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Senior High

Semua

數學

三角比性質常見的經典考題

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華爾街之狼

2021.08.16 公開

學測 111學測指考分科測驗數學三角函數

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ページ1:

3
三角比性質
It cos Acosxsin, cosBe sin Asin C để sinh Arsin Bt sinh
* sin² a + cos³d=1
7
=) sin + A+ sin+B+ sin²C = (1-cos³A) + Cl-Cos<B) + sin³ C
⇒ 2 - [cos³ X sin ²C + sin²xsin³c) + sin'c
C
⇒2 - sin² ( [cos' x + sin x] tsin >( => 2-sik?<+ ship?c
(2) EM A sinzo + sin³ 0=1, #It sin? Q+ sin 50+ sin 6 Q = ?
*sin² = 1-cost'>
A=2
= 2
Cos?Q= sin 3 Q = cos 4Q= sin ba
⇒ | + |- cos³d + (1-cos³d) sin³ @ + custa
R
= 2 - Cos² + cos <Q- Cos 4 α + Cus 40=2
A: 2
(3)設fm:
=CosnQ + Sin"Q,neN)求3f(4)-2f(6)+f(2)之值為何?
fill = cos a+ sind. f(2) = cos =Q + sin 3Q = |
f(4) = cos40+ sin4Q = (cost @ + sin²)² 2 custsin³d
=1-2003² sin²
f(1) = cosb@+ sin 6 Q = (cos²d + sin 2013 - 3 Cos=@sin+Q - 3c54d sin³
=
= |- 3 Cos² sinta - 3 C0340 sin² Q
(cos² + sin + 0) = = |` = |
7 Cus 4Q + sin 40 + 2 sin?d cos?R=|⇒ sin a cos Q=0
73f(4) - 2 f (6) + f(z) = 3-6 cos=@sind -> + 6 cos² (sin 4d + 6cos 40 sin?
= 2-
• 6 cost sind [1 + sin² Q + cos²0] => 2-6x0×2=2
A: 2
+1

ページ2:

(4) ƒ x² (tand + =) X+2=0,TA-AR 3-252, te sind cos (=?
tand
X = 3-2 X-6x+1=0
» qß-6x+1= x²- (tand+ tang 1x+2=> [tand+ Fond
+6)X=1
X X=3-252 => tanQ+
=9+2√2
* sin Q=Q, cosQ=b. sin & cus&= ab. fand+:
=
+1 =9+252
7
92+62
+ a
ab
=9+2527
alb =9+252 = ab=
9+2√2
=> ab= sind cosα =
9-252
9+2√2
73
A
9-25
73
((5) 設日為第二象限角,若兩方程式+3(csQ)X+1=0與x=(simo)x-6=0
恰有一相同解,則tan Q= ?
sind 76
***= tx
2²+3 cos &α +1=0 = cosα =
tando
21
d = sindα-6=0 = sind =
d² 66
tand = sind
32
=
-323718
CuSQ
變
.x sin a+ costa-11+> A
tan Q====
A: -==
| (6) 272 90° ≤ 0 ≤ 180°, 1974 cos-α +2√6 sind -5=0, EQ=?
JE
(註:sn15°= 18+, sin 18° = √51/1) 90° ≤ @ ≤180°
4
* Cos³ + sin = cost Q = \-sin²
41 1-sin) +
-4 sin+sina -1=0
56452
寸
5 SixQ=
SinQ = 56-52
⇒ sind = -258±√F=
√(±52
8
4
{d=180°-15=105°
0-180-15=165
A=105°165°

ページ3:

(7)試求
(8;
135
135
Σ sin² (zk°) ¿ 11 =?
K=1
Σsin (zk°) = sin 32°+ sin 24° + sin ² 6° +.. + sin' 268° + sin² 270°
|=|
= (sin '20+ sin 2288°) + (sin² 4°+ sin 266°) +.. + sin 2270°
=) (sin ³ 20+ sin ³ (90x3-2)°) + (sin³ 4° + sin² (90×3-41°) + ·· + sin³ 270°
=) (sin 22°+ cos²2°) + (sin 24° + cost 4°) + ... + (-1)²
71x
134
1 1/2 +1 = 67+1 = 68
A=88
1/ 100°< 0 < 225°, 14, 16 19 √1 + zsin@cusa + VT-2sin@cos& = ?
√ sinzQ+ cus² & +asina cosa +
Sih² + cos - 2sin & CuSQ
-=) | sin at cos @| + | sind-cosa |
⇒ -sina-cosa estina- cosa
⇒-2005
*180°CQ< 225°
= cos &co, sih aco
> cos & <sin Q
A=-2005
(9)將半徑1之半圓弧分成180等分,設等分點為只、屋、屋...
Pinq,$ { APK ZTE = ?
7
-2
Σ APK = AP₁ + AP₂² + AP3 + .. + Aping
⇒ BPing ⇒
AP₁ = Bir AP₁ + APin = AB = 2² = 4
7AP₁² + B₁₁ = AB² = 4
=
B
AP90 + BP₁ = 4 = 2 AP₁ = AP₁ = √2
1179-1
(AP₁² + AP₁m) + (AP² + AP³ ³) + ... + AT 9 = 4 × 12-14 + (2)² = 358
1=358
A
FA

ページ4:

(10) a = sin 1198° b = cos(-460°). C= tan>09% d= tan 430°,
試比較a、b、c、d的大小?
A = sin ( 360° x 3 + 1180°)
b=cos (780°-480") #d>>c>b
= sin 118°
= Cos (+100°)=C05100
140
=5in62070
=> Sin 62° sin 60°
7562=97
= Cos(100°-90°)
=-sin 10° <o
C= tan (90°x2 + 290) = tan 29°70
=> C < tan 30° =) << ==
(1)
* AB = ? +
d=tan (360+70°)
=
tan 70°>0
= sih '70'sin 60°
>
cosque cosgoo A÷d>a>c>b
A (cosd, sind). B (>cosß-2 sip), Bd=126°
Aǝrt
B=1=2
70
2
A(cosol, sind)"
A
1722-2×2×1× (-)
=5+2=1=AB
→ AB = √7
BCZcosp.-2sinß)
A:√7
(2)如图所示,矩形ABCD中,=3,死=2,若將此矩形放在距離為了
的兩平行線山、L2之間,且使A.C分別落在2.山上,
X
tan Q = ?
(26)²+ (a)²=9
a+b²=4
f=⇒28=36
* 0+x=3 = 2y=34
Lz
· x + y = 9
(畢氏定理)
13
07624
= tanh
=
a
=
a =
A
42

ページ5:

(3) △ABC中,三邊長9.b.c對應的高為ha-kb.hc,如图所示。
abc_
2 tanA=1. tanB = 2. tan C=3, *
hahshc
之值為何?
B
*tan A==
C
a
⇒ sinA=亢
tan B=2
= sin B =
A
1
tanc = 3 ⇒
b
3
=) sin C = √
X SinA = k sin B = ha sin c = he
a
> abc
:
Tahihc
sinA sinB sinc
1 2 3
归
A:f
A:
(14) ABC BC. CA. ±ha-6. hb = 4. hc = 3.
求COSA及△ABC的面積為何?
a=b=C= | | a
邊長比》4K,3k,水
=
= = = 2=3=4
4K
⇒ COSA =
2x44x3k
= 34 =
C
B
>K
16k²+9k34k²
△ABC = = xaxa = zxzkx6=6k = 海龍公式
8
6k = k*k* kxk ⇒ k² (1 ≤ > k=
√ * * -
=
(S=
24+4+4K
= 2;
=
A: T

ページ6:

(15) B-C
ĀDA 1 ¿+ Q L = Rk, AD=4, AB=BC=1;
求印與COSLADC為何?
* AC=Y. CD = x
C
180-
y
Q
2
0
2
1³²+1² - 2005 (180-0) = y²
x+4-4xcusα=4
+8=1600530
⇒ 2+2000=42
x²+ y²= (2+2) = 16 = 16c0540 + 2003 (+2= 16 = 16cos³0+2cus f-14=0
=> ({cos(-1) (cos(+1)=0 = cos(=}}, \| (1997)
CD = ADxcosQ = 4x. = = =
√ CD =
A: [
COSLADC=
}
(16)圓內接四邊形ABCD中,B=3. =2.6=3.LABC=120,求雨長=?
A!
B
AC = 9+4 + 2 x x x 3 x (+) = 19 = AC = √19
** 10 = y = y + 9 = £y x = = 19 7 y² 38 -10 = 0;
-
→) (y-5) (4+2)=0=y=5
*圓內接四邊形公式 > 3×3+2×5=5斤x師
=> 9+10=√19 × BD => BD = √19
A-519
(17)如右图,在△ABC中,AB=8,死=7,CA=6,∠BAE=LCAE>
E點在△ABC的外接圓上,求線段之長二三
B
AB = AC = BD = PC = 4=3=71471/7
*ADAB XAC-BOX DC
· → AD² = 8x6 - 4× 3 = 36 = AD = 6
:
⇒ 6xy = 4x3 y=2» AE = 6+2 = 8
=> y=2=⇒
A-8

ページ7:

(18) & ABC, a=3. b=4. tanA=4, * C=?
* SSA = ÁTF4 = c TA FFA Á‡ :
B
5
3
=) Cus A = = =) (=5
A
C
A
C
3
3 = 4²+C² - 2x 4xcx Cu SA
=) 9 = 16+C² 8xCx*
= 50-320 +35=0
ac=7.5
A = 7.5
(19) # SABC#, AB = 10, AC=9, COL BAC = XP-Q5981tz
,
邊丽、死上使得△APQ之面積為△ABC面積之一半,
求權之最小可能值為何?
P
10-x
B
x
A
y
19-y
*AP=x, AQ=Y, COSLBAC=
\² = x²+ y² – zx yxz² = x=y² &xy
△APL
△ABC
IXysinA
zx9xbxsinA ⇒ y=45
**£= x+ y² > > √xy² = x + y² = 90 =2xy
2
⇒版:y=x45 ==些
(20) 8 ABC $, AB=5, AC=2, BC=x (3<x<7), LABC=0,
則cosQ之最小值為何三
A
5
B.
Q
C
25+X-10XC05Q=4
= x² 10% cosQ + f = 0 = P >0 (¥) ¾)>0)
⇒ 100 cost - 84 >0 ⇒ cos2d 1
{cos acoso
恤
A = √1
5

ページ8:

(21) 412 13 559 55-7 AB = 30, LCAD = LCBD = 45°, ADX BC 730₤ LAOB= 75°,
*共圓
0
A
30
*/ CD=X
450
B
在△CDA中:
在△ABD中:
x
30
=2R
25°
=2R
Sin45°
57730°
B
30
x
A
30
=
sin30°
= CD=30√2
A=30√2
5i445°
(22)如图所示,在河岸一邊的兩點A.B.測得對岸兩點PQ,
得LPAQ =60°,∠Q AB=45", <PBA=30", ∠QBP=750,且AB=2公里
則PQ兩點間的距離為多少公里?
Q
河流
②
sin4jo
BP
Sin1050
BP = √3+1
BQ
54450
=
2
Sin300
BP
4
33Q=252
B
=
A
1450
B
→ P₁ = (√3+1) + (2/2)² = 2x (25)([3+1) Cos15°
⇒ 瓜²= 12+23-4⇒ā-18+255
A:S8+鸡公里

ページ9:

(23)從平面的A、B、C三點,測量山頂的仰角均為30;
且LABC=450,AC=300公尺,求山高為多少公尺?
A
山頂→
300
=2R = R = 150√2
sin45p
1300m
R=15052
| 150√2×53=
A:506公尺
(24)一船向東370南航行,速度60浬/時,於上午9時,
測得一島的方向為東 530北,至中午12時,再測得
該島的方向在船的北23西,則中午時船與島
的距離為何?
9時
30°
西
1530
60x3=180
30°
180
d = 180×2 = 120√3
√3
A-1205331
(25)從地面一直線上三點A、B、C,測得一山頂的仰角分別
為30°、45、60°,若山頂的垂足與A-B.C不天線,
且曆=300公尺,死~200公尺,求山高為多少公尺
1h =
A
130° h
300
1450
200
B
E
E
4.4.4
130° D
⇒ Stewart 公式:
(h)200+ (+)’300 =k㎡(500).
⇒ ¥h=5h²+ 300000
⇒ h²=15000
+ 300 000x500
>h=100515 A=100515/är

ページ10:

(26)從地面上共線A、B、C三點,測得一山頂的仰角分別為
30.45.60°,老山頂的垂足與A、B、C不共線,且=2008公尺,
既=200g公尺,求山高為多少公尺
D
300
53h
D
2008
130°
h
B
心
D
B145°
E
2008
55h
A
LU
⇒中線定理
=> (5³h) + ( 1 )² = 2 [h²+ > auf]> h=100456 A=100456/GR
(27)
B
A
ABC, AB = 3. BC=4, LA= 3LC, & sin² (=?
3
ㄌ
A
D
A
= cos Q =
E
⇒ BD=3, PC = 1. AD = 1
CD
Stewart 公式求:派
3³x|+ AC³× 3 = 1*+ (3+1+³×1×4
#AC
CE
=
⇒ Cos²Q+sm²Q=1
A = 12
| > | - cos 40 = sin ² Q = 1 - 1/72 = 1/2 = sin³ (

ページ11:

(28) △ABC中,∠A的内角平分線交於D,AC=4,AD=3.DC=2,
* AB = ?
ABX BD=4
x
Q
4
*內分比⇒ x=4=y=232y=x
B
y
2
C+Stuart定理解未知報:
⇒ xx z + 4xy = 3² × (y+²) + yxxx (y+z) +&x=zyst>
⇒ 2x (ay) + 3y - 2y ² 48 = 6y²+39-18=0 ⇒ (y+2) ( zy− 3) =ð
y = = AB = × 2 = 3
A= 3
(29) 如图所示,△ABC中,>花,過A的高=12,LA的內角平
分線段AD=13,過A的中線丽=15,則既 = ?
M=X MD=4. DH=5
B
ㄈㄣ
7-9
C
M D
H
AB = √12 (79)
FC=x-9
AB
*內分比:麗=
AC = √12²+17-912, BD=x+4. FC=x-4
:
= √12²+(749)2
X+4
⇒合分比:号
c-d
=
√12717912
⇒
古
⇒ 12²+(x+9)² - (1+4)2
12717-912
=
=> 91x-4)²= 4[12²+ (x-9)2]
⇒ =
=) BC= 2x =
=
2x18x
2x8X
⇒
12717912
[π =
1756
5
125105
5
A=
125105
5

ページ12:

(30)在△ABC中,CA.LB.LC的對邊長分別為9.13.6,若
tan 1/2 x tan ₂ = 1, 1 C=?
x
*內切圓:
z
√x+y=c
3
X+2=13
7
x+y+8=>
= 22+C
Y+z=q #S= Zz+C ( )
tan 12 = 1, tan — 1 = 1/1 7
48
=
=
FDA (rxs) = s(s-a) (s-b) (s-c) = r²(x+y+ z) = xyz
海龍
内切圆
7
竖(x+y+z)=8
⇒x=1+2
A-14
* X=YTZ = 2x+Y+7= ( +13 ⇒ >X+ 9 = C+13 ⇒ 2x=c+4
x+7+7=22+6=9= 1 + 2 = 16 7 18 - C+14= C-14
(31) 說&ABC三邊長分別為7.8.9,若△ABC内切圓切三星於
D.E.F,求△DEF面積?
B
S=
*ODEF =OIEF +OIED + DIFD
E b
180fy
a
a+b+c= 1+1+9 = 12
z
ABC= rs = 12r
JABC= abc 7x8x9
41 =1255
4R
⇒R
ODEF=sin(-A)+ y² sin (180-13)+
154 (180-c)
=
±³ (sin A+ sin B + sin c)
= r² +
7
2R
氣
ra+b+c)
X
=(rs)
2
1 (rs) = 1/2 <ABC= DEF
>R
*ABC= √12x 3x4x5=1255
=
√5
汞
=

ページ13:

(3)若AABC内接於半徑為1的圓且△ABC的面積為1,
則 sinAsinB, sinC 為三邊長的三角形面積為何?
正弦定理:
a
b
C
=
=
sin B
SinC
=2R 7 0 ABCN & Sin AsinB sin ()
SinA
=2R = 1x2
a
b
ShA
sin B
~
⇒面積比: ㄧㄤ=
=
C
(ShAJ
621=4
sinc
SinAsin B sinc
<ABC
4
C
D
3
(33) #2 } FF ÁF, SABCIF, LBAC=90%; Œ ÑD = DE=BE = 2 = 3=5, 1
<BAE= 2. LDAE = B. LDAC=ɣ, ste
sinß
sind-sinɣ
= 2
* AABC= SACD + SADE + CABE
a
A
E
9
b = x + xxy + xybsing
y
12
B
b
2
(5+3+2) × 1 × ab = ±≤ xax sind⇒ sind = 3
3
(5+3+2) x = {xab =
==× xysinß = sinß=
Toxy
5+1+2) x = xab = = = = x y b sind = sinɣ = a
(573+2)
sin ß
398
Taxy
司
=
= 3
Sindsinɣ
A = 3
ky
(34) 如图所示,△ABC中,兩平分LBAC,M為死的中点,即上丽,
B
AB=14. AC = 20, * MW = ?
14
W+
14
M
*
BN = ND =|=|
=) CD = NM = 2 =|=6=3
A-3

ページ14:

(35)如图所示,△ABC中,∠C=90°,且=2DE=BE>已知ACD=d
sinß
LDCE=B. LECB=ɣ, ife sind sing
= ?
A
2 D
3
b
*ABC APC+ CDE + ACEB
⇒正信定理,面積比求解
SABC= ab
B & ADC = = axsind
ACDE=xysm B
> CEB = 4b sinɣ
0+x+ab = ±axsind sind
Bab
③.
-X
-x=-=-=ab = = = yb sinɣ
11x+ab === xy sin ß= sinß = Tray sinɣ=
sinß
=>
sind-sinf
=
3ab
TAXY
69
=
4
xxx xiy
69
A₁ 44
(36) * A FI ABCD 4, AD = 5, BC = 10, EL AD // BC, LADC=120°,
若梯形ABCD 有一個內切圓(與四個邊都相切的圓)
→則梯形ABCD的面積為何?
5
*DC=X
D
AB=5-X
15-2753x
ABCD = (5+10) × √xx
2X
= 55 x
B:
?-X A' 10-2X D'x'
* ${/}} #s⇒ ÃD + EBC = AB +CD = 5+10= >X+AB
AB=15-2X
= (√39) + (5-7)² = (15-2x)² ⇒ x=4
□ ABCD = 1553 × = 15√3x4 = 30√3
Z
A=3053

ページ15:

(37)已知銳角三角形ABC外接圓的半徑為8,且圓心到限
的距離為3,到的距離為6,則
* AL = X DE =y
二!
B
D
3
SS
15
D
BE = √64-36=257
AD = √64-9 = √55
*餘弦定理求yor圓冪求解
⇒ B.O.DE共圓⇒圓冪(托勒密定理)
6 x√53 + 3x2√7 = 8y = y = 3√7 + 3√55
180-
B
E
⇒ y=> x=
2
4
3√7+3555
A=
2
(38) 設銳角三角形ABC的外接圓半徑為8,已知外接圓圓心
到存的距離為7,到死的距離為2,則△ABC的面積
為何
B
215
28
AF = [64-49 = JTB
BD = 264-4 = 2555
1455 *B. F. O.D# 117 BF XOD + OFX BD = JB XDF
⇒ Jī5x2+7x2575 = 8XPF ⇒DF = 25TF
D
2515
C
AC
4575
=) AC = 4√15
Z√TS
*SABC= FAX BC x sin B = = x²JTS x 4JTSX sin B = 60 sin B
✓0B=2R
DF
*
=2R =)
=875
SinB
sin B
⇒ SinB =
⇒△ABC=60xsinB = 60x4+ = 15515
匹
A=15515

ページ16:

(7)如图所示,為正方形ABCD內部一點,且丽=2.803.死=4,則
求防反正方形ABCD的面積為何!
A
D
D
AP² + PC²= PD² +PB²
⇒ 4+16 = PO²+ 9
B
⇒ = T
D
冷AB=X ⇒ DABCD = x2
x
B
P
190-0
△APB中:
¥5
6X
0 PBC4:
C
x
⇒ sih20+cos2d=1
x79-6xcosQ=4⇒ cosQ=
x+9-cu (90-0) = 16 = cos(90-0)=
Cos(90-0)= 54
DABCD
10-35722
⇒ (17) + (x²-1) = 1 > x²-20x²+37=0>> X² 10±357 1 ½
SFD=√TT
A:
[面積:10+3斤
(40)在△ABC中,a-b.c是三個內角A、B、C的對邊長,若
△ABC的面積 S 滿足S=a²(b-c)²,且b+c = 34,
求△ABC面積的最大值為何?
*算几不等式:
JABC = bc sin A
S=α²= b²-c²+2bc
|cos A =
6+0=92
26c
⇒4 sinA=2bkcosA+ 264
⇒ ==sinA=2cosA+2
4cosA-sinA+4=0
' => > bc cosA = b² + c = a² πt'xs
⇒2bccosA+26C
sin³A+ Cos³A=1
⇒ sin A=0.
⇒ △ABC=XX员
sin² A+ cos² A = 1
:68
A:68