Luonggiac-Chuong10

# Luonggiac-Chuong10 - C HNG X H E TH C L N G TRONG TAM GIA C I NH LY HA M SIN VA COSIN C ho ABC c o a b c la n l t la ba ca n h o i die n cu a A B C

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CHÖÔNG X: HEÄ THÖÙC LÖÔÏNG TRONG TAM GIAÙC I. ÑÒNH LYÙ HAØM SIN VAØ COSIN Cho ABC Δ coù a, b, c laàn löôït laø ba caïnh ñoái dieän cuûa ± ± ± A,B,C, R laø baùn kính ñöôøng troøn ngoaïi tieáp ABC Δ , S laø dieän tích ABC Δ thì === =+ =+− 22 2 2 2 2 2 2 222 abc 2R sin A sin B sin C ab c 2 b c c o s A b c 4 S . c o t g bac2 a c c o s B ac4 S . c o t g B cab2 a b c o s C ab4 S . c o t g A C Baøi 184 ABC Δ . Chöùng minh: A 2B a b bc =⇔ = + Ta coù: 2 2 2 a b bc 4R sin A 4R sin B 4R sinB.sinC =+⇔ = + () () () ⇔−= ⇔− −− = + = ⇔+ = −= += > ⇔−=∨−= π = sin A sin B sin B sin C 11 1 cos 2A 1 cos 2B sin Bsin C cos 2B cos 2A 2sin B sin C 2sin B A sin B A 2sin B sin C sin B A sin A B sin B sin C sin A B sin B do sin A B sin C 0 ABBAB B l o a ï i A Caùch khaùc: + = +− + ⇔= sin A sin B sin B sin C (s in A sin B) (s in A sin B) sin B sin C AB AB AB 2 cos sin .2sin co s sin B sin C 2 2 = = + = > π sin B A sin A B sin B sin C sin A B sin B do sin A B sin C 0 B l o a ï i A

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Baøi 185: Cho ABC Δ . Chöùng minh: ( ) 22 2 sin A B ab sin C c = Ta coù −− = 2 4 R s i n A 4 R s i n B c4 R s i n C () == −+ +− += > 2 11 1 cos 2A 1 cos 2B sin A sin B sin C sin C 2sin A B sin B A cos 2B cos 2A 2sin C sin A B .sin A B sin A B sin C sin C do sin A B sin C 0 Baøi 186: ABC Δ bieát raèng A B1 tg tg 3 =⋅ Chöùng minh 2 c Ta coù : ⋅= = A A B A B tg tg 3sin sin cos cos 3 A B do cos 0,cos 0 ⎛⎞ >> ⎜⎟ ⎝⎠ A BA 2sin sin cos cos sin sin AB cos cos cos 2 cos 2cos * ⇔= + ⎡⎤ ⇔− = ⎢⎥ ⎣⎦ B Maët khaùc: ab2 R s i n As i n B + = ++ = =+ = = A B 4R sin cos AB AB 8R sin cos do * 4R sin A B 4R sinC 2c Caùch khaùc: ⇔+ = c 2R sin A sin B 4R sin C
+− ⇔= −+ + ⎛⎞ = = ⎜⎟ ⎝⎠ A BA B CC 2sin cos 4sin 22 2 2 A BC A B A B cos 2 cos do sin cos 2 2 C 2 ⇔+= A B A B A cos cos sin sin 2cos cos sin 2 AB 3sin sin B 2 ⇔⋅= A B1 tg tg 3 Baøi 187: Cho ABC Δ , chöùng minh neáu taïo moät caáp soá coäng thì cotgA,cotgB,cotgC 222 a,b,c cuõng laø caáp soá coäng. Ta coù: () cot gA,cot gB, cot gC laø caáp soá coäng cot gA cot gC 2 cot gB * Caùch 1: ( ) ()() [] 2 2 2 2 sin A C 2cosB Ta coù: * sin B 2sin A sinCcosB sin A sinC sin B s inB c o sA C c o sA C c o sin B cos A C cos A C cos A C 1 sin B cos B cos2A cos2C 2 1 sin B 1 sin B 1 2sin A 1 2sin C 2 2sin B sin A sin C + = + −− + ⎡⎤ ⎣⎦ + + ⇔=− + + + 22 2 2 2b a c 4R 2b a c a , b ,c laø caâùp soá coäng =+ + ⇔• Caùch 2: ⇔=+ = == 2 2 2 2 22 2 Ta coù: a b c 2ab cos A 1 ab c 4b c s i n A . c o t g A 2 c 4 S c o t g A bca Do ñoù cotgA 4S acb abc Töông töï cotgB ,cotgC 4S 4S Do ñoù: * 2 4S 4S 4S 2b a c

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Baøi 188: Cho ABC Δ coù 22 sin B sin C 2sin A += 2 Chöùng minh ± 0 BAC 60 . () 2 2 22 2 2 Ta coù: sin B sin C 2sin A bc2 a 4R a * ⇔+= A Do ñònh lyù haøm cosin neân ta coù 2 ab c 2 b c c o s =+ ( ) ± +− ⇔= = + =≥ = 0 2b c b c bca cos A ( do * ) 2bc 4bc bc 2 b c 1 do Cauchy 4bc 2 Vaïây : BAC 60 .
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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-Chuong10 - C HNG X H E TH C L N G TRONG TAM GIA C I NH LY HA M SIN VA COSIN C ho ABC c o a b c la n l t la ba ca n h o i die n cu a A B C

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