SKKN học HÌNH học KHÔNG GIAN QUA 15 PHÚT CỦNG cố với ANH NGỮ

HỌC HÌNH HỌC KHÔNG GIAN QUA 15 PHÚT CỦNG CỐ VỚI ANH NGỮ Lời ngỏ : Hình học không gian phân môn học toán hay Tuy , để học sinh thích học ,chịu tìm hiểu hết chương trình không dễ Thực tế , thấy, học nào, tạo tình mới, tự nhiên, không áp lực, bọn trẻ tỏ thích học toán, học toán giỏi, kể học sinh chọn thi khối C, D Hưởng ứng nhu cầu bước đầu, dạy học toán Anh ngữ nhà trường đề ra, năm nay, em khối 11 có phần củng cố hình học không gian tiếng Anh, khoảng 5_ 15 phút sau hết, tiết tập, học, dịp tạo yêu cầu cho em vui mà học Có thể chia tài liệu nầy làm phần : _ Phần : Những toán quan hệ song song Bài toán đơn giản, số lượng ít, em cần phải làm quen khái niệm , thuật giải phân môn , từ vựng Những toán quan hệ vuông góc Bài toán phức tạp dần, phong phú , chuẩn bị nhiều hơn, học sinh viết tự tin _ Phần : Trình bày tình tiết học thực _ Phần : Một số đề tập tham khảo, thay đổi, cho tiết học sau Có toán học sinh tập viết hoạt động nhóm, xin đưa vào đây, ghi nhận lực tính hiếu học em Một số tập đề nghị nữa, chủ yếu lấy từ chương trình hành, mục đích học toán Rất tiếc thời gian dành cho vấn đề eo hẹp Tôi e rằng, nhiều lỗi biên dịch Mấy lời trình bày Kính mong quí Thầy dẫn thêm Giáo viên Nguyễn Kim Thoa Unit : PARALLEL RELATION Problem : Given pyramid S.ABCD M ∈ AB, N ∈ CD (α ) ∋ M , (α ) // SA a/ Find the intersections of (α ) and (SAB) and (SAC) b/ Find the intersection of pyramid S.ABCD and (α ) c/ Find the MN’s condition in order the intersect plane of pyramid S.ABCD and (α ) is a trapezium Solution S a Is there any common point of (α ) and (SAB) ? P _ Direction of (α ) ? b Means of the intersection of pyramid and a plane ? Q A D M R B N C a/ Find the intersections of (α ) and (SAB) and (SAC): α // SA   SA ⊂ ( SAB )  M ∈ α ∩ ( SAB )  ⇒ α ∩ ( SAB ) = MP ( MP // SA) On the analogy, call { R} = MN ∩ AC , we have α ∩ ( SAC ) = RQ // SA b/ Find the intersection of pyramid S.ABCD and (α ) : In turn MP,PQ,QN,NM are intersections of (α ) and (SAB), (SBC), (SCD), (ABCD) Thus the intersection of pyramid S.ABCD and (α ) is a quadrangle MNPQ c/ Find the MN’s condition in order the intersection of pyramid S.ABCD and (α ) is a trapezium :  MP // QN (1) MNPQ is a trapezium ⇒   MN // PQ (2) (1) ⇒ SA // QN (as SA//MP) ⇒ SA //( SCD) ((!), as S int er sec tion po int) (2) ⇒ MN // BC ,  BC = ( ABCD ) ∩ ( SBC )  as  MN ⊂ ( ABCD)  PQ ⊂ ( SBC )   PQ = α ∩ ( SBC ) (agreed)  MN ⊂ α & BC ⊂ ( SBC ) Oppositely, if MN//BC then MN//PQ, as  Thus MN//BC is the MN’s condition in order the intersection of pyramid S.ABCD and (α ) is a trapezium Problem : On a plane (P), given parallelogram ABCD Draw the parallel rays : Ax // By // Cz // Dt ( in the same direction of (P)) A plane ( Q ) intersects with Ax, By, Cz, Dt at A’, B’, C’, D’ a/ Prove (Ax,By)//(Cz,Dt) b/ Which shape is tetragon A’B’C’D’? c/ Prove AA’ + CC’ = BB’ + DD’ z y C' t x B' O' D' A' B C O A D Solution: a/ (Ax,By)//(Cz,Dt): Ax // Cz   ⇒ ( Ax, By ) //(Cz, Dt ) AB // DC  b/ The shape of tetragonal A’B’C’D’?  A ' B ' = (Q) ∩ ( Ax, By )  C ' D ' = (Q) ∩ (Cz , Dt ) ⇒ A ' B '// C ' D ' (1) ( Ax, By ) //(Cz , Dt )  On the analogy, we also have A ' D '// B ' C ' (2) (1)&(2) ⇒ A’B’C’D’ is a parallelogram c/ AA’ + CC’ = BB’ + DD’ : Put O = AC ∩ BD; O ' = A'C ' ∩ B ' D ' , then : O is the middle point of AC , O’ is the middle point of A’C’, so: AA ’ + CC’ = 2OO’ On the analogy, we also have BB’ + DD’ = 2OO’ Therefore : AA’ + CC’ = BB’ + DD’ Problem : Given a uu pyramid S.ABC Call I a central point of SB uur ur JS = −2 JC , O is the centre of ∆ABC a/ Find the intersection of the pyramid and (OIJ) b/ M ∈ (α ) , ( α ) // (OIJ) Call BC = x ( x > ) Find x in order ( α ) intersect with pyramid S.ABC c/ Find the intersection of the pyramid and ( α ) base on x Solution : a/ Find the section of the pyramid and plane(OIJ) : Call P0 the midlle point of BC • In (SBC): IS JS IS JC DB ≠ ⇒ IJ ∩ BC = D : = (Ménélaus) IB JC IB JS DC DB ⇒ (−1).(− ) = ⇒ DB = DC , DC ie: C is central point of BD ( thus : DB = DP0 ; BD = BP0 ) • In (ABC): OD ∩ AB = L; OD ∩ AC = K  In ∆ ABP: ⇒ LA DB OP0 = ( by Ménélaus theorem) LB DP0 OA LA    − ÷ = ⇒ LB = − LA (thus BA = BL ) LB    In ∆ ABC: LA KC DB = (by Ménélaus theorem) LB KA DC KC = ⇒ KA = −3KC KA Thus: −  Thus the section of plane (OIJ) and the pyramid is IJKL b/ Find x in order ( α ) intersect with pyramid S.ABCD : ( α ) // (OIJ) or ( α ) ≡ (OIJ) ⇒ the section of ( α ) and each plane of the pyramid, if existing, must either parallel or coincide with the side of the plane section IJKL ( α ) ∋ M , ( ABC ) ∋ M ⇒ (α ) ∩ ( ABC ) ≠ ∅ The direction of the intersection of ( α ) and (ABC)is the direction of LK, not the direction both of AB and AC, so certainly, ( α ) intersect with both AB and AC Call the intersection of ( α ) and AB, AC : N, R, the intersections of SB, SC, SA and ( α ) : P, Q, E, if existing, then : BM BN = x > then ≥ 0, , and : BC BA BN BL BA BN BP = < ⇒0< < BP BI BS BA BS CR CK CA CR CQ CR BN = < ⇒0< < < , moreover, CQ CJ CS CA CS CA BA If Thus, if N ∈ AB then ( α ) intersect with the pyramid S.ABCD M S Q P D J I j R A O C K N L B BM BM > and →0 BC BC BM =0 N≡B⇔ M ≡ B ⇔ BC As N → B ⇔ M → B ⇔ N → A ⇔ M → M0 N ≡ A ⇔ M ≡ M0 : BM BA = =5 BC BL Thus, if N move on AB then M move on BM0 , ( α ) intersect either BC or AC Thus : ( α ) intersect the pyramid ⇔ < x < c/ Confute the intersection of the pyramid and ( α ) base on x 0< x < ⇔ < BM < ⇔ M ∈ BC: the plane section is ∆ MNR BC • x = ⇔ M ≡C section is ∆ CNR • < x < ⇔1< : the BM < ⇔ M ∈ CM0 and BM = BC : the section BC is NPQR • Espeacially, x = 2: M ≡ D : the section is KLIJ x = 4: P ≡ Q ≡ S : the section is ∆ SNR ( M ≡ M ) • • < x < ⇔4< BM < ⇔ M ∈ M0M1: the section is ∆ QNE BC S D J I A P L N K O j M C B M S Q P D J I R A O N L B K C M S T D J I A O R N K C L B Cho hình hộp ABCD A’B’C’D’ Tâm I: tâm bình hành ABCD.A’B’C’D’ uuur uuuuur uur Chứng minh: vectơ BC , B ' D ', IK đồng phẳng Given a parallelepiped ABCD A’B’C’D’ Call I centre - Unit : PERPENDICULAR RELATION Program : Given a regular tetrahedron ABCD Call M central point of BC Calculate the angle between lines : a/ AM and CD b/ AB and CD A A P P N N B D B H M D H M C C Hình a Hình b Solution : a/ Calculate the angle between AM and CD : Draw MN//CD then ·AMN = ( AM , MN ) = ( AM , CD) ∆AMN is isosceles at A: MN · cos AMN = = AM a = 3 a · , CD = arccos Therefore : AM ( ⇒ ·AMN = arccos ) b/ Calculate the angle between AB and CD : Call N,P central points of BD, AC  a   ⇒ PN ⊥ BD, N : central po int BD  PB = PD =  a   a 2 a2 PN = PB − BN =  ÷÷ −  ÷ =    2 · MP // AB; MN // CD ⇒ PMN = ( MP, MN ) = ( AB, CD ) 2 PM + MN − PN · · In ∆ MPN : cos PMN = = ⇒ PMN = 900 PM MN 20/ Given a pyramid S ABCD whose bottom ABCD is a parallelogram with centre O In tun call I, J central points of SA, SD a/ Prove (OIJ ) //( SDC ) b/ Call M, N central points of AB, OJ in turn Prove MN // (SBC) 21/ Given a pyramid S ABCD whose bottom is a quare of side a with centre O I is the cental point of BC SO ⊥ ( ABCD), SA = 2a a/ Prove BC ⊥ ( SIO) b/ In turn call OH, OK heights of ∆SOI , ∆SOC Prove SC ⊥ HK Calculate HK c/ In turn call M, N, P, Q central points of SA, SB, SC, SD Calculate MNPQ’s area base on a 22/ Given a pyramid S.ABCD whose bottom ABCD is trapezium ( AB // DC, AB =DC ) SA ⊥ ( ABCD ) a/ Find the intersection of (SAC) and (SBD) b/ M ∈ SD Find N = SC ∩ ( ABM ) c/ I = AM ∩ ( BN ) Find the set of I when I move on SD d/ E ∈ DC Draw AH ⊥ BE at H Prove SH ⊥ BE Find the set of H when E move on DC 23/ Given a regular prism ABC.A’B’C’ of side a Call D central point of CC’ a/ Calculate the angle between DB and A’B’, and the angle between (DAB) and (ABC) b/ Prove a pyramid D ABB’A’ is a regular pyramid BẢNG TRA TỪ VỰNG : Acute Angle Area góc nhọn góc diện tích Center Central point Circle Chord Common tâm trung điểm hình tròn dây cung chung Diameter Diamond Direction chu vi hình thoi phương, hướng Intersect Intersectton Intersectton (point ) giao ( tương ) giao giao điểm Height đường cao Line Line of intersection đường giao tuyến Midpoint trung điểm Parallel Parallelepiped Parallelogram Plane intersec Prism → Regular prism Pyramid → Regular pyramid song song hình hộp hình bình hành Thiết diện thẳng Hình lăng trụ → lăng trụ hình chóp → hình chóp Quadrangle Tứ giác Ray Rhombus Rectangle Relation tia hình thoi hình chữ nhật quan hệ Space Square không gian hình vuông Surface of one side mặt bên Tetrahedron → Regular tetrahedron Theorem Triangle → Right-angled triangle tứ diện → tứ diện định lý tam giác Tam giác vuông → Isosceles triangle → Equilateral triangle Trapezium → Tam giác cân → Tam giác hình thang I/ THE INTEGRAL CALCULUS 1/ Trigonometric function: Given a trigonometrical function: y = sin x + cos x a/ Find the determination of this function b/ Draw the graph of this function c/ Find the maxima and minima of this function 2/ Simple trigonometric equations Solving trigonometric equations: 1/ sin x = 2 / cos x = sin π 10 π )= 3 / cot x = tan x / cot(3 x + / sin x − cos x = 3/ Regular trigonometric equations: Solving trigonometric equations 1/ sin x + cos x = 2/ sin x + cos x = −1 / 3(sin x + cos x) = 2sin x / sin x + sin 2 x + sin 3 x + sin 4 x = / + sin x + cos x + sin x + cos x = EXERCISES 1/ Given a trigonometrical function: y = cos x a/ Find the determination of this function b/ Draw the graph of this function c/ Find the maxima and minima of this function in (−π ,5π ) 2/ Solving trigonometric equations: 1/ s in3x=sin 5π / cos(3 x − 150 ) = − 2 x π  / tan  − ÷ =  12  π  / cos  x − ÷+ = 6  / sin x = cos + 3(sin x − 1) 2 / sin x cos x cos x − sin xcos x + sin x = 6/ cos x − sin x = π x x / sin x − sin ( x + ) = 4sin cos cos x 2 / 3.cos x + sin x = cos x sin x =0 + cos x π  11/ cos x + sin  x + ÷ 4  12 / tan x.cot x = 10 / 3/ Solving trigonometric equations 1/ cos x + 4sin x − = / tan x + cos x = 3/ sin x + cos x = sin x cos x     /  sin x + ÷+  sin x + ÷= sin x   sin x   π π   / sin  x − ÷− cos  x − ÷ = 3 3   / sin x + cos x = 2sin x / cos x + 4sin x + 3sin x = / sin x = 3sin x.cos x − / + sin x.sin x = 3cos x 10 / tan x − 2sin x = sin x sin x 12 / 2sin17 x + cos x + sin x = 11/ cot x − tan x + 4sin x = 13 / 5sin x − = 3(1 − sin x) tan x π   14 / 2 sin  x − ÷cos x =  12  15 / cos x + sin x = 1 16 / + = sin x cos x sin x 17 / 2sin x + cos x = 13 sin14 x 18 / 12 cos x + 5sin x + 4/ Given this equation: 1/ Prove x = +8 = 12 cos x + 5sin x + 14 sin x + cos 3x = cos x cos x − sin x π + kπ is a solution of this equation 2/ Solve this equation with the way call tan x = t ( x ≠ π + kπ ) 5/ The value of angle in a right triangle ∆ABC is solution of this equation: sin x + sin x sin x − 3cos3 x = Prove that ∆ABC is an isoscele triangle 6/ Given a equation: cos x − (2m + 1) cos x + m + = a/ Solve this equation with m =  π 3π  b/ Find the value of m to make this equation have solution x ∈  ; ÷ 2  7/ Find solutions x ∈ ( 0, 2π ) of a equation: 9π  sin  x +  15π   ÷− 3cos  x −   8/ Given a equation: m sin x + (m + 1) cos x = a/ Solve this equation with m =  ÷ = + 2sin x  m cos x b/ Find the value of m to make this equation have solution 9/ Find solutions x ∈ ( 0, 2π ) of each of the following: sin x = cos x − sin x sin x − sin x b / = cos x + sin x − cos x a / II/ SETS AND PROBABILITY 1/ Counting rules 1/ Every card in the deck of 52 has an equal chance of being selected Find the probability of selecting: a/ a red card b/ a black Ace 2/ What is the a priori probability of c/ the Jack of diamonds (a) a given number showing on a roll of fair, six-sided die (b) selecting a male from 100 2/ Rearrangment 3/ Arrangement 4/ Combinatory III/.INDUCTION, ARITHMETIC PROGRESSION 1/ Prove a / 12 + 22 + 32 + + n = n(n + 1)(2n + 1) b / (13n − 1) contain (∀n ∈ ¥ ) (∀n ∈ ¥ ) 2/ Find the first tem and the arithmetical ratio of each of the following: u5 = 19 a /  u9 = 35 u6 = b /. 2 u2 + u4 = 16 3/ Find the first tem and the denominator of each of the following: u1 + u5 = 51 a /  u2 + u6 = 102 u1 + u2 + u3 = 21 b /  2 u1 + u2 + u3 = 189 EXERCISES 1/ Prove that 1+ 1 + + + > n n (∀n ≥ 2) 2/ Prove that + cos x + cos x + + cos nx = 3/ Prove that sin (n + 1) x nx cos 2 x sin ( x ≠ kπ ) 1 + + + >1 n +1 n + 3n + 1 2n + 1 b / < 2n + 3n + (∀n ∈ ¥ *) a / (∀n ∈ ¥ *) c / 1.2 + 2.5 + + n(3n − 1) = n ( n + 1) (∀n ∈ ¥ *) d / + + + + (3n − 1) = n.(3n − 1) (∀n ∈ ¥ *) e / n(2n − 3n + 1) contain (∀n ∈ ¥ *) f / 11n +1 + 122 n −1 contain 133 (∀n ∈ ¥ *) 4/ Find the first tem and the arithmetical ratio of each of the following: u2 − u3 + u5 = 10 a /  u4 + u6 = 26 u6 + u7 = 60 b /. 2 u2 + u12 = 1170 u3 − u7 = −8 c /. u2 u7 = 75 u12 + u22 + u32 = 155 d /   S3 = 21 5/ Find the first tem and the denominator of each of the following: u2 + u4 + u6 = −42 a /  u3 + u5 = 20 u1 + u2 + u3 = 108 b /  u4 + u5 + u6 = 21  1 1 1 u1 + u2 + u3 + u4 + u5 = 49  + + + + ÷ c /   u1 u2 u3 u4 u5  u + u = 35  ( uu ≠ ) 6/ Calculate a / S1 = 105 + 110 + 115 + 120 + + 995 b / S = 100 − 99 + 982 − 97 + 96 − 952 + + 2 − 12 c / S3 = + 77 + 777 + + 77 { n letters 7/ Given arithmetic progression a,b,c Sum of them is 21 Turn in add 2,3,9 to a,b,c, we have a denominator progression Find a,b,c IV/ CONTINUITY AND LIMITS 1/ Find the limits: 1/ lim (2n + 1)(2n5 + n) n7 + 2 / lim( n3 − 2n − n / lim (2 x + 3) x→2 / lim ( x + + x ) x →1 / lim x →∞ x2 + x3 − x + 2/ Muster the continuity of this function in ¡ 3  f ( x) =   x +1 −1  x + − ( x ≤ 0) ( x > 0) EXERCISES 1/ In problem to prove each of the relationships with the aid of a series whose general term is the given function an 1/ lim =0 x→ ∞ n! nn / lim =0 x→ ∞ (2 n )! ( n !) n =0 nn (2n) n / lim =0 ( a >1) x→ ∞ a n! nn / lim =0 x→ ∞ ( n !) / lim x→ ∞ 3/ In problem to find the limits sin x x →0 x tan x / lim x →0 x cos x − sin x / lim π cos x x→ 1/ lim   / lim  − ÷ x →0 sin x tan x   − cos x / lim x →0 x2 4/ In problem to 11 find the limits 1/ 2/ 3/ 4/ 5/ x +5 lim x →2 x −3 x lim x→ 1 −x x −2 x +1 lim x→ x3 −x ( x −1) − x lim x→ x −1  3x (2 x −1)(3 x + x +2  lim  − ÷ x →∞ x −1 4x2     / lim  − ÷ x→ 1 − x −x  x +1 + x / lim x →∞ x3 + x −x x +1 − − x / lim x →0 x x −1 / lim x → x −5 x −1 10 / lim( x →∞ 11/ lim x →∞ x +a − x ) + x3 − 3 +x x −1 5/ Muster the continuity of this function in ¡ 5  f ( x) =  x −   x −1 ( x = 1) ( x ≠ 1) 6/ Muster the continuity of this function at x0 =  x − 3x + f ( x) =  2 x − ( x < 1) ( x ≥ 1) V/ DERIVATIVE Find the derivative : f ( x) = sin(2 x − 1) EXERCISES VI/.THE DIFFERENTIAL CALCULUS 1/ Find the derivatives of each of the following a / f ( x) = x − x b / f ( x) = 12 x c / f ( x) = x cos x d / f ( x ) = tan x e / f ( x) = x cos(1 − x) f / f ( x) = cos x g / f ( x) = sin(2 x − 1) [...]... B’C’ ⇒ II ' = BB ' = AA ' -Thus: o AA ' II ' is a rectangle and AA ' ⊥ ( ABC ) oBB ' C ' C is a square C' A' I' B' J A C I B -AI: the height of equilateral triangle ABC with AB= a a 3   2  ⇒ AI ⊥ ( BB ' C ) AI ⊥ BB ' ( as (1) )  ⇒ AI ⊥ BC , AI = (*) ⇒ AI ⊥ B ' C (1) -B’C’, C’B are diagonals of a square BCC’B’ Thus, C ' B ⊥ B ' C  IJ // C ' B ⇒ IJ ⊥ B ' C (2)  -Call J central point of CC” : ... tetrahedron OABC OA ⊥ OB, OA ⊥ OC , OC ⊥ OB HO ⊥ ( ABC ), O ∈ ( ABC ) a/ Prove ∆ABC is acute angled b/ Prove H is a orthocentric of ∆ABC c/ Prove 1 1 1 1 = + + 2 2 2 OH OA OB OC 2 15/ Given a pyramid S.ABCD whose bottom is a square with centre O SO ⊥ ( ABCD) I is the central point of BC a/ Prove AC ⊥ SD, BC ⊥ ( SOI ) b/ In turn call OH, OK the hieghts of ∆SOI , ∆SOC Prove SC ⊥ (OHK ) 16/ The base of... triangle with right angle O 2 OA a 2 =1 ¶ ⇒ cos S AO = = AS 2 a 2 ⇒ S ¶AO = 45o Thus ( AS· , ( ABCD)) = 45o b/ Calculate the angle between SB and AC AC ⊥ SO ( as SO ⊥ ( ABCD )) AC ⊥ BD ( as ABCD is a square ) ⇒ AC ⊥ ( SBD) ⇒ AC ⊥ SB ⇒ ( AC , SB ) = 90o Problem 4 : On a plane (P), given rhombus ABCD From 4 points A,B,C,D, draw the rays Ax, By, Cz, Dt that all are perpendicular to ( P) and in the same... ⊂ ( SAB ), BK ⊥ SA  ⇒ (( S ·AB ), ( SAD)) = ( BK = 900 KD ⊂ ( SAB ), KD ⊥ SA ⇒ ( SAB ) ⊥ ( SAD) ( Ngọc Hà, Hoàng Dịu-11 Văn ) SOME SIMILAR PROBLEMS 5/ Given a pyramid S.ABCD whose bottom ABCD is a square SA ⊥ ( ABCD) a/ Prove CD ⊥ ( ABJ ) and AB ⊥ (CID) b/ Prove IJ ⊥ CD c/.Calculate IJ base on a 6/ Given tretrahedron ABCD ∆ABC is a equilateral triangle of side a AD ⊥ BC and AD = a , the value od... BC, D is a symatric piont of A to I SD ⊥ ( ABC ), SD = a 6 2 a/ Prove ( SBC ) ⊥ ( SAD) b/ Find the distance between I and SA c/ Prove ( SAB) ⊥ ( SAC ) 8/ Given a pyramid S,ABCD whose bottom ABCD is a square with center is O SA=SB=SC=SD=AB=a M is SC’s midpoint a/ Calculate SO b/ Prove ( MBD) ⊥ ( SAC ) c/ Calculate OM and the angle between (MBD) and (ABCD) 9/ Given a retangular parallelepiped ABCD.A’B’C’D’... ∆SBD is an right-angled triangle O = AC ∩ BD As AS = AB, ·AOS = ·AOB = 1v ⇒ ∆ABO = ∆ASO ⇒ BO = SO SO = BO = OD ⇒ ∆SBD is right angled at S Problem 5 : Given a regular pyramid S.ABCD whose bottom of is a square a/ Calculate the angle between SA and (ABCD) b/ Calculate the angle between SB and AC Solution: S A B O D C a/ Calculate the angle between SA and (ABCD) BD ∩ AC = { O} ⇒ SO ⊥ ( ABCD ) (as S.ABCD... paralellogram ABCD, M is the midpoint of the edge SA , N is the midpoint of the edge BC, I is the midpoint of the edge CD Prove S,J,O are in the straight line 17/ Given a pyramid S.ABCD whose bottom is a square ABCD of side a SA ⊥ ( ABCD), SA = a · , ( ABCB)) a/ Call M the central point of CD Calculate cos( S M b/ Calculate the value of the distance between A and (SBC) · ), ( SBD)) c/ Calculate (( SC B... parallelogram with centre O In tun call I, J central points of SA, SD a/ Prove (OIJ ) //( SDC ) b/ Call M, N central points of AB, OJ in turn Prove MN // (SBC) 21/ Given a pyramid S ABCD whose bottom is a quare of side a with centre O I is the cental point of BC SO ⊥ ( ABCD), SA = 2a a/ Prove BC ⊥ ( SIO) b/ In turn call OH, OK heights of ∆SOI , ∆SOC Prove SC ⊥ HK Calculate HK c/ In turn call M, N, P, Q ... song hình hộp hình bình hành Thiết diện thẳng Hình lăng trụ → lăng trụ hình chóp → hình chóp Quadrangle Tứ giác Ray Rhombus Rectangle Relation tia hình thoi hình chữ nhật quan hệ Space Square không. .. 7/ Find solutions x ∈ ( 0, 2π ) of a equation: 9π  sin  x +  15   ÷− 3cos  x −   8/ Given a equation: m sin x + (m + 1) cos x = a/ Solve this equation with m =  ÷ = + 2sin x  m cos... kπ is a solution of this equation 2/ Solve this equation with the way call tan x = t ( x ≠ π + kπ ) 5/ The value of angle in a right triangle ∆ABC is solution of this equation: sin x + sin x sin