Luonggiac-Chuong11

# Luonggiac-Chuong11 - CHNG XI NHAN DAN G TAM GIAC I TNH CA C...

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CHÖÔNG XI: NHAÄN DAÏNG TAM GIAÙC I. TÍNH CAÙC GOÙC CUÛA TAM GIAÙC Baøi 201: Tính caùc goùc cuûa ABC Δ neáu : ()()() () 3 sin B C sin C A cos A B * 2 ++ += Do A BC + π Neân: 3 *s i n A s i n B c o s C 2 +−= +− ⎛⎞ ⇔− ⎜⎟ ⎝⎠ = + = −− = −+ + = = = == 2 2 2 2 2 2 2 = A BA B C 3 2sin cos 2 cos 1 22 2 CA B C 1 2cos cos 2 2 CC A B 4cos 1 0 2 B A B 1 cos 0 2 B A B 2 cos cos sin 0 2 B AB sin 0 2 C cos0 1 2 A 2 π = ⎪⎪ ⎨⎨ = = π π = C 23 B 0 2 6 2 C 3 Baøi 202: ABC Δ bieát: 5 cos2A 3 cos2B cos2C 0 (*) 2 + = Ta coù: 2 5 *2 c o s A 1 2 3 c o s B C c o s B C 2 0 + + = ⎡⎤ ⎣⎦

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( ) () ⇔− + = ⎡⎤ −− + ⎣⎦ + = −= ⎪⎪ ⇔⇔ ⎨⎨ = =− = == 2 2 2 2 2 0 0 4cos A 4 3cosA.cos B C 3 0 2cosA 3cos B C 3 3cos B C 0 3sin B C 0 sin B C 0 BC0 3 3 cos A cos A cos B C 2 2 A3 0 BC7 5 = Baøi 203: Chöùng minh ABC Δ coù neáu : 0 C 120 = A BC sin A sin B sin C 2sin sin (*) 22 2 ++− = Ta coù A B A C A (*) cos cos sin 2 2 CA B CC A B A 2cos cos cos sin 2 2 2 2 B C AB cos cos sin cos cos 2 2 2 B A B A B cos cos cos cos cos 2 2 2 CAB cos cos cos cos 222 +− ⇔+ = −+ = + ⎛⎞ = ⎜⎟ ⎝⎠ = ⎢⎥ ⇔= 2 B 2 + C1 cos (do A cos 0 2 > vaø B cos 0 2 > A B 0; 22 2 π < < ) 0 2 0 Baøi 204 : Tính caùc goùc cuûa C ΔΑΒ bieát soá ño 3 goùc taïo caáp soá coäng vaø 33 sin A sin B sin C 2 + ++= Khoâng laøm maát tính chaát toång quaùt cuûa baøi toaùn giaû söû A << Ta coù: A, B, C taïo 1 caáp soá coäng neân A + C = 2B Maø A π neân B 3 π = Luùc ñoù: sin A sin B sin C 2 +
33 sin A sin sin C 32 3 sin A sin C 2 AC AC 3 2sin cos 22 2 BA C 3 2cos 2 3A C 3 2. cos 2 CA 3 cos cos 6 π+ ⇔++= ⇔+= +− ⇔= ⎛⎞ ⎜⎟ ⎝⎠ −π = Do C > A neân coù: C ΔΑΒ π = = ππ ⎪⎪ += = ⎨⎨ == C 26 2 2 A 36 BB Baøi 205: Tính caùc goùc cuûa ABC Δ neáu ( ) () +≤ ++= + 2 bca 1 sin A sin B sin C 1 2 2 AÙp duïng ñònh lyù haøm cosin: cos A 2bc = 2 2 Do (1): neân co s A 0 Do ñoù: A A 24 ≤< π ⇔≤ < 2 2 π Vaäy A2 cos cos 2 π =∗ Maët khaùc: sin A sin B sinC ++ BC BC sin A cos + =+ A BC sin A cos 2 12 1 2 +⋅ do * vaø cos 1 2 Maø sin A sin B sin C 1 2 do (2) +

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Daáu “=” taïi (2) xaûy ra = ⇔= = sin A 1 A 2 cos 22 BC cos 1 2 π = π = = A 2 4 Baøi 206 : (Ñeà thi tuyeån sinh Ñaïi hoïc khoái A, naêm 2004) Cho ABC Δ khoâng tuø thoûa ñieàu kieän ( ) cos2A 2 2cosB 2 2cosC 3 * ++= Tính ba goùc cuûa ABC Δ * Caùch 1 Ñaët M = 2 2cosB 2 2cosC 3 + +− Ta coù: M = 2 BC BC 2cos A 4 2 cos cos 4 + 2 A 4 2sin cos 4 Do A sin 0 2 > vaø B - C cos 1 2 Neân 2 A M2 c o sA42 s i n 4 2 Maët khaùc: ABC Δ khoâng tuø neân 0A 2 π < ⇒≤ 2 0c o s A1 cos A cos A Do ñoù: A c o s A42 s i n 4 2 ≤+ 2 2 2 A A M1 2 s i n 4 2 s i n AA M4 s i n 4 2 s i n 2 A 2 s i n 1 0 2 ⎛⎞ ⇔≤− + ⎜⎟ ⎝⎠ ⇔≤ + 4 Do giaû thieát (*) ta coù M=0 Vaäy: 2 0 0 cos A
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## This note was uploaded on 11/26/2011 for the course MATH 1002 taught by Professor Chuck during the Spring '11 term at University of Western States.

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Luonggiac-Chuong11 - CHNG XI NHAN DAN G TAM GIAC I TNH CA C...

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