Trig Practice 08 and Specimen Papers

IB Math High Level Year : Trig: Practice 08 and Spec Papers Trig Practice 08 and Specimen Papers. In triangle ABC, AB = 9 cm, AC = cm, and Bˆ is twice the size of Ĉ. Find the cosine of Ĉ.. In the diagram below, AD is perpendicular to BC. CD =, BD = and AD =. CÂD = and BÂD =. (Total 5 marks) Find the exact value of cos ( ).. The diagram below shows the boundary of the cross-section of a water channel. x The equation that represents this boundary is y = 6 sec where x and y are both measured 6 in cm. The top of the channel is level with the ground and has a width of cm. The maximum depth of the channel is 6 cm. Find the width of the water surface in the channel when the water depth is 0 cm. Give your answer in the form a arccos b where a, b.. A system of equations is given by cos x + cos y =. sin x + sin y =.. (a) For each equation express y in terms of x. () Hence solve the system for 0 x <, 0 < y <. () C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM of

IB Math High Level Year : Trig: Practice 08 and Spec Papers 5. Find, in its simplest form, the argument of (sin + i ( cos )) where is an acute angle. (Total 7 marks) 6. In triangle ABC, BC = a, AC = b, AB = c and [BD] is perpendicular to [AC]. 7. (a) Show that CD = b c cos A. Hence, by using Pythagoras Theorem in the triangle BCD, prove the cosine rule for the triangle ABC. (c) If ABˆ C = 60, use the cosine rule to show that c = a b a. () () (7) (Total marks) The above three dimensional diagram shows the points P and Q which are respectively west and south-west of the base R of a vertical flagpole RS on horizontal ground. The angles of elevation of the top S of the flagpole from P and Q are respectively 5 and 0, and PQ = 0 m. Determine the height of the flagpole. (Total 8 marks) 8. The depth, h (t) metres, of water at the entrance to a harbour at t hours after midnight on a particular day is given by t h (t) = 8 + sin,0t. 6 (a) Find the maximum depth and the minimum depth of the water. () Find the values of t for which h (t) 8. 9. Consider triangle ABC with BÂC = 7.8, AB = 8.75 and BC = 6. Find AC. 0. (a) Sketch the curve f(x) = sin x, 0 x. Hence sketch on a separate diagram the graph of g(x) = csc x, 0 x, clearly stating the coordinates of any local maximum or minimum points and the equations of any asymptotes. (c) Show that tan x + cot x csc x. () (Total 7 marks) C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM of () (5) ()

IB Math High Level Year : Trig: Practice 08 and Spec Papers (d) Hence or otherwise, find the coordinates of the local maximum and local minimum points on the graph of y = tan x + cot x, 0 x. (e) Find the solution of the equation csc x =.5 tan x 0.5, 0 x. (5) (6) (Total marks). In a triangle ABC, Â = 5, BC = cm and AC = 6.5 cm. Find the possible values of Bˆ and the corresponding values of AB. (Total 7 marks). Solve sin x = cos x, 0 x. 5. The obtuse angle B is such that tan B =. Find the values of (a) sin B; () cos B; () (c) sin B; () (d0 cos B. (). Given that tan θ =, find the possible values of tan θ. (Total 5 marks) 5. Let sin x = s. (a) Show that the equation cos x sin x cosec x + 6 = 0 can be expressed as 8s 0s + = 0. () Hence solve the equation for x, in the interval [0, ]. (6) (Total 9 marks) 6. (a) If sin (x α) = k sin (x + α) express tan x in terms of k and α. () Hence find the values of x between 0 and 60 when k = and α = 0. (6) (Total 9 marks) 7. The angle θ satisfies the equation tan θ 5 sec θ 0 = 0, where θ is in the second quadrant. Find the value of sec θ. C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM of

IB Math High Level Year : Trig: Practice 08 and Spec Papers 8. The lengths of the sides of a triangle ABC are x, x and x +. The largest angle is 0. (a) Find the value of x. 5 Show that the area of the triangle is. (c) p q Find sin A + sin B + sin C giving your answer in the form r where p, q, r. () (Total marks) 9. A farmer owns a triangular field ABC. The side [AC] is 0 m, the side [AB] is 65 m and the angle between these two sides is 60. (a) Calculate the length of the third side of the field. () Find the area of the field in the form p, where p is an integer. () Let D be a point on [BC] such that [AD] bisects the 60 angle. The farmer divides the field into two parts by constructing a straight fence [AD] of length x metres. (c) (i) 65x Show that the area o the smaller part is given by and find an expression for the area of the larger part. (ii) Hence, find the value of x in the form q, where q is an integer. (8) (d) BD 5 Prove that. DC 8 (6) () (6) (Total 0 marks) C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM of

IB Math High Level Year : Trig: Practice 08 and Spec Papers - MarkScheme Trig Practice 08 and Specimen Papers MarkScheme. 9 sin C sin B 9 sin C sin C Using double angle formula 9( sin C cos C) = sin C 6 sin C ( cos C ) = 0 or equivalent (sin C 0) cos C = M (). METHOD AC = 5 and AB = (may be seen on diagram) () cosα and sin α () 5 5 cos β and sin β () Note: If only the two cosines are correctly given award ()()(A0). Use of cos ( ) = cos cos + sin sin = (substituting) M 5 5 7 = 7 5 65 N METHOD AC = 5 and AB = (may be seen on diagram) () Use of cos ( + ) = AC AB 9 sin C sin C cosc BC ACAB 5 6 = 5 5 Use of cos ( + ) + cos ( ) = cos cos cosα and cos β 5 () cos 7 α β 5 5 5 65. x 0 cm water depth corresponds to 6 sec 6 6 () x Rearranging to obtain an equation of the form sec k or 6 equivalent ie making a trignometrical function the subject of the equation. M 7 N [5] C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM of 8

IB Math High Level Year : Trig: Practice 08 and Spec Papers - MarkScheme x 8 cos 6 () x 8 = arccos 6 M 6 8 x = arccos Note: Do not penalize the omission of. 7 8 Width of water surface is arccos (cm) R N Note: Candidate who starts with 0 instead of 6 has the potential to gain the two M marks and the R mark.. (a) y = arccos (. cos x) y = arcsin (. sin x) The solutions are x =.6, y = 0.6 x = 0.6, y =.6 5. (sin + i ( cos)) = sin ( cos) + i sin ( cos) M Let be the required argument. sin θ cosθ tan = sin θ cosθ M sin θ cosθ = cos θ cosθ cos θ sin θ cosθ = cosθ cosθ = tan = 6. (a) CD = AC AD = b c cos A RAG METHOD BC = BD + CD a = (c sin A) + (b c cos A) () = c sin A + b bc cos A + c cos A = b + c bc cos A METHOD BD = AB AD = BC CD () c c cos A = a b + bc cos A c cos A a = b + c bc cos A (c) METHOD b = a + c ac cos 60 b = a + c ac c ac + a b = 0 M c = a a a b a b a a b a = [7] C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM of 8

IB Math High Level Year : Trig: Practice 08 and Spec Papers - MarkScheme = a b a AG Note: Candidates can only obtain a maximum of the first three marks if they verify that the answer given in the question satisfies the equation. METHOD b = a + c ac cos 60 b = a + c ac c ac = b a c a ac + a c a c b b c a b a a b a a a M 7. PR = h tan 55, QR = h tan 50 where RS = h M Use the cosine rule in triangle PQR. 0 = h tan 55 + h tan 50 h tan 55 h tan 50 cos 5 00 h tan 55 tan 50 tan55 tan50 cos5 () = 79.9... () h = 9.5 (m) 8. (a) t Either finding depths graphically, using sin 6 or solving h(t) = 0 for t h (t) max = (m), h (t) min = (m) N t Attempting to solve 8 + sin 8 algebraically or graphically 6 t [0, 6] [, 8] {} N 9. METHOD Attempting to use the cosine rule i.e. BC = AB + AC AB AC cos BÂC 6 = 8.75 + AC 8.75 AC cos 7.8 (or equivalent) Attempting to solve the quadratic in AC e.g. graphically, numerically or with quadratic formula M Evidence from a sketch graph or their quadratic formula (AC = ) that there are two values of AC to determine. () AC = 9.60 or AC =. N Note: Award M(A0)A0 for one correct value of AC. METHOD BC AB Attempting to use the sine rule i.e. sin BÂC sin AĈB 8.75sin 7.8 sin C 6 0.898... C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM of 8 () AG () [] [8]

IB Math High Level Year : Trig: Practice 08 and Spec Papers - MarkScheme 0. (a) C = 6.576... C = 6.6... and B = 78.8... or B = 5.5576... EITHER AC 6 Attempting to solve or sin 78.8... sin 7.8 AC 6 sin 5.5576... sin 7.8 M OR Attempting to solve AC = 8.75 + 6 8.75 6 cos 5.5576... or AC = 8.75 + 6 8.75 6 cos 78.8... M AC = 9.60 or AC =. N Note: Award ()A0MA0 for one correct value of AC. [7] Note: Award for shape. for scales given on each axis. A A5 Asymptotes x = 0, x =, x = Max,, Min, Note: Award for shape A for asymptotes, for one error, A0 otherwise. for max. for min. (c) sin x cos x tanx + cot x cos x sin x M sin x cos x sin x cos x sin x C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM of 8

IB Math High Level Year : Trig: Practice 08 and Spec Papers - MarkScheme csc x AG (d) tan x + cot x csc x Max is at 8, Min is at, 8 (e) csc x =.5 tan x 0.5 tan x cot x tan x M tan x + cot x = tan x tan x tan x = 0 M tan x tan x = 0 ( tan x + )(tan x ) = 0 M tan x = or x = Note: Award A0 for answer in degrees or if more than one value given for x.. sin B sin 5 6.5 M Bˆ = 68.8 or Ĉ = 76. or.8 (accept ) AB BC sinc sina AB sin76. sin 5 AB = 6.77 cm AB sin.8 sin 5 AB =.88cm (accept.90). sin x cos x cos x = 0 cos x ( sin x ) = 0 () cos x = 0 sin x = x = x =,. (a) 5 sin B = cos B = (c) sin B = sin B cos B 5 = C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM 5 of 8 [] [7]

IB Math High Level Year : Trig: Practice 08 and Spec Papers - MarkScheme 0 = 69 (d) cosb = cos B = 69 9 = 69. tan Using tan θ = tan tan tan tan θ + 8 tan θ = 0 Using factorisation or the quadratic formula tan θ = or 5. (a) ( s ) s s + 6 = 0 M s 8s + 6s = 0 8s 0s + = 0 AG Attempt to factorise or use the quadratic formula sin x = or sin x = () sin x = x or x sin x = x or x Note: Penalise if extraneous solutions given. 6. (a) sinx cos α cosx sin α = k sin x cos α + k cos x sin α tan x cos α sin α = k tan x cos α + k sin α M ( k ) sin ( k ) tan x = tan ( k ) cos ( k ) sin 0 cos0 tan x = Now sin 0 = sin 0 = and cos 0 = cos 0 = tan x = x = 60, 0 7. tan θ 5 sec θ 0 = 0 Using + tan θ = sec θ, (sec θ ) 5 sec θ 0 = 0 C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM 6 of 8 [5] [9] [9]

IB Math High Level Year : Trig: Practice 08 and Spec Papers - MarkScheme sec θ 5 sec θ = 0 Solving the equation e.g. ( sec θ + )(sec θ ) = 0 sec θ = or sec θ = θ in second quadrant sec θ is negative (R) sec θ = N 8. (a) 9. (x + ) = (x ) + x (x ) xcos0 M x + x + = x x + + x + x x 0 = x 0x 0 = x(x 5) x = 5 Area = 5 sin 0 M = 5 5 = AG (c) sin A = 5 5 7 sin B sin B M 5 Similarly sin C = 5 sin A + sin B + sin C = [] (a) Using the cosine rule (a = b + c bc cos A) Substituting correctly BC = 65 + 0 (65) (0) cos 60 C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM 7 of 8

IB Math High Level Year : Trig: Practice 08 and Spec Papers - MarkScheme = 5 + 0 86 6760 = 88 BC = 9m N Finding the area using = bc sin A Substituting correctly, area = (65) (0) sin 60 = 690 (accept p = 690) N (c) (i) Smaller area A = (65) (x)sin 0 65x = AG N0 Larger area A = (0) (x) sin 0 M = 6x N (ii) Using A + A = A 65x Substituting 6x 690 69x Simplifying 690 690 Solving x = 69 x 0 (accept q = 0) N (d) Using sin rule in ADB and ACD BD 65 BD sin 0 Substituting correctly sin0 sinadˆ B 65 sin ADˆ B DC 0 DC sin 0 and sin0 sin ADˆ C 0 sin ADˆ C Since ADˆ B ADˆ C = 80 R It follows that sin ADˆ B sinadˆ C BD DC BD 65 65 0 DC 0 BD 5 DC 8 R AG N0 [0] C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM 8 of 8

IB Math High Level Year : Trig: Practice 08 and Spec Papers - MarkScheme C:\Users\Bob\Documents\Dropbox\Desert\HL\ Trig\HLTrigPractice08.docx on 0/7/06 at 9:5 AM of