CHÖÔNG XI: NHAÄN DAÏNG TAM GIAÙC
I. TÍNH CAÙC GOÙC CUÛA TAM GIAÙC
Baøi 201: Tính caùc goùc cuûa
A
BCΔ neáu :
()()()()
3
sin B C sin C A cos A B *
2
++ ++ +=
Do
A
BC
+
+=π
Neân:
()
3
*sinAsinBcosC
2
⇔
+−=
+−
⎛⎞
⇔−
⎜⎟
⎝⎠
−
⇔−=
−
⇔− +=
−−
⎛⎞
−=
⇔
−+−
⎜⎟
⎝⎠
−−
⎛⎞
⇔− + =
⎜⎟
⎝⎠
−
⎧=
⎪
⎪
⇔⎨−
⎪=
⎪
⎩
==
⇔
2
2
2
2
2
2
2
=
A
BAB C 3
2 sin cos 2 cos 1
22 2
CAB C1
2cos cos 2cos
22 22
CCAB
4cos 4cos cos 1 0
222
CAB AB
2cos cos 1 cos 0
22 2
CAB AB
2 cos cos sin 0
22 2
CAB
2cos cos
22
AB
sin 0
2
C
2cos cos0 1
2
A
2
⎧π
⎧
⎪=
⎪⎪
⇔
⎨⎨
−
⎪⎪
=
=⎩
⎪
⎩
π
⎧==
⎪
⎪
⇔⎨π
⎪=
⎪
⎩
C
23
BAB
0
2
AB6
2
C3
Baøi 202: Tính caùc goùc cuûa
A
BC
Δ
bieát:
()
5
cos2A 3 cos 2B cos 2C 0 (*)
2
+++=
Ta coù:
() ()()
25
*2cosA123cosBCcosBC20
⇔
−+ + − + =
⎡⎤
⎣⎦
)
() ()
() ()
()
()
⇔− −+=
⎡⎤
⇔
−−+−−
⎣⎦
⎡⎤
⇔− −+ −=
⎣⎦
−=
⎧−=
⎧
⎪⎪
⇔⇔
⎨⎨
=
=−
⎪⎪
⎩
⎩
⎧=
⎪
⇔⎨==
⎪
⎩
2
22
22
0
0
4cos A 4 3cosA.cos B C 3 0
2cosA 3cos B C 3 3cos B C 0
2cosA 3cos B C 3sin B C 0
sin B C 0 BC 0
3
3cos A
cos A cos B C 2
2
A30
BC75
=
Baøi 203: Chöùng minh
A
BCΔ coù neáu :
0
C 120=
A
BC
sin A sin B sin C 2sin sin 2sin (*)
22 2
++− ⋅ =
Ta coù
A
BABCC ABC
(*) 2sin cos 2sin cos 2sin sin 2sin
22 2222
CAB CC AB A
2cos cos 2sin cos 2cos 2sin sin
22 22 2 2
CAB C AB
cos cos sin cos cos
22 2 22
CAB AB AB
cos cos cos cos cos
22 2 22
CAB AB
2cos cos cos cos cos
222 22
+−
⇔+=
−+
⇔+=+
−
⎛⎞
⇔+=⋅
⎜⎟
⎝⎠
−+
⎡⎤
⇔+=
⎢⎥
⎣⎦
⇔=
2
B
2
+
C1
cos 22
⇔=
(do
A
cos 0
2> vaø B
cos 0
2> vì
A
B
0;
22 2
π
<
<)
⇔= 0
C120
Baøi 204: Tính caùc goùc cuûa C
Δ
ΑΒ 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
BC<<
Ta coù: A, B, C taïo 1 caáp soá coäng neân A + C = 2B
Maø
A
BC++=π
neân B3
π
=
Luùc ñoù: 33
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
222
BAC3
2cos cos
222
3AC3
2. cos
222
CA 3
cos cos
22 6
π+
⇔++=
⇔+=
+−
⇔=
−
⇔=
⎛⎞ −
⇔=
⎜⎟
⎜⎟
⎝⎠
−π
⇔==
Do C > A neân coù:
CΔΑΒ
−π π
⎧⎧
==
⎪⎪
⎪⎪
ππ
⎪⎪
+= ⇔ =
⎨⎨
⎪⎪
ππ
⎪⎪
==
⎪⎪
⎩
⎩
CA C
26 2
2
CA A
36
BB
33
Baøi 205: Tính caùc goùc cuûa
A
BCΔneáu
)
()
⎧+≤
⎪
⎨++=+
⎪
⎩
22 2
bca 1
sin A sin B sin C 1 2 2
AÙp duïng ñònh lyù haøm cosin:
22
bca
cos A 2bc
+−
=
2
2
Do (1): neân co
22
bca+≤ s A 0
≤
Do ñoù:
A
A
24
ππ
≤<π⇔≤ <
22
π
Vaäy
()
A2
cos cos
242
π
≤
=∗
Maët khaùc:
sin A sin B sin C++ BC BC
sin A 2sin cos
22
+
−
=+
A
BC
sin A 2 cos cos
22
−
=+
2
12 1
2
⎛⎞
≤
+⋅
⎜⎟
⎜⎟
⎝⎠
()
−
⎛⎞
≤
⎜⎟
⎝⎠
BC
do * vaø cos 1
2
Maø sin A sin B sin C 1 2 do (2)++=+
Daáu “=” taïi (2) xaûy ra
⎧=
⎪
⎪
⎪
⇔=
⎨
⎪−
⎪
=
⎪
⎩
sin A 1
A
2
cos 22
BC
cos 1
2
π
⎧=
⎪
⎪
⇔⎨
π
⎪
=
=
⎪
⎩
A2
BC4
Baøi 206: (Ñeà thi tuyeån sinh Ñaïi hoïc khoái A, naêm 2004)
Cho
A
BCΔ khoâng tuø thoûa ñieàu kieän
(
)
cos2A 2 2cosB 2 2cosC 3 *++=
Tính ba goùc cuûa
A
BCΔ
* Caùch 1: Ñaët M = cos2A 2 2cosB 2 2cosC 3
+
+−
Ta coù: M = 2BC BC
2cos A 4 2 cos cos 4
22
+
−
+−
⇔ M = 2
A
BC
2cos A 4 2sin cos 4
22
−
+−
Do
A
sin 0
2> vaø B - C
cos 1
2
≤
Neân 2
A
M2cosA42sin 4
2
≤
+−
Maët khaùc:
A
BCΔkhoâng tuø neân 0A 2
π
<
≤
⇒≤ ≤
⇒≤
2
0cosA1
cos A cos A
Do ñoù:
A
M2cosA42sin 4
2
≤+ −
2
2
2
A
A
M12sin 42sin
22
AA
M4sin 42sin 2
22
A
M22sin 1 0
2
⎛⎞
⇔≤− + −
⎜⎟
⎝⎠
⇔≤− + −
⎛⎞
⇔≤− − ≤
⎜⎟
⎝⎠
4
Do giaû thieát (*) ta coù M=0
Vaäy:
2
0
0
cos A cos A
A90
BC
cos 1
2BC45
A1
sin 22
⎧
⎪=
⎪⎧=
−
⎪⎪
=⇔
⎨⎨
==
⎪
⎩
⎪
⎪=
⎪
⎩
* Caùch 2:
()
* cos2A 22cosB 22cosC 3 0⇔+ + −=
()
()
()
()
2
2
2
2
2
2
2
BC BC
cos A 2 2 cos cos 2 0
22
ABC
cos A cos A cos A 2 2 sin cos 2 0
22
AABC
cos A cos A 1 1 2sin 2 2 sin cos 2 0
222
ABC BC
cos A cos A 1 2 sin cos 1 cos 0
22 2
ABC B
cos A cos A 1 2 sin cos sin
22
+−
⇔+ −=
−
⇔−++ −=
−
⎛⎞
⇔−+−+ −
⎜⎟
⎝⎠
−−
⎛⎞⎛
⇔−−−−−
⎜⎟⎜
⎝⎠⎝
−−
⎛⎞
⇔−−−−
⎜⎟
⎝⎠
=
⎞
=
⎟
⎠
C0(*)
2=
Do
A
BCΔ khoâng tuø neân vaø co
cos A 0≥s A 1 0
−
<
Vaäy veá traùi cuûa (*) luoân
≤
0
Daáu “=” xaûy ra
cos A 0
A
BC
2sin cos
22
BC
sin 0
2
⎧
⎪=
⎪
−
⎪
⇔=
⎨
⎪−
⎪=
⎪
⎩
⎧=
⎪
⇔⎨==
⎪
⎩
0
0
A90
BC45
Baøi 207: Chöùng minh
A
BCΔcoù ít nhaát 1 goùc 600 khi vaø chæ khi
sin A sin B sin C 3(*)
cos A cos B cosC
+
+=
+
+
Ta coù:
()
)
)
(*) sin A 3 cos A sin B 3 cosB sin C 3 cosC 0⇔− +− +− =
sin A sin B sin C 0
333
AB AB
2sin cos sin C 0
23 2 3
πππ
⎛⎞⎛⎞⎛⎞
⇔−+−+−=
⎜⎟⎜⎟⎜⎟
⎝⎠⎝⎠⎝⎠
+π − π
⎛⎞ ⎛
⇔−+−
⎜⎟ ⎜
⎝⎠ ⎝
⎞
=
⎟
⎠
CABCC
2sin cos 2sin cos 0
22 3 2 26 26
CABC
2sin cos cos 0
26 2 26
⎡π π⎤ − π π
⎛⎞ ⎛⎞⎛⎞
⇔−− +− −
⎜⎟ ⎜⎟⎜⎟
⎢⎥
⎝⎠ ⎝⎠⎝⎠
⎣⎦
π⎡ − π⎤
⎛⎞ ⎛⎞
⇔−− +−=
⎜⎟ ⎜⎟
⎢⎥
⎝⎠ ⎝⎠
⎣⎦
=
π− ππ
⎛⎞ ⎛⎞⎛
⇔−=∨ =−=−
⎜⎟ ⎜⎟⎜
⎝⎠ ⎝⎠⎝
CABC
sin 0 cos cos cos
26 2 26 3 2
+
⎞
⎟
⎠
AB
π−π+−+π+
⇔=∨ =− ∨ =−
CABABABA
26 2 3 2 2 3 2
B
ππ
⇔=∨=∨=CAB
33
π
3