第二十五章锐角的三角比(讲义)(解析版)_2026语数英7-9年级上《举一反三系列》多版本_2026《初中数学•举一反三系列》9上多版本_2026《初中数学•举一反三系列》9上(沪教)

                                                                           
                                                                       
第二十五章  锐角的三角比（举一反三讲义）全章题型归纳

 ...................................................................................................................................................................4
 1   !" ....................................................................................................................4
 2  #$% !" ........................................................................................................................8
 3  &'()*+,-./ 0 ......................................................................................................12
 4  .1/ 0 ..................................................................................................................................16
 5  2340%./5 0 ..................................................................................................................22
67 .................................................................................................................................................................29
 6  89/ 0 !" ......................................................................................................29
 7  :;< !" ..............................................................................................................33
 8  =>+./ 0 ..................................................................................................................40
 9  89/ 0?@ABCDEF ..............................................................................................47
 10 ./ 0DG) ..........................................................................................................................53
知识点1  锐角的三角比
1. HIJKIJHLJKLDMN
OPQRS2Rt  %S	C = 90°STUV A DW4XY4D!Z[	ADHIS\]sin S^
sin  =

_
V A D`4XY4D!Z[	ADKIS\]cos S^cos  =

_
V A DW4X`4D!Z[	ADHLS\]tan S^tan  =

_
V A D`4XW4D!Z[	ADKLS\]tan S^cot  =

_
2.  A DHLJKLJHIJKIab A D !_
3. cd/ 0DY4edfgh,/4SQij0 < sin  < 1k0 < cos  < 1Stan  > 0_
知识点2  特殊角的三角比的值
                                                                           
                                                                       
1. lmD !DMNn/ 0Dopqrst
30°
45°
60°
sin 
1
2
2
2
3
2
cos 
3
2
2
2
1
2
tan 
3
3
1
3
cot 
3
1
3
3
知识点3  锐角的三角比间的关系
2Rt  %S	C = 90°S	，	S	DW4uvw aSbSc_cxyMzqr!" + $" = %"_
{1|} !~D
sin" + cos" = ( !
% )
" + ( $
% )
"
= !"($"
%"
= 1_
{2|tan Xsin Scos ~D
sin 
cos  =
!
%
$
%
= !
$ = tan 
知识点4  解直角三角形的概念
h€S/ 0%S‚/ƒS„j…†‡ˆS^ ,4n‰†_c/ 0%D*+‡ˆS
Š‹K>+‡ˆDŒSZ[./ 0_
知识点5  解直角三角形的依据
2Rt  %S	 = 90°S	，	S	DW4uvw aSbSc_
{1| 4Ž~D
) + ) = ){xyMz|
{2|‰Ž~D
	* + 	+ = ,-{‰K|
{3|4Ž~D
sin  =

Scos  =

Stan  =

_
知识点6  解直角三角形的基本类型及解法
./ 0j3‘’“”
•*+Y4nh/4–*+‰/4—*+Y4nh˜*+
h/4nh_‹.™š›tOs
.
 !
                                                                           
                                                                       
*+”
*+,-
.™š
PR
Y4 cJh/
4{O a|
{1|$ =
%"/!"
{2|csin  = !
%S	
{3|	 = 
/	
‰4
‰/4
{aSb|
{1|% =
!" + $"
{2|ctan  = !
$S	
{3|	 = 
/	
Y4 cJh
{O	|
{1|	 = 
/	
{2|csin  = !
%Sr! = % 0 sin 
{3|ccos  = $
%Sr$ = % 0 cos 
h4
h
h/4J
h{O
aJ	|
{1|	 = 
/	
{2|ctan  = !
$Sr$ =
!
tan 
{3|csin  = !
%Sr% =
!
sin 
Rt  %S	C = 90°S	，
	S	DW4uvw aSbS
cSOPQR
知识点7  解直角三角形在实际问题中的应用
1. œ)./ 0.žŸ D™š
{1|¡žŸ ¢£w S^¤Š¥¦P0S;§w./ 0D 
{2|lm %D,-S¨©ª) !:./ 0
{3|r« D¬­
{4|r«žŸ D¬­_
2. ®¯”
{1|°J±
©²A2³¥A´µ¶S²AX³¥AQ·DZ[°©²A2³¥Asµ¶S²AX³¥AQ·D
Z[±_
OP{1|QRSOC w³¥ASOD w¸¹ASOASOB w²ASTUV	1ºw°S	1ºw±_
P
1
P{2|
                                                                           
                                                                       
{2|µ»
H¼µ½CH¾µ½X¿Àµ½Q0·DÁd 90°DZ[µ»_
OP{2|QRSOA QtRDµ»b¼Âà 55°SOB QtRDµ»b¾Âà 45°SOC QtRDµ»
b¾ÂÄ 70°SOD QtRDµ»b¼ÂÄ 30°_
{3|ÅÆJÅ
ŦD¸/ÇÆn³¥ÈÆD!Z[ŦDÅÆ{CÅ!|S\] iS^2 =
3
4SÅÆɮʷ5 6 7D0Ë_
ŦX³¥¦DÌZÅ{CÍY|S\]8Sdbj2 = 9:; <_
【培优篇】
【题型1  设参数法求锐角三角比值】
Î 1{24-25 ÏÐÑ´·ÒÓÔ³·ÕÖ|OPS2×0=%S=> ? Ø AB dÙ ESÚ5=SÛ
> =
1
3SÜcos	=>D"b          
_
¬­
6
6
uÝ“qÞlm×0DopŠ4ßSàáâ*+,-rŠABßÆSãäœ) !DMN.
	=>D"_“åæçèé×0DopJxyMziê !DMNSëìíîïð+ñb.D
ò_
ó..
 = @S
A 340=b×0S
B = =  = @S
                                                                           
                                                                       
A > =
1
3S
B > = @S
A => ? S
B 2Rt  =>%S> = /> = @/@ = @S
cxyMzqr=> =
="/>" =
&@)"/& @)" =
5@S
2Rt  =>%S= =
=>" + >" =
( 5@)" + @" =
6@S
B 	=> =
>
= =
@
6@ =
6
6 S
ô¬­w
6
6 _
õË 1-1{2025·ö¾¥÷ø· ù|OPS2ú0=%S = 3S= = 5SÙ>2=´S¡ú0=
û>üýSÙ=þÿĀ24´DÙCāSĂătan	>C =      
¬­
4
3
uÝ“çèéú0DopSüýDopS !:Scú0nüýDopqr = = = 5S
= =  = 3S	 = 	 = 90°SC = = = 5S>C = =>S^rC =
C"/" = 4Sr«
C = /C = 1S> = @S2Rt  >C%œ)xyMzqr> =
4
3S?ĄlmHLDMN^q.Sí
îi´+ñÙb.Dò_
ó..
A340=wú0S
B = = = 5S= =  = 3S	 = 	 = 90°S
cüýqrSC = = = 5S>C = =>S
B2Rt  C%SC =
C"/" =
5"/3" = 4S
BC = /C = / = 1S
> = @SÜ>C = => = =/> = /@S
A2Rt  >C%S>" +C" = >C"S
                                                                           
                                                                       
B@" + 1" = &/@'"S
.r@ =
4
3S
B> =
4
3S
Btan	>C =
>
C =
4
3S
ô¬­w
4
3_
õË 1-2OPS2:4  %S= ? S¹ąw=Si=S=w`4]ú0=>SÚ5>Ø=4
dÙCSÜcos	>D"w{    |
A_
5
14 7
B_
2
7 7
C_
1
14 21
D_
1
7 21
¬­A
uÝ:4  D4ßw aSÜ =  =  = !_lm:4 0Dopqr= = = =
1
2!SĆ
ĄqcxyMzŠ= =
3
2 !_lmú0DopćqrŠ> = = =
3
2 !S= = > =
1
2!S	> = 90°S
^ćqœ)xyMzŠ> =
7
2 !_ŒÙ C ]E ? >dÙ GScFGHI =
1
2> 0 E =
1
2> 0 >SqrŠ
E =
21
14 !S?ĄcxyMzqŠE =
5 7
14 !SãäcKIDMN^q._
ó..
:4  D4ßw aSÜ =  =  = !_
A= ? S
B= = = =
1
2 =
1
2!S	= = 	= = 90°S
B= =
"/=" =
3
2 !_
A340=>bú0S
B> = = =
3
2 !S= = > =
1
2!S	> = 90°S
                                                                           
                                                                       
B> =
" + >" =
7
2 !_
OPSŒÙ C ]E ? >dÙ GS
AFGHI =
1
2> 0 E =
1
2> 0 >S
B
7
2 ! × E =
1
2! 0
3
2 !S
BE =
21
14 !S
BE =
"/E" =
5 7
14 !S
Bcos	> =
E
 =
5 7
14 !
!
=
5 7
14 _
ôª A_
ÙĈ“çè:4 0DopSxyMzSú0DopSDKI"_Hĉ]ŠĊċAb.
ò_
õË 1-3{2025·3Č¾č·hù|OPSVú0=ûWAĎüSÙ B Ā2ÙKāSKØ=d
Ù ESÛ
=
= =
1
2SÜsin	=>D"w{    |
A_
5
5
B_
1
2
C_
3
5
D_
2 5
5
¬­C
uÝ“çèú0XüýS./ 0SlmüýDopSú0DopSr«> = >S
= = @L= = @S> = > = MSÜ
=> = @/MSxyMzŠ@LMŽ~DďSàlm 0
DMNS?@.^q_
                                                                           
                                                                       
ó..
Aú0=S
=
= =
1
2S
B	= = 90°S N =S= = @L= = @S
B	= = 	S
AüýS
B	> = 	S
B	> = 	>S
B> = >S
> = > = MSÜ
=> = @/MS
2Rt  =>%S>" = =" +=>"S
BM" = @" + ( @/M)"S
.r
@ = 0{Đđ|C@ =
4
5MS
B=> = @/M =
3
5MS
Bsin	=> =
=>
> =
M
5
M =
3
5
ôª C_
【题型2  网格中求锐角三角比值】
Î 2OPS #$%DÙ AJBJCJD a2ÁHµ0÷Ù´SÚ5J=ØdÙ PSÜ	ODHL"
b{    |
A_2
B_
3
2
C_
5
2
D_
2 3
3
¬­A
uÝ“çèéHLĒSxyMzSHµ0Dop:SÚ5>J>S	= = 	=> 
= 	>= = 	>= = 45°Sc¥@ADopr	O = 	>ScxyMzŠ>J>DßScHLĒ
Štan	>D"íîHLĒDMNS]ŠĊċAēr	O = 	>S8Ĕ/ 0.b.D
ò_
                                                                           
                                                                       
ó..
OPSÚ5>J>S 
cHµ0Dopr
	= = 	=> = 	>= = 	>= = 45°S
B >N=S	> = 90°S
B 	O = 	>S
B > =
2" + 2" = 2 2S
> =
1" + 1" =
2S
B tan	> =
>
> =
2 2
2 = 2S 
B tan	O = 2
ôª
A_
õË 2-1OPS2 4×4 D#$%Sĕ†ÁHµ0D4ßw 1SÙ ASBSC Ė2$Ù´SD b AB X#$
ADØÙSÜP2Q
	=
2 D"b      
_
¬­
5
5
uÝlmxyMzėMzqrABC b/ 0Sàlm/ Y4´D%A:dY4DhĘq
r CD=AD=DBSáâ:ę 0Dopn 0DƒDopqr	 =
1
2	=ScĚqrsin
	=
2 =sin_
ó..
lmgcxyMzr
 =
2" + 1" =
5L =
3" + 4" = L =
4" + 2" = 2 5,
                                                                           
                                                                       
BAB2=AC2+BC2S
BAC?BCS	C=90°S
áâ#$q+ D uvw AB D%ÙS
BCD=AD=DBS
B	B=	DCBS
ćA	B+	DCB=	ADCS
B	 =
1
2	=,
Bsin
	=
2 =sin =

 =
5
5    S
ô¬­w
5
5 _
ÙĈ“çè./ 0SxyMzDėMzS:ę 0DopS 0ƒDop_òbr
Š	 =
1
2	=_
õË 2-2{2025·ěĜĝĞ·ùğĠġ|OPbc 6 †0ĢJeÁģĤĥ}D×0Ħ·D#$S×0D÷
Ùºw$ÙS*+×0Dh†	1 = 60°SÙLLa2$Ù´SÜcos	D"b       
_
¬­
2 7
7
uÝ“çè×0DopS !JħĨ 04:+ñS.DòbĩĪĊċA89/
 0. _
OPSÚ5>S>SīĬ	> = 90°S>JCJB „ASàlmcos	 =
>
.^q_
ó..
OPSÚ5>S>S
×0D4ßw!Scgr	>C = 30°S	>C = 60°S> =
3!S> = !S
B	> = 90°S
                                                                           
                                                                       
Ü =
7!S
A	> = 	E = 	E = 60°S
B	> = 180°S
B>JJ„AS
2Rt  >%S
cos	 =
>
 =
 !
7! =
2 7
7 _
ô¬­w
2 7
7 _
õË 2-3{2025·ĜĭĮį·İù|OPS24 × 3D#$P%SÙJJJ=a2ÁHµ0D÷Ù´S
J=ĥØdÙ>SÜsin	>=D"b      
_
¬­
7 2
10 /
7
10 2
uÝ“çèé./ 0êxyMzSlm¿D*+,-ıáâP0ĩĪ¨©DĊċAb.
Dò_Ú5=CSCSŒÙC]CE ? =S¹ąwESޜ)xyMzŠ=n=CDßSàœ)¦IJ
ŠCEDßSijä2Rt  =CE%Sœ) !DMNŠsin	C=ED"Sãälmgqr
 N C=S
ĆĄqr	>= = 	C=ES^q.¬_
ó..
OP
Ú5=CSCSŒÙC]CE ? =S¹ąwES
cgr
= =
1" + 2" =
5S
=C =
1" + 3" =
10S
 =CD¦IJ = 3 × /
1
2 × 3 × /
1
2 × 2 × /
1
2 × 3 × 1
                                                                           
                                                                       
= /// 3
2
=
7
2S
B 
1
2= 0 CE =
7
2S
5CE = 7S
CE =
7 5
5 S
2Rt  =CE%Ssin	C=E =
CE
=C =
7 5
5
10 =
7 2
10 S
cgr
 N C=S
B 	>= = 	C=ES
B sin	>= = sin	C=E =
7 2
10 S
ô¬­w
7 2
10
【题型3  灵活运用已知条件解直角三角形】
Î 3{24-25 ÏÐÑ´·ĴĵßĶ·ÕÖ|OPS2  %S	 = 45°S	 = 60°S= = 1_ķis™
š]P
①uviÙnÙwĸĹJed
1
2DßwĘĺ]ĸĻS‰ĻĥØdÙSnÙT②]/ASTØ
dÙ=SÜ=Dßw{    |
A_ 6
B_2 3
C_ 3 +1
D_ 3
¬­D
uÝ“çè]P—’“]PJAB¹/¥uADopS.Dòbz.gS&'()Q+ñ.
 _c]PŒq+S/ASTwABD¹/¥uASqr= = =SÜ	 = 	= = 45°SrŠ
	= = 	 + 	= = 90°S2Rt  =%Sqr= = =Utan60° =
3S^qr= =
3_
ó..
c]PŒq+S/ASTwABD¹/¥uAS
B= = =S
B	 = 	= = 45°S
                                                                           
                                                                       
B	= = 	 + 	= = 90°_
2Rt  =%Stan	 = tan60° =
=
= =
=
1 =
3S
B= =
3S
B= =
3_
ôª
D_
õË 3-1{24-25 ÏÐÑs·ļÄÄĽ·Õ%|OPS2   %S  = 5 S 	 =
1
2 S 	 =
1
3 SÜ  Dßw{    |
A_2 5
B_3
C_2
D_ 10
¬­D
uÝ“çèé./ 0SxyMzSŒÙ C ]= ? dÙ DS= = @SÜ= = /@Slm
	 =
1
2 S 	 =
1
3 Sr«= =
1
2@ =
1
3(/@)SŠ@ = 2S?Ąr«= = L= = 3Sœ)xyMz
^q._
ó..
ŒÙ C ]= ? dÙ DS
= = @SÜ= = /@S
A	 =
1
2 S 	 =
1
3 S
B
=
= =
1
2,
=
= =
1
3S
B= =
1
2@ =
1
3(/@)S
B@ = 2S
B= = L= = 3S
B =
=" + =" =
10S
ôª
D_
õË 3-2{2025·ľĿĜeŀ·%çŁ|OPSRt  %S	 = 90°S	 = 60°S = 2_2
                                                                           
                                                                       
n´uvłŃSSTSēS = T_uvi MSN wĸĹJied
1
2STDßwĘĺ]ĻS‰Ļ2	
ńØdÙ F_]ŅACØdÙ DSÜÙ D «DņŇw     _
¬­
2 3
3 /
2
3 3
uÝ“çèé¥uAD]n¥uADopS./ 0:+ñÙ_c]Pq+S=¥u
	Sr=E = =S	= =
1
2	 = 30°S./ 0^q._
ó..
]=E ? dÙESÜÙ D «DņŇw=EDßS
c]Pq+S=¥u	S
A	 = 90°S
B=E = =S
ARt  %S	 = 90°S	 = 60°S
B	= =
1
2	 = 30°S
A = 2S
B= =  0 tan	= =
2 3
3 S
B=E = = =
2 3
3 _
ô¬­w
2 3
3 _
õË 3-3{2025·ļÄňʼn·ùğĠġ|OPS2  %S	 = L = 3,tan =
1
2SÜ%A=Dß
w{    |
                                                                           
                                                                       
A_ 5
B_2
C_
3
2
D_
5
2
¬­D
uÝ“çèé./ 0DG)SxyMzSHĉĩĪĊċAb.Dò_
ŒÙ A ]> ? dÙ>Sr«> = >Sijä.Rt  >Sr«> = >SijälmABnŊiê 0
%Ar«= = = =
3
2SÜ=> = =/> =
1
2SàWRt  =>()xyMz.^q_
ó..
ŒÙ A ]> ? dÙ>S
A 	 = 45°S
B >b:ę/ 0S
B > = >S
Atan =
1
2S
B
>
> =
1
2S
B > = >S
B  = > = 3S
B > = > = 1
A =b%AS
B = = = =
3
2S
B => = =/> = 1
2
B Rt  =>%S= =
>" + =>" =
5
2 _
ôª
D_
                                                                           
                                                                       
【题型4  解双直角三角形】
Î 4{24-25 ÏÐÑ´·øÃŋŌ·Õ%|OPS¡ ōn ō=>CýŎ2hŏS/4X=>
ģĤŐâS*+ßw16 cmSÛ ō=>Cûµ½őŒS̶ġr1ßb 6cmSÜőŒņŇ=b
{   |
A_2cm
B_5 3 cm
C_(5 3/' cm
D_(8 3/' cm
¬­C
uÝcg+S>C N SÜ	=1 = 	C = 45°SOPS]1S ? dSSÜ =  0 cos30°S
S = 1 0 cos30°S1S = 1 0 sin30°S=S =
1S
tan45°Slm= = /=S/SSEF.^q_
ó..
cg+S>C N S
B	=1 = 	C = 45°S
OPS]1S ? dSS
B =  0 cos30° = 16 ×
3
2 = 8 3cmSS = 1 0 cos30° = 6 ×
3
2 = 3 3cmS1S = 1 0 sin30° = 3cmS
B=S =
1S
tan45° = 3cmS
B= = /=S/S = 8 3// 3 = (5 3/)cmS
ôª
C_
ÙĈ“çèé¥őDopS¥@ADopSHLSHISKI:+ñ_ëìíî¥őDopS¥@
ADopSHLSHISKIb.Dò_
õË 4-1{24-25 ÏÐÑ´·ļÄÄĽ·ÕÖ|OPS2Rt  %S	 = 90°SD¹/¥uAØ
XÙ ESÛtan	=> =
4
3S = 6SÜ>Dßw     _
                                                                           
                                                                       
¬­
7
4
uÝ“åæçèé./ 0êAB¹/¥uADopSlmgrŠ	 = 	=>S?ĄrŠ	
DHL"SàáâDß^qŠDßS?h™rŠDßÆS?ĄrŠ=DßSãä2Rt  =>%S
Š>Dß^q. _
ó..
c+S
A	 = 90°S=> ? S
B	 + 	 = 	=> + 	 = 90°S
B	 = 	=>S
Btan = tan	=> =
4
3S
ćA = 6S
B = 8S
B =
6" + 8" = 10S
A=>¹/¥uS
B= =
1
2 = 5S	>= = 90°S
2Rt  =>%Stan	=> =
=
=>S
B
5
=> =
4
3S
B=> =
15
4 S
B> =
5" +
15
4
"
=
25
4 S
B> = /> = R/
25
4 =
7
4_
ô¬­w
7
4_
õË 4-2{2025·øÃŋŌ·%çŁ|OPS2 0œŔ%S	 = 57°S	 = 38°S¡œŔûŕ
ŒÙ A D/AüýSēÙĀ24´DÙ>āSüŖ=ØdÙ=à¡œŔûŕŒÙ>D/AüýSē
                                                                           
                                                                       
ÙĀ24´DÙEāSüŖ>CØdÙC_s›áŗ·ŘDb{   |
A_=E = >E
B_E> ? >
C_	=> = 42°
D_=> = EC
¬­A
uÝ“çèé 0DĎü S¹/DMNS:ę 0DřMXopiê/ 0%HI"
D.S2ĎüŒ%c4ßnÆŚõSq.ĎüśäDÆb.“Dò_lm  =>bc
 =Ďür«q.	=>DÆScĚřŜ C ªŝlmĎüśäÆD.Sq.	>=EX	=>E
DÆSc“:W:4”qřŜ A ªŝS.	>EDÆqřŜ B ªŝŞáŗ·ŘSlm/ 0
%DHI".4ß^qřŜ D ªŝ_
ó..
C ªŝS2  %S	 = 57°S	 = 38°S
B	 = R
/D/R = 85°S
A  =>bc  =Ďür«S
B	=> = 	= =
85°
2 = 42.5°Sô C ªŝşŠ
A ªŝSA  =>bc  =Ďür«S	=> = 	= = 42.5°S
B	>= = 	 = 57°S
B	=> = 	= = R
/D/ W = 80.5°S
B	>=E = R
/	=>/	= = R
/R
W × 2 = 19°S
A  >CEbc  >CĎür«S
B	>EC = 	 = 38°S
B	>E= = R
/	>EC = R
/R = 142°S
2  >E=%S	=>E = R
/ / = 19°S
A	>=E = 	=>E = 19°S
B=E = >ESô A ªŝHĉ
B ªŝSA	>= + 	=>E = 57° + 19° = 76°S
                                                                           
                                                                       
^	>E = 76°S
BE>X>Ś¹/Sô B şŠ
D ªŝSŒÙ G ]ES ? =>Ø=>dÙ MSOPS
Ş=> = ECS
A  >CEbc  >CĎür«S
B	>C = 	>CE = 90°S
A=E = >ES
B  =E>w:ę 0S
AES ? =>S
B=S = >SS^=> = >SS
BEC = >SS
2Rt  >SE%Ssin	=>E = sin19° =
SE
>ES
2Rt  >CE%Ssin	>EC = sin38° =
>C
>ES
Asin19° X sin38°S
BSE X >CS
ćA>S =
>E"/SE" X EC =
>E"/>C"SX*+ŚšSô D ªŝşŠ_
ôª
A_
õË 4-3{2025·ĽŢţŤ·hù|OPS2  %S =  = 12S = 10SÙ=w%ÙSÙOi
ĕť 1 †Ŧ»DŧÆĆŠŨûYY(Œ_©  O=w:ę 0¶SZD"w{   |
                                                                           
                                                                       
A_
25
6 C 18
B_
25
6 C 18 C 19
C_
25
6 C 18 C 19 C
119
6
D_
25
6 C 18 C 19 C 20
¬­C
uÝ“çèé:ę 0DopS./ 0G):+ñSuÙ P 2´n´ũŗSijälm
:ę 0Dopn./ 0DG).^q_
ó..
Ú5=S
A = S = 10S
B= ? S= = = = 5S
①©Ù P 2´¶S	O= > 	= = 90°S
B  O=w:ę 0¶SŪjO= = =S
BO= = =S
ΠD ]=[ ? Od QS
                                                                           
                                                                       
ÜO = [S
Acos =
[
= =
=
S
B
[
5 =
5
12S
B[ =
25
12S
BO =
25
6 S
BZ =
25
6 ÷ 1 =
25
6 
②©Ù P 2´¶S
A  O=w:ę 0S
B= = OC=O = OC= = =OS
©= = O = 5¶SOPS
Z = (12 × /) ÷ 1 = 19
©=O = O¶SOPSŒ P ]O[ ? =d QS
Ü[ =
1
2= =
5
2S
Acos =
=
 =
[
OS
B
5
12 =
5
2
OS
.rO = 6S
                                                                           
                                                                       
BZ = (2 ×  /) ÷ 1 = 18
©= = =O¶SOPSŒ D ]=[ ? Od QS
ÜO = [S
Acos =
=
 =
[
=S
B
5
12 =
[
5 S
.r[ =
25
12S
BO =
25
6
BZ = 2 ×  /
25
6
÷ 1 =
119
6 
ū´St D"w
25
6 C 18 C 19 C
119
6 S
ôª
C_
【题型5  在四边形中解直接三角形】
Î 5{2025·Ŭ¾ŭŮ· ù|OPS2ú0=%S= =  = RL>LCuvw=L4´DÙSk
C = 3S¡ú0=û/A>CüýSr«340>CTSSÙLDWGÙuvwÙSLT{ÙSĀ2=´
µ|SÚ5TS©LTLS لA¶S>Dßw{    |
A_2
B_
4
3
C_
16
3
D_1
¬­D
uÝ“åæçèéú0DopSüý SxyMzS./ 0:+ñÙSëìíî‹opS
                                                                           
                                                                       
âzĩĪĊċAb.ĚDò_OPS\SX=DØÙw]SůßCTØ=d[SáâC = CT = 3S
ÜC = 5SqrT = 4Sáâ
T[
][ = sin	T][ = sin	TC =
CT
C =
3
5ST[ = @SÜ][ = @S]T =
[]"/T[" = @Sqr] = @ + 4S.@ =
2
3Sà?h™.^q_
ó..
A2ú0=%S= =  = 8S
B= =  = 8S = = = 4S	 = 	 = 	= = 	= = 90°S=NS
cWüqr
C = TC = 3S	STC = 	 = 90°SST =  = 4S
B	CT = 90°S
BC = 5S
OPS\SX=DØÙw]SůßCTØ=d[S
BT =
5"/3" = 4S
cWüqr
	C> = 	[C>S
A=NS
B	=>C = 	C>S	=] = 	CTS
B	[>C = 	[C>S
B[> = [CS
A	STC = 90°S
B	]T[ = 90°S
B
T[
][ = sin	T][ = sin	TC =
CT
C =
3
5S
T[ = @SÜ][ = @S]T =
[]"/T[" = @S
B] = @ + 4S
Bsin	=] =
3
5 =
=
] =
4
@(4S
.r
@ =
2
3S
                                                                           
                                                                       
BT[ = 3 ×
2
3 = 2S][ = 5 ×
2
3 =
10
3 S
B>[ = C[ = 3 + 2 = 5S
}z
tan	]= = tan	TC =
3
4S
B
=
=] =
3
4S^
4
=] =
3
4S
B=] =
16
3 S
B[= = ]=/][ =
16
3 /
10
3 = 2S
B> = =/[>/=[ = R// = 1_
ôª
D_
õË 5-1{2025·øÃŰŤ· ù|OPS2340=%S> = >S=C = CS=C ? >SCN=S
tan	= =
1
4S>C = 2SÜDßw{   |
A_3 2
B_4 2
C_2 5
D_4 5
¬­D
uÝ“çèé 0%»AMzS./ 0S¥@340DřMnopSxyMz_Þr>C
b  =D%»ASr= = >C = 4SÜtan	= =
1
4Sr=C = C = 8SàīĬ340=Cb¥@
340Sr= = C = 4S2Rt  C%Sœ)xyMz.^q_
ó..
A> = >S=C = CS
B>Cb  =D%»AS
B>CN=S= = >C = 4S
A=C ? >S
B	C> = 90°S
Atan	= =
1
4S
B
>C
C =
1
4S
                                                                           
                                                                       
A>C = 2S
B=C = C = 8S
ACN=S>CN=S
B340=Cb¥@340S
B= = C = 4S
2Rt  C%S =
C" + C" = 4 5S
ôª
D_
õË 5-2{2025·ľĿĜűŲų·ùğĠġ|OPS340=w4ßw 4 DHµ0SÙ E w=D%ÙS
Ú5>ıůßŴÙ FSÚ5CSÚ5C=ıůßØůßAdÙ GSÛtan	C> =
4
5¶SÜEDßw      
_
¬­1
uÝ“çèéHµ0DopS./ 0SŒC]C^ ? =dÙ^SÜ	^C = 	C^= = 90°Slm
gqC^ = @S^ = @Slm./ 0›µr@S^qr^=S?Ą.^qSëì./
 0b.Dò_
ó..
 A 340=wHµ0S
B  = = = =S	= = 	= = 90°S= N S
A >w=%ÙS
B > = >= =
1
2= = 2S
B 2Rt  >%Stan	> =

> =
4
2 = 2S
OPSŒC]C^ ? =dÙ^SÜ	^C = 	C^= = 90°S
                                                                           
                                                                       
2Rt  ^C%Stan	C^ =
C^
^ =
4
5S
C^ = @S^ = @S
B >^ = @/ S
A 	C>^ = 	>S
B tan	C>^ = tan	> = 2S
B 2Rt  >^C%Stan	C>^ =
C^
>^ = 2S
^
@
@/ = 2S
ŵŒŶŷ.r@ =
2
3S
B C^ = @ =
8
3S=^ = /@ = /
10
3 =
2
3S
B 2Rt  C^=%Stan	C=^ =
C^
=^ = 4S
A =NES
B 	E = 	=CS
B tanE = tan	C=^ = 4S
2Rt  =E%StanE =
=E
E = 4S
B E = 1S
ô¬­w
1_
õË 5-3{2025·ö¾ŸŹ·İù|*+
OPS2  %S	 = 
L=wAB´hÙSk = =L>
wABD%ÙSÚ5=J>=Sůß«ÙCSērC = SÚ5=CSŒÙ]=CD¥@AØ=Dů
ßAdÙE_
                                                                           
                                                                       
(1)340=CEbHµ0źŻżŽĬzc
(2)Ûtan	 =
3
5L= = 2SAB=>DßÆ_
¬­(1)bSzc¯.Ý
(2)
10
2
uÝ{1|c¥@ADopr«	E = 	=CSàc‰† 0Ĥ:DřMXopSáâ¥@340
DřMJ×0DřMêHµ0DřM^qr«¬­
{2|cHLĒtan	 =
3
5S2Rt  %SŠABßS?Ą2Rt  =C%S}zŠABßS2Rt
 =>C%ScxyMz.^qr«¬­_
ó.{1|.
bS
zcOs
A E N =CS
B 	E = 	=CS
A 	 = 90°S
B 	E = 	C= = 90°S
2  En  C=%S
	E = 	=C
 = C
	E = 	C= = 90°
 
B E_  C=(ASA)S
B E = C=SE = =S
A E N =CS
B 340=CEb¥@340S
A =E ? S
B 340=CEb×0S
                                                                           
                                                                       
A  = CL	 = 90°S
B =¹/¥uCSÜ= = =CS
B 340=CEbHµ0
{2|.
2Rt  %S	 = 90°Stan	 =
3
5SÜ

 =
3
5S
 = @SÜ = @S= =  = @S
B = = /= = @/@ = @ = 2S
.r@ = 1S
B  = L = 5S
2Rt  %ScxyMzqr =
" + " =
34S
ŒÙ=]=C ? SOPQR
2Rt  =C%Stan	 =
3
5L= = 2SÜ
=C
C =
3
5S
=C = MSÜC = MS
B
34M = 2S.rM =
34
17 S
B =C =
3 34
17 LC =
5 34
17 S
A >wABD%ÙS
B > =
1
2 =
34
2 SÜ>C = >/C =
34
2 /
5 34
17 =
7 34
34 S
2Rt  =>C%ScxyMzqr=> =
=C" + >C" =
10
2 _
ÙĈ“çè¥@340êħĨ¥@340ūâSžê¥@AopJĤ:DřMXopJ¥@340
řMJ×0řMJHµ0řMJ./ 0JxyMz:+ñSë\ĥſƀopXřMb. D
ò_
                                                                           
                                                                       
【拔尖篇】
【题型6  构造直角三角形求锐角三角比值】
Î 6{2025·ľĿĜƁ§·ùğĠġ|2¥@340=%SF b=D%ÙSÙ E 2ŅA´Sk
> =
1
2SÚ5>C_Û = L= = L	 = 60°SÜtan	C>D"w     _
¬­ 3C
3
2
uÝ©Ù E 2AB´¶Sœ)¥@340DřMnopSáâħĨ !EF^q©Ù E 2A
BDůßA´¶SŒÙ F ]CS ? dÙ MSœ) !.¬^q_
ó..
©Ù E 2AB´¶S
A¥@340=%SF b=D%ÙS
B = =L= = L= N SC = C= =
1
2= =
1
2S
A>a =
1
2S
B>a =
1
2S
B>a = CL>a N CS
B340>aCb¥@340S
B N >aCS
B	C>a = 	 = 60°S
Btan	C>a = tan	 = tan60° =
3
©Ù E 2ABDůßA´¶S
ŒÙ F ]CS ? dÙ MS
A	C>a = 	 = 60°S
Btan	C>a = tan	 = tan60° =
3
BCS = >aCsin60° = 2 3S>aS = >aCcos60° = 2S
A = = = 6S
B>a = >a = > =
1
2 = 3S
B>S = > + >a/>aS = 3 + 1 = 4S
Btan	C> =
CS
>S =
2 3
4 =
3
2 
                                                                           
                                                                       
ū´QƂStan	C>D"w 3C
3
2 S
ô¬­w
 3C
3
2 _
ÙĈ“çèé¥@340DřMnopS !DG)SħĨD !DEFSu”ƃƄSëì
íîřMnopS !DG)b.Dò_
õË 6-1{24-25 ÏÐÑ´·øÃƅƆ·ÕÖ|OPS2  %S	 = 90°S = 2 5S = 6S=
b4´D%ASÜcos	=D"b{   |
A_
2
7
B_
5
7
C_
5
7
D_
3 5
7
¬­A
uÝ“çèé./ 0SxyMzS/ 0Y4´D%A:dY4DhĘSëìíî/
 0Y4´D%A:dY4DhĘb.Dò_œ)xyMzŠ = 2 14Sàlm/ 0Y4
´D%A:dY4DhĘSŠ= = = =
14SŒÙ]b ? =dÙ HSœ) 0¦IJËŠbS
àŠ=bScKIDMNƇƈmEF^q_ 
ó..
A2Rt  %S	 = 90°S = 2 5S = 6S
B =
" + " =
(2 5)
" + 6" = 2 14S
A=b4´D%AS
BFcdH =
1
2FcGH =
1
2 ×
1
2e = 3 5S= = = =
1
2 =
14S
ŒÙ]b ? =dÙ HS
B
1
2=eb = 3 5S
                                                                           
                                                                       
Bb =
3 70
7 S
2Rt  b=%S=b =
="/b" =
/
630
49 =
2 14
7 S
Bcos	= =
=b
= =
2 14
7
14 =
2
7S
ôª
A_ 
õË 6-2{24-25 ÏÐÑ´·3ČƉʼn·ÕÖ|OPS2¥¦/ƊÀ%S3401D412 x Ƌ´S
12 y Ƌ´k N 1SAB1SDßuvbµ@"/@ + 20 = 0D‰†l{1 < |SPJQ uvw
1J1´‰ÙS1[ = 5S¡  O1[ĎüSēÙ O Ā24´DÙ D āSÜtan	O[= =        
_
¬­
1
2/0.5
uÝޜ)ƌËu..µ@"/@ + 20 = 0qr«1 = 4S = 5SrŠ3401[wú0SÜ
[ = 1 = 4SlmxyMzŠ1K = 3SÜ= = 2Scüýr«1O = =OS	O[= = 	O[1Sijäœ)
xyMzŠO= =
5
2S?Ą.^q_
ó..
@"/@ + 20 = 0
(@/) (@/) = 0
r@a = 4S@" = 5_
A 1 < S
B 1 = 4S = 5S
Ú5[S
A  N 1S1[ =  = 5S
B 3401[w¥@340_
A 	1[ = 90°S
B 3401[wú0S
                                                                           
                                                                       
B [ = 1 = 4S	[ = 90°S
B = =
=["/[" = 3S
B= = /= = / = 2S
c  O1[ĎüSēÙ1Ā2´DÙ D āS
B1O = =OS	O[= = 	O[1
BO = 1/O1 = /O1 = /O=
A	O= = 90°
BO" +=" = O="
B(/O=)" + 2" = O="
BO= =
5
2
BO1 = O= =
5
2
Btan	O[= = tan	O[1 =
O1
[1 =
5
2
5 =
1
2_
ÙĈ“çèéüýDopS.h‡İƍµSxyMzSú0DřMXopS./ 0Sz.
ƊÀXP0opSëìíîú0DopXřMb.Dò_
õË 6-3{2025·Ĝĭ¾É·ùğĠġ|OPSHµ0=%S¡4ƎÙė¶ƏƐ;Ŵ>SÚ5>S
=>SÛ	>= = 90°SÜsin	>=D"b{   |
A_
1
2
B_
3
2
C_
5
5
D_
2 5
5
¬­C
uÝŒ]C ? >S¹ąwCSlm‰† 0Ĥ:DřMMzSĉM  Cf  =>(AAS)SĆĄl
mĤ: 0Dopr«C = =>Sàlm¡4ƎÙė¶ƏƐ;Ŵ>SĉM  >w:ę 0Sá
â“ Aâh”r«Cb>4´D%AS?Ą>C = C = =>S^> = =>S2Rt  =>%S> = =>S
=> = !SÜ> = !ScxyMzr«= =
5!S œ)HI"MN.^qr«¬­_
ó..
Œ]C ? >S¹ąwCSOPQR
                                                                           
                                                                       
B 	C = 90°S
2Hµ0=%S	= = 90°S = =S
B 	C + 	=> = 90°S
A 	C + 	C = 90°S
B 	=> = 	CS
2  Cn  =>%S
	C = 90° = 	>=
	C = 	=>
 = =
 
B Cf  =>(AAS)S
B=> = CS
A ¡4ƎÙė¶ƏƐ;Ŵ>S
B > = S
A C ? >S
B c“ Aâh”qrCb>4´D%AS^>C = C = =>S
B > = =>S
2Rt  =>%S> = =>S=> = !SÜ> = !S
cxyMzr«= =
=>" + >" =
5!S 
B sin	>= =
=>
= =
!
5! =
5
5 S
ôª
C_
ÙĈ“çèHI"SžêHµ0DopJĤ:DřMXopJ:ę 0DřMXopJxyM
z:+ñSëìíîĥſƀƑƒJřMXopb. Dò_
【题型7  等角转换法求锐角三角比值】
Î 7{2025·3ČƓø·hù|OPSÙ>bú0=%=4´DhÙS >û>üýw  C>SÙC
Ā2=´_Û	=C>DeÁwSkƔą
sin/cos
cos(2sin = /
1
10SÜtan	>D"w       
_
                                                                           
                                                                       
¬­2
uÝ“çèéú0DopSüýDopS !DEFSíî !DEFµbò_
lmú0SüýDopr« = CL> = >CS	C> = 	 = 90°Sc !DEFr«
=>
=C =
3
4SÜ
tan	C = tan	=C> = tan =
=>
=C =
C
 =
3
4ScĚr«> = @L= =  = 
@SáâHL"DEF^q
._
ó..
A340=bú0S
B = =L= = L	 = 	 = 	 = 	= = 90°S
AüýS
B = CL> = >CS	C> = 	 = 90°S
2Rt  =>C%Ssin =
=>
>C,cos =
=C
>CS
B
sin/cos
cos(2sin
=
=>
>C / =C
>C
=C
>C + 2 × =>
>C
=
=>/=C
=C( =>S
B
=>/=C
=C( => = /
1
10S
ƕzrS
=>
=C =
3
4S
Btan =
=>
=C =
3
4S
A	C + 	C = 	C + 	=C> = 90°S
B	C = 	=C> = S
Btan	C = tan	=C> = tan =
=>
=C =
C
 =
3
4S
=> = @L=C = @SÜ>C =
=>" + =C" =
(@)" + (@)" = @S
B> = >C = @S
                                                                           
                                                                       
B= = => + > = @ + @ = R@S
A
C
 =
3
4S
BC =
3
4 =
3
4 × R@ = @S
B= = C + =C = @ + @ = 
@ = S
Btan	> =

> =

@
@ = 2S
ô¬­w
2 _
õË 7-1{2025·øÄƖƗ· ù|OPS2340=%S = S ? =S=¥u	=SX=
ĥØdÙ>SÛ= = 3S = 4SÜAB>Dßw        
_
¬­
5
2
uÝ“ŐÙçè:ę 0DopJ¥uADopJĤ: 0DřMXopJxyMzJ./
 0:+ñSHĉĩĪĊċAb.Dò_
]>g ? =dÙgSĄ ? =S= = 3S = 4Sr= = 5Sc=¥u	=Sr> = >gScsin	= =
>g
> =
=
= =
3
5Sr> = >g = 
3
5>SÜ> +
3
5> = 4Sr> =
5
2SÜ> =
3
2S]b ? =dÙbLC ? =
Ø=DůßAdÙCSÜb = CSqīĬRt  bfRt  CSr	b = 	CSƘƙŠ
	= + 	 = 	=> + 	 = 180°Sc = Sr	> = 	>SÜ 	> + 	 = 180°Sq
īĬ	> = 	=>SÜ	> = 	>= = 90°Sc
>
 = tan	> = tan	=> =
>
= =
1
2Sr = >SQ
i> =
5> =
5
2Sr> =
5
2 Sdbr« D¬­_
ó.]>g ? =dÙgSÜ	g> = 90°S
                                                                           
                                                                       
A ? =S= = 3S = 4S
B	= = 90°S
B= =
=" + " =
3" + 4" = 5S
A=¥u	=SX=ĥØdÙ>Sk> ? =S>g ? =S
B> = >gL	= = 	=>S
Asin	= =
>g
> =
=
= =
3
5S
B> = >g =
3
5>S
B> +
3
5> = 4S
.r> =
5
2S
B> =
3
5 ×
5
2 =
3
2S
]b ? =dÙbSC ? =Ø=DůßAdÙCSÜ	b = 	C = 90°Sb = CS
2Rt  bn Rt  C%S
 = 
b = C
 
,
BRt  bfRt  C(HL)S
B 	b = 	CS
A 	CD +	C = 180°S
B 	= + 	b = 180°S
B 	= + 	 = 
/&	= + 	b' = 180°S
B 	=> + 	 = 180°S
A  = S
B 	> = 	>S
B 	=> + 	 = 180°S
                                                                           
                                                                       
B 	> + 	 = 	=> + 	S
B 	> = 	=>S
B 	> = 	>=/	> = 	>=/	=> = 	>= = 90°S
B
>
 = tan	> = tan	=> =
>
= =
3
2
3 =
1
2S
B  = >S
A > =
" + >" =
& >'" + >" =
5> =
5
2S
B > =
5
2 _
ô¬­w
5
2 _
õË 7-2OPS2Rt  %S	 = 
L = SÙ=wY4´hÙSk= = =S¡  =
û/A=ĎüSÙDWGÙwKSÜsin	K= =     _
  
¬­
10
10
uÝ“çèéĸń5340D+ñSHIĒSüýDopiêxyMzS.¬“DòbĬĉ
gSƚŠQ ƛæD,-_ÞīĬJKJJ=3لĸSƘŠ	k= = 	=SŒÙ=]=> ? 
dÙ>Sœ)¥@AuAB·!ÎMzr«> = >ScxyMzr«= =
10!SàcHIĒ^q._
ó..
 A 	 = 90°S = S
B 	 = 	 = 45°S
cüýopr	K= = 	 = 45°S
B 	K= = 	= = 45°S
B JKJJ=3لĸS
B 	K= = 	=S
ŒÙ=]=> ? dÙ>S
                                                                           
                                                                       
  
A 	 = 90°S
B => N S
B = = =S
B > = >S
A 	 = 45°S
B =>b:ę/ 0S
B => = >S
=> = > = !SÜ> = > = !S
2Rt  =>%S= =
=>" + >" =
!" + (!)" =
10!S
B sin	K= = sin	= =
=>
= =
!
10! =
10
10 S
ô¬­w
10
10
õË 7-3{2025·Ŭ¾Ɯʼn·hù|OPS*+Ùb/A7ƒhMÙSb/A7´DŒABS = 5SÚ
5JSFcGH = 15_© + ŃãÁ"¶sin	D"_ÁƝ2.Œ%Ũƞ
“ƟċƠzơ
DĥW(ŒƃƢSÛ¡ƣ]ƤABSÜÙ2¥@d/A7D/A´(Œ”_żƥçÁƝDƃƦ©
 + ŃãÁ"¶sin	 =           
_
¬­
12
13
uÝŒÙ]7aN7S]Ùd/A7aDWºÙKSØ/A7adÙ>SÚ5KØ/A7adÙOSÚ5OS
ŒÙO]O= ? dÙ=SÚ5KS©JJK لA¶S^Ù(Œ«ÙOā¶S + ŃãÁ"_S
ÞŠO=nKD"SàɌxyMzŠK = 13SɌÆDƇ<Sīr	O = 	OSɌsin
                                                                           
                                                                       
	 = sin	O = sin	O =
K
K^q._
ó..
OPSŒÙ]7aN7S]Ùd/A7aDWºÙKSØ/A7adÙ>SÚ5KØ/A7adÙOS
Ú5OSŒÙO]O= ? dÙ=SÚ5KS
A ÙKbÙd/A7aDWºÙS
B /A7a¹/¥uKS
B OK = OSK> = >SK = S
B  +  =  + K l KS
B ©JJK لA¶S^Ù(Œ«ÙOā¶S + ŃãÁ"_
A  = 5SFcGH =
1
2 0 O= = 15S
B O= =
FcGH
1
2  = 6_
A 7aN7SkO= ? S7a ? KS
B 	O= = 	=O> = 	O> = 90°S
B 340O=>bú0S
B O= = > = K> = 6S	>= = 90°S
B K = > = 12S	>O + 	O = 90°S	K +	O = 90°S
A OK = OS
B 	K = 	>OS
B 	O = 	OS
A K = 12S = 5S
2Rt  K%SK" = " +K"S
B K =
5" + 12" = 13S
                                                                           
                                                                       
B sin	 = sin	O = sin	O =
K
K =
12
13_
B © + ŃãÁ"¶Ssin	 =
12
13_
ô¬­w
12
13_
ÙĈ“åæçèéŒÙã"D¡ƧƨƩùSAB¹/¥uADopSú0DřMXopS !S
xyMzSëìíîŒÙã"”ùD.ƃƦb.ò_
【题型8  巧设未知数解直角三角形】
Î 8{2025·ƪÃƫƬ·ùğĠġ|2:ę  %S = SD b BC ´hÙSŒÙ D ]=> ? =Ø
ůßAdÙ ESÛtan	 =
24
7 S
=
 =
2
5SÜ

>D"w        
_
¬­
13
4
u݌٠A ]O ? dÙ PSŒÙ B ]b ? dÙ HSŒÙ E ]>C ? Ø BC DůßAdÙ FSc
HLĒrtan	 =
b
b =
24
7 Sb = !Sb = D!Sœ)xyMzuvŠ =  = !S
b = R!S = 
!SÜO = O = !SàŠ= = 
!SÜ= = 
!S=O = !SO = 
!S?Ąr
tan	O =
O
O =
4
3S tan	>C =
>C
CSlm	O = 	>Cr
>C
C =
4
3S>C = mSC = mSÜ> = mS
=C = 
! + mScHLĒtan	=O =
=O
O =
1
4Stan	>=C =
>C
=C =
m
20(mS^q._
ó..
ŒÙ A ]O ? dÙ PSŒÙ B ]b ? dÙ HSŒÙ E ]>C ? ØDůßAdÙ FS
OPQR
                                                                           
                                                                       
 
B ON>CS
2Rt  b%Stan	 =
b
b =
24
7 S
Bb = !Sb = D!S
 =
b" + b"
=
& !'" + &D!'"
= !S
B  =  = !S
B b = /b
= !/D!
= R!S
B  =
b" + b"
=
& !'" + &R!'"
= 
!S
A O ? S
B O = O = !S
A
=
 =
2
5S
B = =
2
5 = 
!S
B = = /=
= 
!/
!
= 
!S
=O = O/=
                                                                           
                                                                       
= !/
!
= !S
B O =
"/O"
=
& !'"/&!'"
= 
!S
B tan	O =
O
O =
 
!
! =
4
3S
Atan	>C =
>C
CS	O = 	>CS
B
>C
C =
4
3S
>C = mSC = mS
> =
>C" + C"
=
&m'" + &m'"
= mS
B =C = = + C
= 
! + mS
2Rt  O=%Stan	=O =
=O
O =
!
 
! =
1
4S
2Rt  =>C%Stan	>=C =
>C
=C =
m
 
!(mS
A => ? =SO ? S
B 	>=C + 	=O = 90°S	=O + 	=O = 90°S
B 	>=C = 	=OS
B
m
 
!(m =
1
4S
.r
m =
 
!
13 S
B > = m =


!
13 S
B

> =
 !


!
13
=
13
4 _
ô¬­w
13
4 _
                                                                           
                                                                       
ÙĈ“çè:ę 0DopSHLĒSxyMzSíîĥƭ 0DřMêopS:ę 0
DopSƮ8Ĕĥƭ 0SıƮëìœ)HLĒnxyMz?@.b.Dò_
õË 8-1{2025·3Čƪ‡·%çŁ|340=%SX=ØdÙ OSO bD%ÙS1 = =1S
*+ = 4S= = 2Stan	= =
3
5 SÜDßw        
_
¬­
8 3
3 /
8
3 3
uÝ“çèéxyMzJ !DMNêĥƭ 0DřMXopS.DòbɌ]¹A89
/ 0Sœ) 0ĥƭn !ƘƙABßÆ_ƯÙ BSD uv]D¹ABSœ)
 1=> n 1Cr«1C = 1>SC = =>Sàœ) = =SƘŠoZ  >= p oZ  CS?Ąr«
C = >S1> = @Sáâ O bD%ÙÜqƘŠ> = @S> = @Sctan	= =
3
5 qtR=> =
3
@S2oZ  =>xyMzĔص^q. xSÜ = R@q_
ó.OPSŒ D ]=> ? d ESŒ B ]C ? d FS
A	1>= = 	1C = 90°S	=1> = 	1S
B  1=> n 1CSÜ
1
=1 =
1C
1> =
C
=> = 2S
1> = @ SÜ1C = @S>C = @
A  = 4S= = 2S
B 
= = 2
B 
= = C
=> = 2
A 	>= = 	C = 90°S
B oZ  >= p oZ  CS
B
C
> =

= = 2S
                                                                           
                                                                       
B C = >S
^> = >C = @S
B 1 = > + 1> = @S
AO bD%ÙS
B 1 = 1 = @S
B > = 1 + 1> = @S
A tan	= =
3
5 S
B
=>
> =
3
5 S
B => =
3@S
2oZ  =>%S= = 2S
cxyMz
>" +=>" = ="S^&@'" + ( 3@'
"
= 2"S
.r
@ =
3
3 S
B  = 1 = R@ =
8 3
3 _
ô¬­w
8 3
3 _
õË 8-2{24-25 ưÐÑs·¼Ʊ·ÕÖ|OP 1S2  %S	 = 90°S = 5S¡‹uƲ·ⅠJ
ⅡJⅢ ƳuSijäàƴ·OP 2 D×0O[ST{ŚŐýJĮƵƶ|SÛTb/OE = 1SÜ[bDßw      
_
¬­
7
5
uÝlmgS	 = 90°S = 5S×0O[STSTb/OE = 1S[b = => = @S
> = ST = [S = [b + bS = = + => =
1
2 =
5
2STb = SOE = C = C =
1
2S
= = O[ = OT = ST =
5
2ScTb/OE = 1Sr/
1
2 = 1^ = / SlmxyMzSô = 3S
 = 4SŒÙ=]=[ ? dÙ[SlmgSrtan =
=[
[ =

 =
4
3S=[ = @S[ = @SÜ= =
                                                                           
                                                                       
[" + =[" = @S[ = /[ = /@Sô(@)" + (/@)" =
5
2
"
S.¬^q_
ó..
lmgS	 = 90°S = 5S×0O[STSTb/OE = 1S
[b = => = @S> = ST = [S = [b + bS = = + => =
1
2 =
5
2S
Tb = SOE = C = C =
1
2S= = O[ = OT = ST =
5
2S
cTb/OE = 1Sr/
1
2 = 1^ = / S
lmxyMzSr" +" = "S
ô( / )" +" = 25S
.r = 3S = /
7
5{Đđ|Sô = 4S
ŒÙ=]=[ ? dÙ[S
lmgSrtan =
=[
[ =

 =
4
3S
=[ = @S[ = @S
Ü= =
[" + =[" = @S[ = /[ = /@S
ô(@)" + (/@)" =
5
2
"
S
.r@ =
1
2S@ =
11
50S
ô@ =
11
10S@ =
5
2{Đđ|S
ô[b = => =
5
2/@ =
7
5S
ô¬­w
7
5_
ÙĈ“çèéxyMzS×0DopS.µS !DG)SƴPDſƀgNSëìíîopn
 !DG)b.Dò_
õË 8-3{24-25 ÏÐÑs·ƷƸƷʼn·Õ%|OPS2×0=%SŒ÷Ù D ]=> ? S=C ? S¹
ąuvw ESFSÚá>C_Ûcos =
1
3S  >CD¦IJw 4SÜ×0=D¦IJw        
_
                                                                           
                                                                       
¬­18
u݌٠F ]CE ? dÙ GSīĬ  =>f  =C(AAS)Sr> = CSÜ> = CS
> = C = !Sàc¥@ADopr	CE = 	S?Ąc !MNrE =
1
3C = !SÜCE = 2 2
!Sc 0¦IJËŠ3 2!" = 4SijäcxyMzŠ=> = 3 2!S^q. _
ó..
OPSŒÙ F ]CE ? dÙ GS
A=> ? S=C ? S
B 	=> = 	=C = 90°S
A340=b×0S
B= = =S	 = 	S
2  =>n  =C%S
	=> = 	=C
	 = 	
= = =
 
S
B  =>f  =C(AAS)S
B> = CS
A = S
B/> = /CS
^> = CS
> = C = !S
A= N S
                                                                           
                                                                       
B	CE = 	S
Bcos	CE = cosS
Acos	CE =
E
CScos =
1
3S
B
E
C =
1
3S
BE =
1
3C = !S
BCE =
C"/E" = 2 2!S
AFGIq =
1
2> 0 CE =
1
2 × ! × 2 2! = 3 2!"SkFGIq = 4S
B3 2!" = 4S
Acos =
>
= =
>
 =
1
3S
B> =
1
3,
B> =
2
3S
B =
3
2> =
9
2!S
B> =
3
2!S
B=> =
="/>" =
9
2 !
"
/
3
2 !
"
= 3 2!S
BF菱形cGHd =  0 => =
9
2! × 3 2! =
9
2 × 3 2!" =
9
2 × 4 = 18_
ô¬­w
18_
ÙĈ“çèé×0X 0ūâ_ëìíî×0DopSĤ: 0DřMXopS¥@ADopS
 !MNSxyMzS 0¦IJ:+ñSb.Dò_
【题型9  构造直角三角形进行线段或角的计算】
Î 9{24-25 ÏÐÑ´·Őŀ·Õ%|OPSÙ=b  ƒhÙS= = =SX=ĥØdÙ>S
	= = 	SÚ5=SÛ = 4S= =
13Stan	= =
1
2SÜ= =        
_
                                                                           
                                                                       
¬­3 5
uÝ“åæçèé.ƹ/ 0SŒÙ=]=T ? ØůßAdTSÞc	= = 	S
	>= = 	>Sr«	= = 	=S^qr«tan	= = tan	= =
1
2 =
=T
TS=T = @SÜT = @S
T = T/ = @/S2Rt  =T%Sœ)xyMz›µr@ = 3S^qr«=T = @ = 3S
T = @ = 6Sãälm= = = =
=T" + T"EF^q_
ó..
OPSŒÙ=]=T ? ØůßAdTSÜ	T = 90°S
 
A 	= = 	S	>= = 	>S
B 	= = 	=S
Atan	= =
1
2S
Btan	= = tan	= =
1
2 =
=T
TS
B=T = @SÜT = @S
A = 4S
BT = T/ = @/S
2Rt  =T%S= =
13S=T" +T" = ="S
B@" + ( @/)" = ( 13)
"
S
.r@a =
1
5,@" = 3S
AT = @/ > 0S
B@ = 3S
B=T = @ = 3ST = @ = 6S
                                                                           
                                                                       
B= = = =
=T" + T" =
3" + 6" = 3 5S
ô¬­w
3 5_
õË 9-1OPS2  %S =  = 4Scos =
1
4S=b%AS¡  û/A=ĎüäSÙĀ
2Ù>SĂă>Dßw        
_
¬­ 6
uÝ“çè 0DĎüūâEFSžê !S:ę 0S¥@340êxyMzSƮHĉ?
@ABD;<ê]ĊċA.ƹ/ 0b.ò_“ތÙ]S ? dÙSSEFrŠ
= = = = => = SàīĬ340>=b¥@340Sr> = =Sà2  =%.=^q_
ó..
OPSŒÙ]S ? dÙSSŒÙ=]=T ? dÙTS
A =  = 4S
BS = SS
Acos =
S
 =
S
4 =
1
4S
BS = S = 1S
B = 2S
A=b%AS
B= = = =
1
2 = 2S
                                                                           
                                                                       
cĎü+= = => = 2S
B= = = = => = S
B	= = 	=S
	= = S
B	= =
R
/
2
S
B	= = R
/
R
/
2
= 90° +

2S
cĎü+	>= = 	= = 90° +

2S
B	>= = 	>=/	= = 90° +

2/
R
/
2
= S
B	>= = 	=S
B=> N S
B340>=b¥@340S
B> = =S
A=T ? S
Bcos = cos =
T
= =
T
2 =
1
4S
BT =
1
2S
BT = /T = /
1
2 =
3
2S=T =
="/T" =
2"/
1
2
"
=
15
2 S
B= =
=T" + T" =
15
2
"
+
3
2
"
=
6S
B> = = =
6S
ô¬­w
 6_
õË 9-2OPS2oZr%S	 = 90°S = 3S = 5S	n	D¥uAĥØdÙ=SŒ
Ù=]=>//ØdÙ>SĂă=>Dßw        
_
  
                                                                           
                                                                       
¬­
5
3
u݌٠D ] DF?BC dÙ FScgƺr	DBC=45°S	ACB=	DEBSÜjsin	 = sin	=> =
3
5Sij
älm !êABDnŊq._
ó..
ŒÙ D ] DF?BC dÙ FSOPQR
  
A	 = 90°S	n	D¥uAĥØdÙ=S
B	DBC=45°S	DCB=	ACDS
BDFB b:ę/ 0S^ DF=BFS
ADENACS
B	ACB=	DEBS	ACD=	CDES
B	CDE=	DCES
BDE=ECS
AAB=3SAC=5S
BBC=4Ssin	 = sin	=> =

 =
3
5S
 DF=BF=3xSÜj
EF=4xSDE=EC=5xS
B@ + @ + @ = 4S.r@ =
1
3S
B=> =
5
3
ô¬­w
5
3_
ÙĈ“åæçè./ 0Sëìíî !b.Dò_
õË 9-3OP_2ABC %S	ACB=60°SAC=1SD b4 AB D%ÙSE b4 BC ´hÙ_Û DE ¥uABC
DƻßSÜ DE Dßb     _
                                                                           
                                                                       
¬­
3
2 
ó.uÝOPSůß BC Ŵ MSē CM=CASÚ5 AMS] CN?AM d NSlmgr« ME=EBSl
m 0%»AMzr« DE=
1
2AMSlm:ę 0DopŠ	ACNSlmHIDƑƒŠ ANSEF^q_
ó.OPSůß BC Ŵ MSē CM=CASÚ5 AMS] CN?AM d NS
ADE ¥uABC DƻßS AD=DBS
BBE=CE+ACS
BME=EBS
ć AD=DBS
BDE=
1
2AMSDENAMS
A	ACB=60°S
B	ACM=120°S
ACM=CAS
B	ACN=60°SAN=MNS
BAN=AC•sin	ACN=
3
2 S
BAM= 3S
BDE=
3
2 S
ô¬­w
3
2 _
                                                                           
                                                                       
ÙĈ“çèé 0%»AMzJ:ę 0DopJ./ 0Síî 0%»AMzJH
ĉĩĪĊċAb.Dò_
【题型10  解直角三角形的应用】
Î 10{2025·Őŀ·%çŁ|wĪƼƽĵƾƿSǀĵǁǂ)ǃDŽDždžXĮǃLJLj²‰‘µËljġƽĵNJ
Nj_OPSASBSCSD 2}h¥¦ń_A bDždžnjSǀh¶ǍSǎġ«ǏĮǃLJ»d A DHõ½ 10 ǐ
ǑD B āSǒĮǃLJ»d A D¾ÂÄ30°µ½ 20 ǐǑD D ā_‰ĮǃLJ}¶Ǔǔ C āLj²SD »d C DH
ĵ½´SB »d C D¼ÂÄ30°µ½´_{çm
 2 t 1.41S 3 t 1.73S 5 t 2.24S 7 t 2.65|
(1)=DßÆ{áǕǖǗÁÙäh»|
(2)ǏJǒ‰ĮǃLJ}¶uvĆ BSD ŠŨûS=ǔ C ā?@Lj²SǒĮǃLJŧÆwǏĮǃLJŧÆD 2
ǘ_©‰ĮǃLJĥņ 20 ǐǑ¶SǙUqiǚǛĥ5ǜ«	_ż ǏĮǃLJǓŇ B āǝǞǐǑ¶S‰Į
ǃLJqiǚǛĥ5ǜ«	{áǕǖǗÁÙäh»|Ż
¬­(1)26.5ǐǑ
(2)3.8ǐǑ
uÝ“åæçèé./ 0DžŸG)Sú0DopXřMS Hĉ]ŠĊċA89/ 0
b.Dòǟ
{1|ŒÙ A ]> ? =d ESŒÙ B ]C ? =d FScgrS	=> = 30°S.Rt  =>r«> = 10 3
ǐǑS=> = 10ǐǑSīĬ340>Cbú0S r«>C =  = 10ǐǑSC = > = 10 3ǐǑSr«
                                                                           
                                                                       
=C = => + >C = 20ǐǑSàœ)xyMz^qŠ=Dߏ
{2|©ǏĮǃLJ(Œ« MSǒĮǃLJ(Œ« N ¶S̶ƔąST = 20ǐǑ_ŒÙ M ]S] ? =d TSc
grS	C = 
/
 = 60°S.Rt  Cr« = 20ǐǑSC = 10ǐǑSÜ= = =C + C = 30ǐǑS
S = @ǐǑSÜ=T = @ǐǑSS = ( 
/@)ǐǑS.Rt  S]r«] = 
/
1
2 @ ǐǑSS] =
10 3/
3
2 @ ǐǑSÜ]T = 
/
3
2 @ ǐǑScxyMzr20" = 10 3/
3
2 @
"
+ 
/
3
2 @
"
S.µ^q
r«¬­ǟ
ó.{1|.
OPQRSŒÙ A ]> ? =d ESŒÙ B ]C ? =d FS
B	>= = 	C = 90°S
cgrS	=> = 30°S
2Rt  =>%S> = = 0 cos	=> = 20 0 cos30° = 10 3ǐǑS
=> = = 0 sin	=> = 20 0 sin30° = 10ǐǑS
AĮǃLJ»d A DHõ½ 10 ǐǑD B āSD »d C DHĵ½´S
B N =S
B> ? SC ? S
B340>Cbú0S 
B>C =  = 10ǐǑSC = > = 10 3ǐǑS
B=C = => + >C = 20ǐǑS
B= =
=C" + C" =
20" + (10 3)
"
= 10 7 t 26.5ǐǑS
¬
=DßÆǠw26.5ǐǑ
{2|.
OPQRS©ǏĮǃLJ(Œ« MSǒĮǃLJ(Œ« N ¶S̶ƔąST = 20ǐǑ_ŒÙ M ]S] ? =
d TS
                                                                           
                                                                       
cgrS	C = 
/
 = 60°S
2Rt  C%S =
C
sin	C =
10 3
sin60° = 20ǐǑS
C =
C
tan	C =
10 3
tan60° = 10ǐǑS
B= = =C + C = 30ǐǑS
S = @ǐǑSÜ=T = @ǐǑSS = ( 
/@)ǐǑS
2Rt  S]%S] = S 0 cos	S] = ( 
/@) 0 cos60° = 
/
1
2 @ ǐǑS
S] = S 0 sin	S] = ( 
/@) 0 sin60° = 10 3/
3
2 @ ǐǑS
B]T = =/=T/] = 
/ @/ 
/
1
2 @ = 
/
3
2 @ ǐǑS
2Rt  ST]%ScxyMzrST" = S]" +T]"S
B20" = 10 3/
3
2 @
"
+ 
/
3
2 @
"
S
B@ = / 5C@ = 15 + 5 5{̶edDßSĐđ|S
BS = / 5 t 3.8ǐǑS
¬
ǏĮǃLJǓŇ B ā3.8ǐǑ¶S‰ĮǃLJqiǚǛĥ5ǜ«	_
õË 10-1{2025·ö¾·ùğĠġ|ǡÇǢƿLJǣǃbh‘qƇǤǥƾǦǧǃǨ?ƈǩǪǫǬ?@Ǣƿ]
ǭDħ‘LJǣǃ_OP 1 bhǮǡÇǢƿLJǣǃDžƠPSP 2 b‹DŽ]RgPSLJǣǃŮǯqƣ]ú0
=S = 1mSǰDZDz>C = CE = 3mSÙEnÙ>2}h¸¹A´{^E> ? =|S	C = 110°SǰDZDz
EbDãeßÆw9mSP%DÙĖ2}hdz/¥¦ńSEb N >C_©ǰDZDzEbǴ«ãeßƶSǡÇǢƿ
LJǣǃDãÇÙb«¦DņŇ_{çm
sin35° t 0.57Scos35° t 0.82Stan35° t 0.70|
                                                                           
                                                                       
¬­ǡÇǢƿǣLJǣǃDãÇÙb«¦DņŇǠw13.3m
uÝŒÙb]bS ? >ESØ>EDůßAdÙSSŒÙC]CT ? >EdÙTSlm:ę 0Dopn 
0ńnSqr	TEC = 	T>C = 35°SET = T>Sáâ¥@ADopn !qrE> = ET = EC 0 cos
35°SES = Eb 0 cos35°Sàlm>S = ES + E>^q._
ó..
O.PSŒÙb]bS ? >ESØ>EDůßAdÙSSŒÙC]CT ? >EdÙTS
A	>CE = 110°S>C = CES
B	TEC = 	T>C = 35°SET = T>S
BET = EC 0 cos35°S
BE> = ET = EC 0 cos35°S
AEb N >CS
B	CEb = 	>CE = 110°S
B	SEb = R
//
 = 35°S
BES = Eb 0 cos35°S
B>S = ES + E> = Eb 0 cos35° + EC 0 cos35° = (Eb + EC)cos35°S
t (9 + 2 × 3) × 0.82 = 12.3(m)S
A12.3 + 1 = 13.3{m|S
                                                                           
                                                                       
BǡÇǢƿǣLJǣǃDãÇÙb«¦DņŇǠw13.3m_
ÙĈ“åæçèé./ 0DG)S:ę 0Dopn 0ńnS¥@ADopSëì
íîi´+ñb.Dò_
õË 10-2{2025·´ǵßĝ·hù|OPbǀsǶǷDǸ¦PSǀūâžǹÁĦ¡ĮǃLJŎ2ÅǺŏÙ
A āSǻĮǃLJǓ«Ù=āS=XŮǼo¥@Sġr= = 11.6ǑS̶2Ù=āćġrÅǺ´DÙD
±w26.6°_5ŕǻĮǃLJǓ«Ù>āS=> ? =S>XŮǼo¥@Sġr=> = 1.8Ǒ_
 
(1)ÅǺDÅƏ
(2)*+¦[JsǶǷD÷ǼCEaXŮǼo¥@kǙU«ŮǼoDņŇĥ:SĮǃLJĆÙǓ«ÙOāS
O ? =SġrO = 16.4ǑS̶2ÙOāġrÙCD±w45°S2Śçǽ‹ǾƌˆDśǿsSjhȀÇÆ
w 3 ǑDȁǶƮȂ?ƈȃsǶǷŻżŽĬzc_
{çm
sin26.6° t 0.45Scos26.6° t 0.89Stan26.6° t 0.5|
¬­(1)2cG = 1:
40
9
(2)ȃȁǶƮ?ƈȃsǶǷSzc¯.Ý
uÝ“åæçèé./ 0DG)Jú0DřMXopJxyMz:+ñÙSHĉ]ŠĊċAJ
89/ 0b.Dò_
{1|cgr
	=> = 26.6°S	=> = 	> = 90°SOP
ŒÙ]v ? =dÙvSƺī340v=>bú
0SÜ> = v=Jv = => = 1.8wVijä2Rt  =>nRt  v%./ 0^q.¬
{2|cgr
	OC = 45°Sà2Rt  OC%./ 0qrC = 16.4wVSOP
ŒÙC]Cb ? d
ÙbSlmxyMzn./ 0qrCb" +b" = C"SCb = @SÜb = 
@SÜ(@)" + (
@)" =
16.4"S.r@ =
2
5S?ĄrCbSãäX 3 Ǒ!Ȅ^q.¬_
ó.{1|.
cgr
	=> = 26.6°S	=> = 	> = 90°
OP
ŒÙ]v ? =dÙvS
                                                                           
                                                                       
B	v= = 90°
B340v=>bú0_
B> = v=Sv = => = 1.8wV
2Rt  =>%S	> = 90°S=> = 1.8wV
Btan	=> =
=>
> = 0.5  .r
> = v= = 3.6wV_
Bv = 8m
2Rt  v%S	v = 90xS
Btan	= =
v
v =
1.8
8 =
9
40S
B2cG = 1:
40
9 _
{2|.
cgr
	OC = 45°
2Rt  OC%S	OC = 90°SO = 16.4wVStan	OC =
O
C = 1S 
BC = 16.4wVS
OP
ŒÙC]Cb ? dÙbS
2Rt  bC%S	bC = 90°Stan	Cb =
Cb
b =
9
40SCb" +b" = C"S
Cb = @SÜb = 
@S
B(@)" + (
@)" = 16.4"S.r@ =
2
5
BCb = 3.6wVS
Ay < 3.6m  ^ȃȁǶƮ?ƈȃsǶǷ_
õË 10-3wġďȅäøÇÆSȆȇÁĦ'ŒŒOs
                                                                           
                                                                       
  
(1)ġďÅ
OP 1SäøhȈj BĥW¥/DøÅSS=SøDÇÆ^w BŦD¸/ÇÆbS[S=oŽ
nSŦDßÆqi/5ġďr«SæøÅÇÆȉƛæ+ǺÅeÁ_
OP 2S}U¡‰l/ȊSTSSODhȋŎ2ŦŏǛȋ A āS/ȊSOûŦµ½ŎȌS2/ȊST
ȍhȋ N )ȎAÁŐƠ GS©/ȊSTX¸¹ATEŐâ¶Sġr‰ȊÌDÆScĚqrøÅ AB Å
zDÆ_ż/5ʊSzŽ~Dď_
(2)ġďøÇ
}UġrøÅSS=DÅßȏƍw 40 ǑS50 ǑS40 ǑSÅȏƍw24°S30°S45°wbSÁ
Ȑ}2]ǭ“´¤éh†ȑ24°DRt  ]^F{OP 3|Sďr^] t 5cmS]F t 2cm_øÇ
=C_{ 2 t 1.41SáǕȒĉ« 1 Ǒ|
(3)ġďȓ?
cdġďDŽ]ďȄeS}UȔƎOƀ§ġď?@éƫƈȕȖSjéisȗDġďµ_
  
OP 4S5S2ȅȘǁ´S¡/Ȋ NP ȌdSTD÷ȋS©STX¸¹ATEŐâ¶S;Œ/ȊTOSēÙ NS
PSD „ASġr	STODÆSĆĄr«ø÷°zaS½äøµ½ś? 40 ǑSǂ)ĥ}µËSġrø÷
°z"¤h†ȑzaD/ 0SďrȃW4nȍh/4uvw!așǑS$așǑSà¤h†ȑz"D
/ 0SďrȃW4nȍh/4uvw!"șǑS$"șǑ_*+ȊÇ MN w1.6ǑSøÇ=C_{á
Ǖ)ŚȑzaSz"DȚțtR|
¬­(1) + z = 90°
(2)øÇ=Cw 69 Ǒ
                                                                           
                                                                       
(3)øÇ=CDÇw
40!a!"
!"$a/!a$" + 1.6 Ǒ__
uÝ{1|œ)KDop^q.
{2|Þrsin24° = 0.4Sàuv2Rt  bJRt  [JRt  =o%S./ 0^q.
{3|Þrtanza =
!a
$aStanz" =
!"
$"S2Rt  T=gnRt  TK=g%SuvrTgnTKgDßSr«µTg/TK
g = 40SmĚ^q._
ó.{1|.
cgr	TS1 = 90°S
    
B + z = 90°
{2|.
2Rt  ]^F%S^] t 5cmS]F t 2cm_
  
Bsin24° =
]F
^] t
2
5 = 0.4S
2Rt  b%S	b = 24°S = 40ǑS
Bb =  0 sin24° = 40 × 0.4 = 16(Ǒ)S
2Rt  [%S	[ = 30°S = 50ǑS 
B[ =  0 sin30° = 50 ×
1
2 = 25(Ǒ)S
2Rt  =o%S	=o = 45°S= = 40ǑS
B=o = = 0 sin45° = 40 ×
2
2 t 28(Ǒ)S
BøÇ=C = 16 + 25 + 28 = 69(Ǒ)S
¬
øÇ=Cw 69 Ǒ
{3|.
OPScgrtanza =
!a
$aStanz" =
!"
$"S
                                                                           
                                                                       
  
øÇ=C = @ + 1.6SÜ=g = @S
  
2Rt  T=g%S	=Tg = zaS=g = @S
B
=g
Tg = tanza =
!a
$aS
BTg =
$a
!a@S
2Rt  TK=g%S	=TKg = z"S=g = @S
B
=g
TKg = tanz" =
!"
$"S
BTKg =
$"
!"@S
ATTK = SSK = 40S
BTg/TKg = 40S^
$a
!a@/
$"
!"@ = 40S
.r@ =
40!a!"
!"$a/!a$"SøÇ=C =
40!a!"
!"$a/!a$" +1.6
¬
øÇ=CDÇw
40!a!"
!"$a/!a$" + 1.6 Ǒ_
ÙĈ“çèé./ 0DG)SƮȜHĉ8ĔŠ/ 0S¡žŸ §ȝw./ 
0D b.¬Ě”Dò_