P. (D) None of these Solution: Þ Þ (s – b) – (s – a) = (s – c) – (s – b) Þ a – b = b – c So a, b, c are in A. P. Hence (A) is the correct answer. 43. If the angles of a triangle are 300 and 450, and the included side is ( + 1) cm, then the area of the triangle is : (A) (B) (C) (D) Solution: We have Þ Þ b = Area of D ABC bc sin A = bc sin 1050 = Þ Hence (B) is the correct answer. 44. In a right angled ABC, is equal to (A) 0 (B) 1 (C) –1 (D) None of these. Solution: Let C = 90, then . Hence (D) is the correct answer. 45. In any ABC, sin 2B is equal to (A)  (B) 2 (C) 3 (D) 4 Solution: sin 2B . Hence (D) is the correct answer. 46. If in a triangle ABC, , then cos A is equal to (A) 1/5 (B) 5/7 (C) 19/35 (D) None of these. Solution: We have (say) Hence (A) is the correct answer. 47. If are respectively the perpendiculars form the vertices of a triangle to the opposite sides, then is equal to (A) (B) (C) (D) Solution: We have . Hence (D) is the correct answer. 48. If are respectively the perpendiculars from the vertices of a triangle to the opposite sides, then is equal to (A) 1/r (B) 1/R (C) 1/ (D) None of these. Solution: We have . Hence (B) is the correct answer. 49. If where  is the area of triangle ABC, then tan A is equal to (A) (B) (C) (D) Solution: . . . (1) . . . (2) [By Eqns. (1) and (2)] . Hence (B) is the correct answer. 50. If the angles of ABC are in the ratio 1 : 2 : 3, then the corresponding sides are as (A) 2 : 3 : 1 (B) (C) (D) Solution: (1 + 2 + 3)k = 180  k = 30  A = 30, B = 60, C = 90 a : b : c = sin A : sin B : sin C or, Hence (D) is the correct answer. 51. If b + c = 3a, then the value of is equal to (A) 1 (B) 2 (C) (D) Solution: . Hence (B) is the correct answer. 52. If , then 4s (s – a) (s – b) (s – c) is equal to (A) (B) (C) (D) Solution:  is right angled, C = 90 . Hence (D) is the correct answer. 53. If in a ABC, cos B cos C + sin A sin B sin C = 1, then a : b : c is equal to (A) (B) (C) (D) None of these. Solution: cos B cos C + sin A sin B sin C = 1 . . . (1) L.H.S. of (1)  1 and R.H.S.  1 Therefore, (1) is true iff B – C = 0 and 1 – sin A = 0 B = C and A = /2  Thus, a : b : c = sin A : sin B : sin C = Hence (B) is the correct answer. 54. In a triangle ABC, O is a point inside the triangle such that OBC = OCA = OAB = 15, then value of cot A + cot B + cot C is (A) (B) (C) (D) Solution: cot A + cot B + cot C = cot 15 = . Hence (D) is the correct answer. 55. In a ABC if , then triangle is (A) right angled (B) obtuse angled (C) isosceles (D) None of these. Solution: Consider Similarly and either a = b or b= c or c = a. Hence (C) is the correct answer. 56. If xy + yz + zx = 1, then is equal to (A)  (B) /2 (C) 1 (D) None of these. Solution: but will not lie in the domain of . So, none of these is appropriate answer. Hence (D) is the correct answer. 57. If , then x is equal to (A) (B) (C) (D) Solution: Since, , therefore . Hence (D) is the correct answer. 58. If and , then the value of A – B is (A) 0 (B) 45 (C) 60 (D) 30 Solution: We have . Hence (D) is the correct answer. 59. If a < 1/32, then the number of solutions of is (A) 0 (B) 1 (C) 2 (D) infinite. Solution: We know that  No solution. Hence (A) is the correct answer. 60. is equal to (A) (B) (C) (D) None of these. Solution: We have . Hence (C) is the correct answer. 61. If , then x is equal to (A) 1 (B) 0 (C) 4/5 (D) 1/5 Solution: . Hence (D) is the correct answer. 62. If , then value of is (A) (B) (C) (D) None of these. Solution: . Hence (C) is the correct answer.