complicated calculations, similar to HKDSE difficult types ** Please answer any 30 questions ** Level 1 In the figure, sin θ =

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Level : single concept, simple calculation, easier than HKDSE types Level : one or two concept, some calculations, similar to HKDSE easy types Level 3: involving high level, logical and abstract thinking skills, or with Level. complicated calculations, similar to HKDSE difficult types ** Please answer any 30 questions ** In the figure, sin θ A. The answer is B. 8 8 7. B.... D.. 7 7 8. Refer to the figure. Find the value of cos θ sin θ. A. The answer is. DF + DE EF cos θ sin θ B. DF EF 3 DE 8 DF 8 4 EF 3 DE 3 EF 3 cos θ sin θ 4 3. D. 0

3. It is given that cos θ and tan θ. A. sin θ. sin θ The answer is A.. Without using a calculator, find the values of sin θ, tan θ 3, tan θ 3 B. sin θ D. sin θ, tan θ 3 3, tan θ As shown in the figure, draw a right-angled triangle AB, where AB and A so that cos θ. By Pythagoras theorem, B A ( ) 4 B sin θ A tan θ B AB AB 4. Simplify tanθ sin θ. A. sin θ B. cos θ. sin θ cosθ D. The answer is A. tanθ sin θ tan θ cos θ sin θ cosθ cosθ sin θ cos θ sinθ

cos θ. Simplify + sin θ. tan θ A. 0 B.. The answer is B. cos θ + sin tan θ sin θ tan sin sin sin cos cos θ + sin θ θ + sin θ cos θ sin θ θ + sin θ θ + sin θ θ θ θ tan θ D. sin θ 6. If α lies in quadrant II and β lies in quadrant IV, then which of the following is/are true? I. sin α tan β > 0 II. cos α sin β > 0 III. tan α cos β > 0 A. II only B. III only. I and III only D. II and III only The answer is A. I: When α lies in quadrant II, sin α > 0. When β lies in quadrant IV, tan β < 0. sin α tan β < 0 I is not true. II: When α lies in quadrant II, cos α < 0. When β lies in quadrant IV, sin β < 0. cos α sin β > 0 II is true. III: When α lies in quadrant II, tan α < 0. When β lies in quadrant IV, cos β > 0. tan α cos β < 0 III is not true. Only II is true. 7. sin (80 θ) cos (70 θ) tan sin 330 A. sin θ B. 3 sin θ 3

. sin θ + cos θ D. sin θ cos θ The answer is A. sin (80 θ) cos (70 θ) tan sin 330 sinθ sinθ sin θ sin θ sin θ 8. Find the minimum value of ( + cos θ)( cos θ). A. 3 B. 4. D. 6 The answer is A. ( + cos θ)( cos θ) 4 cos θ 0 cos θ Minimum value of ( + cos θ)( cos θ) 4 3 9. For 80 θ 360, find the maximum value of 0 + 4 sin (360 θ). A. 4 B. 6. 0 D. 4 The answer is D. 0 + 4 sin (360 θ) 0 + 4( sin θ) 0 4 sin θ For 80 θ 360, sin θ 0 Maximum value of 0 + 4 sin (360 θ) 0 4( ) 4 0. If 3 tan θ 3, where 0 < θ < 360, then θ A. 30 or 0 B. 60 or 40.. 0 or 300. D. 0 or 330. The answer is. 3 tan θ 3 4

tan θ 3 3 tan θ 3 θ 80 60 or 360 60 0 or 300. + sinθ + sin (80 + θ ) sinθ cos ( θ ) A. ( + sinθ ). sinθ cosθ B... tanθ D. The answer is D. + sinθ + sin (80 + θ ) sinθ cos ( θ ) + sinθ + ( sinθ ) sinθ cosθ ( + sinθ )( sinθ ) sinθ cosθ sin θ sinθ cosθ cos θ sinθ cosθ cosθ sinθ tanθ ( + sinθ ). sinθ cosθ. tanθ. 3sin ( θ ) sin (70 + θ ) + tan (80 θ ) cos 60 A. cos θ. B. cos θ.. cos θ. D. 3 cos θ sin θ. The answer is B. 3sin ( θ ) sin (70 + θ ) + tan (80 θ ) cos60 3sinθ cosθ + tanθ

3sinθ cos θ sinθ cosθ 3 cos θ cos θ cos θ 3. A 6 w 6 B In the figure, A 6 and 6, w A. B sin 6. sin 9 B.. A sin 6. sin 6 D. The answer is B. By the sine formula, w B sin 6 sin 6 w B sin 6 sin 6 B sin 6. sin 6 A sin 9. sin 6 4. 6a 0 0 a In the figure, find a, correct to decimal places. A. 0.6 B. 0.93.. D..3 The answer is. By the cosine formula, cos 0 (6a) + (a) 0 6a a 6

6 00 cos 0 60a 60a cos 0 6a 00 a (6 60 cos 0 )a 00 a 00 6 60 cos0 a., cor. to d.p.. A In the figure, AB is an obtuse-angled triangle. A cm, AB 6 cm and AB. Find the area of AB, correct to 3 significant figures. A. 6.9 cm B. 33. cm. 4. cm D. 4.0 cm The answer is B. cm By the sine formula, cm 6 cm sin AB sin sin sin AB 6 AB 80 69.47 0.3, cor. to sig. fig. AB + AB + AB 80 AB + 0.3 + 80 AB 47.47, cor. to 4 sig. fig. Area of AB A AB sin AB 6 sin 47.47 cm 33. cm, cor. to 3 sig. fig. B 6 cm 6. The area of PQR is 9. cm. If PR 0 cm and QR 4 cm, find PRQ, correct to the nearest integer. A. 3 or 7 B. 77 or 03. 80 or 00 D. 89 or 9 7

The answer is B. Area of PQR 9. cm PR QR sin PRQ 9. cm 9. sin PRQ 0 4 PRQ 77.6 or 80 77.6 77 or 03, cor. to the nearest integer 7. A. In the figure, y x sin 48. B. sin96 The answer is. By the sine formula, y x sin 48 sin (80 96 48 ) y x sin 48 sin 36 x sin 48 y sin36 x sin36.. sin 48 x sin 48. D. sin36 x sin96. sin36 8. In AB, AB 0 cm, B 6 cm and B 60. Find sin A. A. 6 3 The answer is. By the cosine formula, B. 7 3 6 A AB + B AB B cos B A 0 + 6 4 cm By the sine formula, 0 6 cos60 cm 8. 4 3 7 D. 4 3

6 cm sin A sin A 4 cm sin 60 6 sin 60 4 3 6 4 4 3 7 9. The area of the triangle in the figure is A. 8x cm. B. The answer is. s (3x + 6x + x) cm 7x cm By Heron s formula, area of the triangle 7x(7x 3x)(7x 6x)(7x x) cm 6x cm 8x cm.. 6x cm. D. 630x cm. 0. 7 3 In the figure, cos θ A. The answer is B. i 7 3. B.... D.. 39 6 4 By the cosine formula, 9

3 + 7 cosθ 3 7 6 0

Level. In the figure, PS A. PQ sin x + QR sin y. B. PQ sin x + QR cos y.. PQ cos x + QR sin y. D. PQ cos x + QR cos y. The answer is. With the notations in the figure, draw QE PS and QF RS. In QPE, PE cos x PQ PE PQ cos x In QRF, QF sin y QR QF QR sin y PS PE + ES PE + QF PQ cos x + QR sin y. In the figure, DB is a straight line. Find the length of B. A. 3 B. 4 3. 6 3 D. 8 3

The answer is B. In AD, D tan 30 AD D 3 6 6 D 3 In ABD, DB tan 60 AD DB 3 6 DB 6 3 B DB D 6 6 3 3 6 3 6 3 3 6 3 3 4 3 3. Which of the following are identities? I. II. sinθ tanθ sin θ cosθ tan θ + tan (90 θ ) sinθ cosθ III. sin (90 θ) + sin θ cos (90 θ) + A. I and II only B. I and III only. II and III only D. I, II and III

The answer is A. I: L.H.S. sin θ tanθ sinθ sin θ cos θ sin θ cosθ sin θ R.H.S. cosθ L.H.S. R.H.S. I is an identity. II: L.H.S. tan θ + tan (90 θ) tan θ + tanθ sinθ cosθ + cosθ sinθ sin θ + cos θ sinθ cosθ sinθ cosθ R.H.S. sinθ cosθ L.H.S. R.H.S. II is an identity. III: L.H.S. sin (90 θ) + sin θ cos θ + sin θ R.H.S. cos (90 θ) + sin θ + sin θ + cos θ sin θ + cos θ L.H.S. R.H.S. III is not an identity. Only I and II are identities. 4. A 6 4 cm D 3 cm B 3

Refer to the figure. Find the area of trapezium ABD, correct to 3 significant figures. A. 7. cm B. 7.6 cm. 83.4 cm D. 84.0 cm The answer is B. With the notations in the figure, draw DE AB. In ADE, AE cos 6 AD AE AD cos6 4 cos 6 cm ED sin 6 AD ED AD sin 6 4 sin 6 cm Area of trapezium ABD ( D + AB) B [ D + ( AE + EB)] B [3 + (4 cos6 + 3)] 4 sin 6 cm 7. 6 cm, cor. to 3 sig. fig.. A P D Q i B In the figure, ABD is a square of side 0 cm. P and Q are the mid-points of AD and D respectively. Find θ, correct to the nearest degree. A. 30 B. 37. 4 D. 3 4

The answer is B. ABD is a square. BQ AB 90 Q D 0 cm cm Q tan QB B cm 0 cm QB 6.7, cor.to the nearest 0.0 Similarly, PBA 6.7. θ AB PBA QB 90 6.7 6.7 37, cor. to the nearest degree 6. For 0 θ < 360, how many roots does the equation sin x tan x have? A. B. 3. 4 D. The answer is. sin x tan x sin x sin x cos x sin x cos x sin x sin x cos x sin x 0 sin x ( cos x ) 0 sin x 0 or cos x 0 sin x 0 or cos x When sin x 0, x 0 or 80 When cos x, x 60 or 360 60 60 or 300 x 0 or 60 or 80 or 300 The equation has 4 roots.

7. For 0 x 360, how many roots does the equation cos x ( sin x + ) 0 have? A. B. 3. 4 D. The answer is. cos x ( sin x + ) 0 cos x 0 or sin x + 0 If cos x 0, x 90 or 70 If sin x + 0, sin x sin x x 80 + 30 or 360 30 0 or 330 x 90 or 0 or 70 or 330 The equation has 4 roots. 8. Which of the following about the properties of y cos x is/are correct? I. The maximum value of cos x is. II. The period of y cos x is 80. III. The minimum value of cos x is. A. I and II only B. I and III only. II and III only D. I, II and III The answer is B. The following are the correct descriptions of y cos x: I. cos x II. cos x is a periodic function, and its period is 360. The answer is B. 9. If 3 sin 3θ + 0, where 0 θ 80, then θ A. θ 70. or 0.3, cor. to d.p. B. θ 48. or 6., cor. to d.p.. θ 73.9 or 06., cor. to d.p. D. θ 0.3 or 67.9, cor. to d.p. The answer is. 6

3sin 3θ + 0 sin 3θ 3 3θ 80 + 4.8 or 3θ.8 or θ 73.9 or 38.9 360 4.8 06., cor to d.p. A + B 30. Find tan A. The answer is D. tan B., where A, B and are angles in a triangle AB.. tan A, B and are the angles of a triangle. A + B + 80º i.e. A + B 80º o A + B 80 tan tan o tan 90 tan tan D. tan 3. P 80 Q S In the figure, PSR 70, SPR 80, PRQ 9, SR cm and RQ 3 cm. Find the length of PQ, correct to 3 significant figures. A. 4.9 cm B.. cm..0 cm D..8 cm The answer is D. In PRS, by the sine formula, PR sin 70 70 cm sin 80 cm 9 R 3 cm 7

sin 70 PR cm sin 80 4.770 9 cm, cor. to sig. fig. In PQR, by the cosine formula, PQ PR + RQ PR RQ cos PRQ PQ 4.770 9 + 3 4.770 9 3 cos 9 cm.8 cm, cor. to 3 sig. fig. 3. P 9 Q cm S In the figure, QPS 9, PQS, QSR 3, SQ cm and SR cm. Find the perimeter of quadrilateral PQRS, correct to decimal place. A. 6. cm B..6 cm..3 cm D. 4.9 cm The answer is D. In SQR, by the cosine formula, QR SQ + SR SQ SR cos QSR QR + cos 3 cm.33 cm, cor. to 3 d.p. In PQS, 3 cm QPS + PQS + PSQ 80 9 + + PSQ 80 PSQ 60 By the sine formula, PQ cm sin PSQ sin 9 sin 60 PQ cm sin 9.739 cm, cor. to 3 d.p. PS cm sin sin 9 sin PS cm sin 9 0.848 cm, cor. to 3 d.p. Perimeter of quadrilateral PQRS R 8

PQ + QR + SR + PS (.739 +.33 + + 0.848) cm 4.9 cm, cor. to d.p. 33. A B 6 In the figure, AB is a vertical flagpole. BD is a straight line on the horizontal ground. The angles of elevation of A from and D are 6 and 30 respectively. and D are 33 m apart. Find the distance between B and, correct to 3 significant figures. A.. m B. 6.0 m. 6. m D. 8.8 m The answer is A. In AD, AD + AD AB AD + 30 6 AD 3 30 33 m By the sine formula, A 33 m sin 30 sin 3 33sin 30 A m sin 3 8.767 m, cor. to sig. fig. In AB, B cos AB A B 8.767 cos 6 m. m, cor. to 3 sig. fig. The distance between B and is. m. D 34. 9

A θ B In the figure, AB and D represent two vertical buildings. B, E and D lie on the same horizontal line. The angles of elevation of A and from E are 4 and 8 respectively. AE m and E m. Find the angle of depression θ of from A, correct to the nearest 0.. A. 9.6 B..6. 3.3 D..4 The answer is D. AE 80 AEB ED 80 4 8 07 In AE, by the cosine formula, A AE + E AE E cos AE A + cos07 m 4.86 m, cor. to 3 d.p. cos EA m m 4 8 E AE + A E AE A + 4.86 4.86 EA 9.4, cor. to 3 d.p. θ 4 9.4.4, cor. to the nearest 0. D 3. 9 P N 0 0 km N Q N In the figure, lighthouses P and Q are 0 km apart. The bearing of R from P is 0. If the bearings of Q from P and R are 09 and 0 respectively, then the distance between Q and R is A. 0 sin 0cos8 km. B. km.. sin8 sin R 0 0cos km. D. sin8 0sin 8 km. sin

The answer is A. 9 P N 0 With the notation in the figure, α 9 80 β 80 0 70 In PQR, 0 km QPR 0 9 PRQ β + 70 + 8 α N β By the sine formula, QR 0 km sin sin 8 0 sin QR km sin8 0 sin The distance between Q and R is km. sin 8 R N Q 36. P 4 m 7 m A 60 4 m In the figure, a flagpole PA of height 4 m is perpendicular to the horizontal ground AB. D is a point on B such that AD B. AB 7 m, A 4 m and BA 60. Let ADP θ, find the value of tan θ. A. B D 37 37 The answer is. B. In AB, by the cosine formula, 37. 3 D. 3

B AB + A AB A cos BA B 7 + 4 7 4 cos 60 m 37 m Area of AB AB A sin 60 B AD AB A sin 60 7 4 sin 60 AD m 37 3 8 m 37 In PAD, PA tan θ AD 4 4 3 37 37 3 3 3 3 4 3 m 37 37. E H F G 4 cm In the figure, ABDEFGH is a cuboid. AB AF 4 cm and B cm. Find the value of cos θ. A. A 3 The answer is A. In ABF, D 4 cm BF AB + AF θ cm B B. 3. 3 D. 3

BF 4 + 4 cm 3 cm In BF, F B + BF F + ( 3) cm 6 cm cos θ B F 6 3 38. A.8 m 30 D B 7 4 In the figure, AB is a vertical tree of height.8 m. B, and D are three points on the same horizontal ground and BD 7. The angles of elevation of A from and D are 4 and 30 respectively. Find the distance between and D, correct to 3 significant figures. A..6 m B..80 m. 4.8 m D. 4.93 m The answer is D. In AB, AB tan AB B.8 B tan 4.8 m In ADB, AB tan ADB BD.8 BD m tan 30 4.849 7 m, cor. to sig. fig. m In BD, by the cosine formula, D B + BD B BD cos BD 3

D.8 + 4.849 7.8 4.849 7 cos 7 m 4.93 m, cor. to 3 sig. fig. The distance between and D is 4.93 m. 39. M P In the figure, MN is a vertical building. P, Q and N are three points on the same horizontal ground. PNQ 90, NPQ 46, MQN 3 and PN 00 m. Find the angle of elevation of M from P, correct to 3 significant figures. A. 3.9 B. 43.. 46. D. 4. The answer is A. In NPQ, QN tan NPQ PN QN 00 tan 46 m In MNQ, 46 tan MQN N 00 m 7.77 m, cor. to sig. fig. MN QN 3 MN 7.77 tan 3 m 36. m, cor. to sig. fig. In MPN, MN tan MPN PN 36. 00 MPN 3.9, cor. to 3 sig. fig. The angle of elevation of M from P is 3.9. Q 40. 4

E H F G D P A The figure shows a cube ABDEFGH. P is the mid-point of BG. Find FPH, correct to 3 significant figures. A. 60.0 B. 78.. 84.6 D. 90.0 The answer is B. Let a cm be the length of each side of the cube. In FGH, FH FG + GH FH a + a cm a cm a GP BG cm In FGP, FP FG + GP FP a + a cm B a cm a HP FP cm In FHP, by the cosine formula, cos FPH FP + HP FH FP HP a a ( a) + a a FPH 78., cor. to 3 sig. fig.

Level 3 4. A 40 7 6 B D In the figure, BD is a straight line. BAD 40, AB 6 and AD 7. Arrange the lengths of AD, BD and D in ascending order. A. AD < BD < D B. BD < AD < D. D < AD < BD D. D < BD < AD The answer is. In AD, AD + AD + AD 80 AD + 6 + 7 80 AD 49 By the sine formula, D AD sin AD sin 6 D ADsin 49 sin 6 sin 49 < sin 6 D < AD In ABD, ABD AD BAD 7 40 3 By the sine formula, BD AD sin 40 sin ABD BD ADsin 40 sin 3 sin 40 > sin 3 BD > AD D < AD < BD 4. 6

A cm 8 cm 8 cm D In the figure, AB A 8 cm, B 7 cm and AD cm. AD is an acute angle. The area of AD is 0 cm. Find BD, correct to 3 significant figures. A. 9.4 B. 93.8. 96.8 D. 98.3 The answer is D. In AB, by the cosine formula, A + B AB cos AB A B 8 + 7 8 8 7 AB 64.06, cor. to sig. fig. Area of AD 0 cm A AD sin AD 0 cm 0 sin AD 8 AD 30 In AD, by the cosine formula, D A + AD A AD cos AD D 8 + 8 cos 30 cm 4.440 cm, cor. to sig. fig. In AD, by the cosine formula, A + D AD cos AD A D 8 + 4.440 8 4.440 B 7 cm AD 34.64, cor. to sig. fig. BD AB + AD 64.06 + 34.64 98.3, cor. to 3 sig. fig. 7

43. V In the figure, VABD is a right pyramid whose base ABD is a square. O is the projection of V on the plane ABD. E is the mid-point of AB. If the angle between the line VA and the plane ABD is 4, find VEO, correct to 3 significant figures. A. 3.3 B. 4.0. 4.7 D. 60.0 The answer is. Let a cm be the length of each side of square ABD. In AB, A AB + B A OA In AOV, a + a a cm cm a A cm tan VAO A OV OA D 4 O E B OV a tan 4 cm a cm AE EB, AO O and EO // B. a EO B cm In VEO, OV tan VEO EO a a VEO 4.7, cor. to 3 sig. fig. 8

44. V A O 6 cm B In the figure, VAB is a regular tetrahedron of side 6 cm. O is the projection of V on the plane AB. Find the volume of the regular tetrahedron, correct to 3 significant figures. A.. cm 3 B. 8. cm 3. 76.4 cm 3 D. 84.3 cm 3 The answer is A. V D A O 6 cm B With the notation in the figure, produce BO to D such that BD A. AB is an equilateral triangle. AD In ABD, AB AD + BD BD A 6 cm 3 cm 6 3 cm 7 cm VD BD 7 cm In VBD, by the cosine formula, VB + BD VD cos VBD VB BD 6 + ( 7) ( 7) 6 7 VBD 4.736, cor. to sig. fig. In VBO, VO sin VBO VB VO 6 sin 4.736 cm 9

4.899 0 cm, cor. to sig. fig. Area of AB AB B sin AB 6 6 sin 60 cm.88 cm, cor. to sig. fig. Volume of the regular tetrahedron area of AB VO 3.88 4.899 0 cm 3 3. cm 3, cor. to 3 sig. fig. 4. V 8 cm A 60 E x cm In the figure, VAB is a right triangular pyramid whose base AB is an equilateral triangle. E is the mid-point of AB. The angle between the line V and the plane AB is 60, and V 8 cm. Find the value of x. A. 3 B. 4 3. 8 3 D. 6 3 The answer is B. In BE, E sin BE B E x sin 60 cm B 3x cm BE AB B VB V 8 cm In VBE, VB VE + BE x cm 30

VE 8 x cm x 64 cm 4 In VE, by the cosine formula, VE V + E V E cos VE 64 x 8 4 3x + 8 64 x 3x 64 + 4 3 x 4 4 4 3 x x x 4 3 x 0 x(x 4 3 ) 0 x 0 (rejected) or 4 3 3x cos 60 46. D E 8 cm F M A 6 cm 6 cm B 8 cm In the figure, ABDEF is a right triangular prism. M is a point on EF such that DM EF. AB B 6 cm, A D 8 cm. Find the angle between the line BD and the plane ABFE, correct to 3 significant figures. A. 33.4 B. 36.6. 7.7 D. 73.3 The answer is B. 3

D E 8 cm F M Join MB. 6 cm 6 cm B 8 cm The required angle is DBM. In BD, BD B + D BD 6 + 8 cm 0 cm DF B 6 cm EF AB 6 cm DE A 8 cm In DEF, by the cosine formula, DF + EF DE cos DFE DF EF 6 + 6 8 6 6 DFE 83.6, cor. to sig. fig. In DFM, DM sin DFM DF DM 6 sin 83.6 cm A.96 9 cm, cor. to sig. fig. In DBM, DM sin DBM BD.96 9 0 DBM 36.6, cor. to 3 sig. fig. The angle between the line BD and the plane ABFE is 36.6. 47. 3

D A cm O B H G Q E P F In the figure, ABDEFGH is a cube of side cm. P and Q are the mid-points of EF and FG respectively. A and BD intersect at O. Find the angle between the lines OP and OQ, correct to 3 significant figures. A. 6.6 B. 36.9. 63.4 D. 7.6 The answer is B. A cm B O R D E H P With the notation in the figure, let R be the mid-point of B. Join OR and QR. The required angle is POQ. PF FQ EF cm cm In PQF, PQ PF + FQ PQ + cm cm AO O, BR R and OR // AB. OR AB cm cm RQ BG cm In ORQ, OQ OR + RQ OQ + cm cm OP OQ cm In POQ, by the cosine formula, cos POQ F Q G OP + OQ PQ OP OQ 33

( ) + ( ) ( POQ 36.9, cor. to 3 sig. fig. The angle between the lines OP and OQ is 36.9. ) 48. A B 00 m N D 38 F E In the figure, a helicopter flies eastwards with a uniform height 00 m along a horizontal path AB., D and E are three points on the same horizontal ground. and D are the projections of B and A on the horizontal ground respectively. The bearing of B from E is N E. The angle of elevation of A from E is 38. F is a point on D such that EF D. If EF 00 m, find the distance between A and B, correct to 3 significant figures. A. 40 m B. 60 m. 3 000 m D. 3 30 m The answer is B. In AED, AD tan AED DE 00 DE m tan 38 3.9 m, cor. to sig. fig. In DEF, DE DF + EF DF 3.9 00 m 4. m, cor. to sig. fig. In EF, F tan EF EF F 00 tan m AB D DF + F E 0.0 m, cor. to sig. fig. 34

( 4. + 0.0) m 60 m, cor. to 3 sig. fig. The distance between A and B is 60 m. 49. 8 m A N B N 6 66 D 9 m 30 Q E P In the figure, AB and D are two vertical buildings, where D 8 m. P, Q and D lie on the same straight line. D is due north of P. Q is 9 m due east of B. The angles of elevation of from P and Q are 30 and 66 respectively. The angle of depression of P from A is 6. Find the height of the building AB, correct to 3 significant figures. A. 3.0 m B. 4.8 m. 3.6 m D. 37.8 m The answer is. In PD, D tan PD PD 8 PD m tan 30 47. m, cor. to sig. fig. In QD, D tan QD QD 8 QD m tan 66 37.844 m, cor. to sig. fig. PQ PD QD (47. 37.844) m 09.376 m, cor. to 6 sig. fig. In BPQ, 3

BP BQ + PQ BP 9 + 09.376 m 4.7 m, cor. to sig. fig. APB 6 In ABP, AB tan APB BP AB 4.7 tan 6 m 0. 3.6 m, cor. to 3 sig. fig. In the figure, PQ is a vertical tree on the horizontal ground. The angle between the inclined plane AB and the horizontal ground QD is 0. A, B,, D, P and Q lie on the same vertical plane. It is given that AB m. The angles of elevation of P from A and B are 0 and 3 respectively. Find the height of the tree PQ, correct to 3 significant figures. A..0 m B. 8.78 m. 0. m D. 3.7 m The answer is. ABP 0 + 3 4 PAB 80 0 0 0 In APB, APB + ABP + PAB 80 APB + 4 + 0 80 By the sine formula, AP m sin 4 sin sin 4 AP m sin APB 36

3.660 m, cor. to sig. fig. In APQ, PQ sin PAQ AP PQ 3.660 sin 0 m 0. m, cor. to 3 sig. fig. The height of the tree PQ is 0. m.. In the figure, VABD is a right pyramid whose base ABD is a square. The angle between the line V and the plane ABD is 60. If the angle between the planes θ VAD and VB is θ, find the value of sin. A. 6 The answer is B. B. 7. 6 D. 7 Let M and N be the mid-points of AD and B respectively, and X be a point on MN such that VX MN. Join VM and VN. θ MVN MVN XVN Let V l. θ In VX, VX sin VX V VX l sin 60 37

3 l X cos VX V X l cos 60 l In XN, XN + N X XN + XN l XN l 4 XN l 8 XN l 8 In VXN, VN VX + XN VN 7 l 8 3 l θ XN sin VN + l 8 7 l 8 l 8 7. In the figure, two square planes ABD and DEF are perpendicular to each other. If M is the mid-point of D, find BME, correct to the nearest degree. A. 78 B. 90. 96 D. 0 The answer is D. 38

Join BF and BE. Let AB l. DM M In BM, D BM B + M BM l l ME BM In BF, + l BF B + F BF l + l In BEF, l BE BF + EF BE l ( l ) + l l 3 l In BME, by the cosine formula, cos BME BM + ME BE BM ME l l ( 3l) + l l BME 0, cor. to the nearest degree 3. 39

In the figure, VABD is a pyramid whose base ABD is a rectangle. VD is perpendicular to the plane ABD. It is given that AB 0 cm, VD 8 cm and VB 7 cm. Find the angle between the planes VAB and ABD, correct to the nearest degree. A. 8 B. 33. 36 D. 44 The answer is. The required angle is VAD. Join BD. In VBD, VB VD + BD BD 7 8 cm cm In ABD, BD AD + AB AD 0 cm cm In VAD, VD tan VAD AD 8 VAD 36, cor. to the nearest degree The angle between the planes VAB and ABD is 36. 4. 40

In the figure, ABDEFGH is a cuboid. It is given that EH cm, EF 6 cm and H 8 cm. M is the mid-point of AB. G and BH intersect at N. If the angle between the line MN and the plane ABD is θ, find the value of tan θ. A. 4 3 The answer is. B. 6. 4 4 D. 6 Let P be a point on B such that NP B. θ NMP MB EH cm 6 cm BP EF 6 cm 3 cm NP H 8 cm 4 cm In MBP, MP MB + BP MP 6 + 3 cm 4 cm In NMP, NP tan θ MP 4 4. 4

In the figure, ABDEFGH is a cuboid, where AB cm, B 4 cm and H 6 cm. X is a point on BG such that BX : XG :. Y is the mid-point of GH. Which of the following forms the greatest angle with the plane ADEF? A. AX B. AY. E D. DX The answer is A. Let P be a point on AF such that XP AF. Join PX, AX, DX and DP. The angle between AX and the plane ADEF is XAP. XP tan XAP AP The angle between DX and the plane ADEF is XDP. XP tan XDP DP BX 6 cm 4 cm + AP BX 4 cm In ADP, DP AP + AD DP 4 + 4 3 cm cm Let Q be a point on EF such that YQ EF. Join YQ, AY and AQ. 4

The angle between AY and the plane ADEF is YAQ. YQ tan YAQ AQ FQ EF 4 cm cm In AQF, AQ AF + FQ AQ 6 + 40 cm Join E. cm The angle between E and the plane ADEF is ED. D tan ED DE DE 6 cm Note that XP YQ D. AP < DP < DE < AQ XP XP D > > > AP DP DE YQ AQ i.e. tan XAP > tan XDP > tan ED > tan YAQ XAP, XDP, ED and YAQ are acute angles. XAP > XDP > ED > YAQ The answer is A. 6. 43

In the figure, a flagpole D stands vertically on the horizontal plane AB. The bearing of B from A is S70 E. The angles of elevation of D from A and B are 8 and 3 respectively. If AB D m, find the bearing of D from A, correct to the nearest degree. A. N6 E B. N78 E. S6 E D. S78 E The answer is A. In DA, D tan 8 A A m tan 8 In DB, D tan 3 B B m tan 3 In AB, by the cosine formula, cos BA AB + A B AB A + tan 8 tan 8 BA 48.40, cor. to d.p. The required angle 80 70 BA 80 70 48.40 tan 3 6, cor. to the nearest degree The bearing of D from A is N6 E. 44

7. In the figure, a rectangular wall ABD stands vertically on the horizontal ground along the east-west direction. The dimensions of the wall are m 4 m. When the bearing of the sun is N60 W, the area of the shadow DPQ of the wall on the horizontal ground is 0 m. Find the angle of elevation of the sun from the ground. A. 30 B. 4. 4 D. 60 The answer is B. With the notation in the figure, let R be a point on D such that PR D. The required angle of elevation is APD (or BQ). Area of DPQ D PR 0 PR m m θ 60 PDR 90 θ 90 60 30 In PRD, PR sin PDR PD PD m sin 30 4 m In ADP, tan APD AD PD 4

4 4 APD 4 The angle of elevation of the sun from the ground is 4. 8. V 4 m A D B The figure shows a right pyramid whose base ABD is a square. If the angle between the planes VAB and VD is θ, find the value of A. m 7 The answer is B. m B.. 7 sin θ. With the notation in the figure, let M and N be the mid-points of AB and D respectively. Let X be a point on MN such that VX MN. Join VM and VN. D. θ MVN MVN θ XVN XN B m m ND D m m In VND, 46

VD VN VN 4 m m In XVN, XN sin XVN VN θ sin + ND 9. In the figure, ABDEFGH is a cuboid. If FBG HBG 60, find FBH, correct to the nearest degree. A. 60 B. 76. 83 D. 90 The answer is B. Let BG l. In BGH, BG cos60 BH BH l GH tan 60 BG GH 3l In BGF, BG cos60 BF BF l tan 60 FG In FGH, FG BG 3l 47

FH FH FG ( FG + GH 3l) + ( 3l) 6l In FBH, by the cosine formula, + GH BF + BH FH cos FBH BF BH (l) + (l) ( 6l) l l 4 FBH 76, cor. to the nearest degree 60. D E F G cm A cm B 3 cm The figure shows a right triangular prism ABDEF whose base is a right-angled triangle. A, B, E and F lie on the same horizontal plane. G and H are two points on the horizontal plane, where G, A, B and H are collinear. If GA cm, AB cm and BH 3 cm, which of the following has the smallest inclination? A. H B. DA. DB D. DG The answer is. Join DB, FH, BE and GE. H Inclination of H HF F tanhf FH Inclination of DA DAE DE tandae AE Inclination of DB DBE DE tandbe BE 48

Inclination of DG DGE DE tandge GE In ABE, BE AE + AB BE AE + AB In FBH, FH BF + BH FH BF + BH AE + BH In EAG, GE AE + GA GE AE + GA AB > BH > GA i.e. AB > BH > GA BE > FH > GE > AE DE F DE DE F DE > > > AE GE FH BE tandae > tandge > tanhf > tandbe DAE, DGE, HF and DBE are acute angles. DAE > DGE > HF > DBE The answer is. 49