专题4.6解三角形九大题型（原卷版）_11.4-2_2026-11-04

      
      
专题4.6 解三角形【九大题型】

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sin𝐴−sin𝐵
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=
sin𝐶
𝑎+𝑏¥
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2 ¥
  
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(2)./
∠𝐶𝐴𝐵= ∠𝐶𝐴𝐷¥
      
      
L3 8-1u2023·Ņņž·ŃŎŏDv„ △𝐴𝐵𝐶0q∠𝐵𝐴𝐶= 60°q △𝐴𝐵𝐶'78’10 3q𝐷’𝐵𝐶'0
Nq𝐷𝐸⊥𝐴𝐶–N𝐸,𝐷𝐹⊥𝐴𝐵–N𝐹.
  
(1)# △𝐷𝐸𝐹'78—
(2)ô𝐴𝐷=
129
2 q#sin∠𝐴𝐵𝐶+ sin∠𝐴𝐶𝐵'>.
L3 8-2u2024·ŋʼnʼnŌ·_Ńv¨eF7Ů(&𝐴𝐵𝐶𝐷'•%ďTń’𝐴𝐶q𝐵𝐷qY0𝐷𝐶=
1
3𝐴𝐵,sin∠𝐵𝐴𝐷⋅
tan∠𝐴𝐵𝐷= sin∠𝐴𝐵𝐷⋅sin∠𝐴𝐷𝐵¥
(1)NJå
 △𝐴𝐵𝐷wڒă%$%&—ôw¥Ŵî/Î-%’ă%qô5wqŴNjāï—
(2)ŷF7Ů(&𝐴𝐵𝐶𝐷'78’ Sqô|𝐷𝐶| = 2qŐ19𝑆< 𝜆q#j
 λ 'Ù>?@¥
L3 8-3u2023·ĦʼnnjÊ·˜Ńvũĉq„F7Ů(&𝐴𝐵𝐶𝐷0q∠𝐴= 135°q𝐴𝐵= 2q∠𝐴𝐵𝐷'FTď
J𝐴𝐷–N𝐸qŐ𝐵𝐸= 2 2¥
      
      
  
(1)#∠𝐴𝐵𝐸X𝐵𝐷—
(2)ô∠𝐵𝐶𝐷= 60°q# △𝐵𝐶𝐷<;'=‘>¥
 9  $%I
),$%&'JK:
ì 9u2023·ţņ·ŃŎŏDv¨eI
𝑓(𝑥) = 2 3sin𝑥cos𝑥−2cos2𝑥.
(1)#I
𝑦= log2𝑓(𝑥)'"Ǎǎ6>ǎ—
(2)¨eŒ% △𝐴𝐵𝐶'$-z%Tń’ AqBqCqô𝑓𝐴
2 = 0q#
𝑏+𝑐
𝑎'=‘>.
L3 9-1u2024·mn·ŃŎŏDv¨eI
𝑓(𝑥) = sin2𝑥
2−3sin
𝑥
2cos
𝑥
2 +1.
(1)#I
𝑦= 𝑓(𝑥)'bǏǐǑƟƗ
(2)„ △𝐴𝐵𝐶0qz% AqBqC '•(Tń’ aqbqcqŐõö𝑎2−𝑏2 = 𝑎𝑐cos𝐵−
1
2𝑏𝑐q#𝑓(𝐵)'Ù>?@.
L3 9-2u23-24 _¬·ŮŰǒ0·ŗǓv¨eI
𝑓(𝑥) = 2sin(𝜔𝑥+ 𝜑) 𝜔> 0,−𝜋
2 < 𝜑< 𝜋
2 '©Tĉĸ
ũĉĬƛ¥
      
      
  
u1v#I
𝑓(𝑥)',U3—
u2v„Œ% △𝐴𝐵𝐶0q%𝐴q𝐵q𝐶Ĭ•'(Tń’𝑎q𝑏q𝑐qô𝑓(𝐴) =
3q𝑏= 2qŐ △𝐴𝐵𝐶'78’3 3
2 q
#𝑎¥
L3 9-3u2024·Ťŧ·$Ńv¨eI
𝑓(𝑥) = 2 3sin𝜔𝑥cos𝜔𝑥+ 2cos2𝜔𝑥,(𝜔> 0)'=û<ŗ’π.
(1)#𝜔'>—
(2)„Œ% △𝐴𝐵𝐶0q% AqBqC Ĭ•'(Tń’ aqbqc.c ’𝑓(𝑥)„ 0, π
2 ě'=‘>qđÁÂ①ÁÂ
②ÁÂ③Ţ$-ÁÂ0…_-§’¨eq#𝑎−𝑏'Ù>?@.ÁÂ①
𝑎cos𝐵+ 𝑏cos𝐴= 2𝑐cos𝐶—ÁÂ
②
2𝑎sin𝐴cos𝐵+ 𝑏sin2𝐴=
3𝑎—ÁÂ③
 △𝐴𝐵𝐶'78’ SqŐ𝑆=
3(𝑎2+𝑏2−𝑐2)
4
.
ũĶ…Š-ÁÂ
Tń,qǔp_-ÁÂRT.
_b
1¥u2024·ŀʼnǕň·˜Ńvŷ △𝐴𝐵𝐶'z% AqBqC '•(Tń’𝑎q𝑏q𝑐qô𝑏= 1q𝑎2−1 = 𝑐(𝑐−1)q
Û Aǖu    v
      
      
A¥
π
3
B¥
2π
3
C¥
π
6
D¥
5π
6
2¥u2024·šňǗǘƙ·$Ńv„ △𝐴𝐵𝐶0q𝐴𝐵= 2q𝐴𝐶= 3q ∠𝐴=
π
3qÛ △𝐴𝐵𝐶ǙĄǚ'ǛĒuǜǜv
A¥ 7
3
B¥ 21
3
C¥2 7
3
D¥2 21
3
3¥u2024·ŤŧĚǝ·˜Ńv„ △𝐴𝐵𝐶0q𝐴𝐵= 4,𝐴𝐶= 5,cos𝐶=
3
4qÛ𝐵𝐶';’u    v
A¥6 4
3
2
B¥6
C¥3 + 3 2
D¥3
4¥u2024·ŭǞǟŰ·$Ńv △𝐴𝐵𝐶'z% AqBqC '•(Tń’ aqbqcqŐ𝑎= 4qsin𝐶=
1
4qô △𝐴𝐵𝐶
9¦,qÛ c 'Ù>¾[’u    v
A¥3
B¥4
C¥5
D¥6
5¥u2024·xǠ·ŃŎŏDvŷ △𝐴𝐵𝐶'z%𝐴,𝐵,𝐶'•(Tń’𝑎,𝑏,𝑐qô𝐵=
2
3π,𝑏= 6,𝑎2 + 𝑐2 = 3𝑎𝑐qÛ △𝐴𝐵𝐶
'78’u    v
A¥9 3
4
B¥
9
4
C¥9 3
2
D¥
9
2
6¥u2024·ŋʼnʼnŌ·ŃŎŏDv„100m'ƗƘ𝐴ƜqD¹ʼngœƖ7ě𝐵、𝐶¦N 𝐵、𝐶 )Ɨč„ƶ_
ƙF7ěv'Ĩ%Tńw75∘615∘qÛ𝐵、𝐶¦N~Æ'AB’u    v¥
A¥200 2
B¥240 2
C¥180 3
D¥200 3
7¥u2024·ŮŰİзŃŎŏDvŘŒ% △𝐴𝐵𝐶'$-z%𝐴,𝐵,𝐶'•(Tń’𝑎,𝑏,𝑐qŐ𝑐= 2,𝐵= 2𝐶qÛ𝑎+ 𝑏
'Ù>?@’ u    v
A¥(2,10)
B¥(2 + 2 2,10)
C¥(2 + 2 2,4 + 2 3) D¥(4 + 2 3,10)
8¥u2024·ĦŖǡŨ·˜Ńvũĉq„F7Ů(&𝐴𝐵𝐶𝐷0q𝐴𝐵= 𝐴𝐷= 2,∠𝐵= 2∠𝐷= 120°qŷ △𝐴𝐵𝐶) △𝐴𝐶𝐷
'78Tń’𝑆1,𝑆2qÛ𝑆2−𝑆1'>’u    v
      
      
A¥2
B¥ 3
C¥1
D¥ 3
2
˜Š
9¥u2024·ŬŭǢž·ŃŎŏDv„ △𝐴𝐵𝐶0q%𝐴𝐵𝐶'•(Tń’𝑎𝑏𝑐qŐ¨e𝑎= 2qÛu    v
A¥ô𝐴= 45∘qŐ △𝐴𝐵𝐶9¦,qÛ𝑏'Ù>?@w(2,2 2)
B¥ô𝐴= 45∘qŐ𝑏= 4qÛ △𝐴𝐵𝐶ǣ9_,.
C¥ô𝑐= 3qŐ △𝐴𝐵𝐶’œ%$%&qÛ𝑏'Ù>?@w( 13,5)
D¥ô𝑐= 3qŐ △𝐴𝐵𝐶’Œ%$%&qÛ𝑏'Ù>?@w( 5, 13)
10¥u2024·ŠĤşň·ŃŎŏDv △𝐴𝐵𝐶0qz% AqBqC Ĭ•'(Tń’ aqbqc¥¨e𝑎= 2q △𝐴𝐵𝐶
'78𝑆=
3
2 𝐴𝐵⋅𝐴𝐶qÛ¿¬îhë'wu    v
A¥𝐴= 30°
B¥ △𝐴𝐵𝐶'<;'=‘>’ 6
C¥ô𝑏𝑐= 4qÛ △𝐴𝐵𝐶’$%&
D¥ô𝐴𝐵(ě'0ď;2–2 3
3 qÛ𝑆=
3
11¥u2024·ŅŤǤǥ·$Ńv¨e △𝐴𝐵𝐶'$-z% AqBqC '•(Tńw aqbqcq78’ 3
4
(𝑎2 + 𝑐2−𝑏2)qÛ¬µîhë'wu    v
A¥cos𝐴cos𝐶'Ù>?@w −1
2 , 1
4
B¥ô𝐷’(𝐴𝐶'0NqŐ𝐵𝐷= 1qÛ △𝐴𝐵𝐶'78'=‘>’ 3
3
C¥ô △𝐴𝐵𝐶wŒ%$%&qÛ
𝑎
𝑐'Ù>?@w 1
2 ,2
D¥ô%𝐵'FTď𝐵𝐸)(𝐴𝐶āJ–N𝐸qŐ𝐵𝐸=
3qÛ𝑎+ 4𝑐'=û>’ 10
$†‡
12¥u2024·Ǧ·$Ńv„ △𝐴𝐵𝐶0q3sin𝐴= 2sin𝐶qcos𝐵=
1
3.Ûsin𝐴=      
.
13¥u2024·ŭǞƯǧĦ·ŃŎŏDvĚǨƁğ–ǟŰſłǠƟqƂĤ–Ťơǩŗqw_LJg&Ɨƥ3ǪƁq
      
      
z9ǫǬ¾ǘǭǮǯƘƭqǰDZDzdzqǴǴǵǶĦqǷǷǸŅƙqǹěŀņǺǻǼƱǽč.ũĉĬƛq’Ƶ
DEĚǨƁ'CqƲƶuǾ 173cmv„N𝐴ƜD¹ƁƘ𝐷'ĩ%’45°qúĂǿN𝐴œƁ'ƼgȀƵ
38m ýĊN𝐵Ɯqè›D¹ƁƘ𝐷'ĩ%’75°q´è¾ƾRĚǨƁ'CӒ      
m.uRiąĶƝëý
0.1v
  
  
14¥u2024·ŋʼnŌō·ŃŎŏDvũĉqF7Ů(&𝐴𝐵𝐶𝐷0q𝐴𝐵= 3,𝐴𝐶= 2𝐵𝐶,𝐴𝐷= 𝐷𝐶,∠𝐴𝐷𝐶= 90∘qÛ
Ů(&𝐴𝐵𝐶𝐷78'=‘>’      
.
Ů,
15¥u2024·ȁņ·ŃŎŏDv„ △𝐴𝐵𝐶0q%𝐴,𝐵,𝐶Ĭ•'(Tń’𝑎,𝑏,𝑐qŐõöcos
𝐴
2
3sin 𝐴
2 −cos 𝐴
2 =
1
2.
(1)#%𝐴—
(2)𝐷’(𝐵𝐶ě_Nq𝐷𝐴⊥𝐵𝐴qŐ𝐵𝐷= 4𝐷𝐶q#cos𝐶.
16¥u2024·ŋʼnŌō·ŃŎŏDv„ △𝐴𝐵𝐶0q%𝐴,𝐵,𝐶'•(Tńw𝑎,𝑏,𝑐,tan𝐶= (𝑎−1)tan𝐵¥
(1)#.
𝑏cos𝐶= 1—
(2)ô𝑎= 2q △𝐴𝐵𝐶78’ 1q#(𝑐';¥
      
      
17¥u2024·ŌŸšȂ·$ŃvũĉqƲƏȃȄ„Ħȅ¬ƙFĆěȆ𝐴œ𝐵ÿȇq„𝐴ƜD¹ĦƘ𝑃Ɯ'ĩ%
∠𝑃𝐴𝑂= 30°qĭȄ¿45km/h'ȈCȉȈÿȇ4TȊĂqýĊ𝐵Ɯqè›D¹ĩ%∠𝑃𝐵𝑂= 45°qŐcos∠𝐴𝑂𝐵= −
3
3 .
(1)#èĦ'𝑂𝑃'>—
(2)#ĭȄ𝐴ý𝐵ÿȇ¸±0ĢD𝑃N'ĩ%ȋ>'=‘>¥
18¥u2024·Ŭŭ·ŃŎŏDv¨e △𝐴𝐵𝐶'z%𝐴,𝐵,𝐶'•(Tń’𝑎,𝑏,𝑐,(𝑐−3𝑏)sin𝐶= (𝑎−𝑏)
(sin𝐴+ sin𝐵)¥
(1)#𝐴—
(2)ô △𝐴𝐵𝐶’Œ%$%&qŐ𝑏= 6q# △𝐴𝐵𝐶'<;𝑙'Ù>?@¥
19¥u2024·Ȍȍŀ‘Ǡ·ŃŎŏDvƲȎRȏƣȐ_ȑŮ(&Ɵǎ ABCD ȒŘȓȔqY0𝐴𝐵= 2ȕȖq𝐵𝐶= 1
ȕȖq𝐴𝐷= 𝐶𝐷q𝐴𝐷⊥𝐶𝐷qȓȔzžOȗȏ 4 ÁƏÿȘ DMDNEMEN ¿X¦ÁșƙȚ ACBDqY
0 MNE Tń’( BCABAC '0N¥
      
      
(1)ô∠𝐴𝐵𝐶=
π
2q#șƙȚ BD ';—
(2)ô∠𝐴𝐵𝐶= 𝛼q𝛼țƛ 4 ÁƏÿȘ'Ļ;C¥