文档文本内容
第二十五章 锐角的三角比·培优卷
120
120
___________
___________
___________
!"# 24 "$%& 10 "$'( 6 "$)* 8 "$ 120 $+ 120 $!",-./01$23
45$&"678$9:;<=>!?@ABCDEFG
HⅠ
IJ&K"L# 10 O"$ 30 $QO" 3 R
1JL3 RL24-25 STU·VWWX·YZ[\RRt △𝐴𝐵𝐶]$∠𝐶= 90°$∠𝐴:∠𝐵= 1:2$^tan𝐴B_
L R
AJ
1
2
BJ
3
2
CJ
3
3
DJ 3
2JL3 RL24-25 STU·`abc·YZ[\RdRt △𝐴𝐵𝐶]$∠𝐶= 90°$𝐴𝐶= 5$𝐴𝐵= 13$^sin𝐴
B_eL R
AJ
5
72
BJ
5
13
CJ
12
13
DJ
13
72
3JL3 RL2025·fg`hij·klmnRdRt △𝐴𝐵𝐶]$∠𝐶= 90°$op A Bqrs.rBtuv∠𝐴
Bwx$yzcot𝐴${|$ cot𝐴=
𝐴𝐶
𝐵𝐶$^cot30°B_eL}}R
AJ
1
2
BJ
3
2
CJ
3
3
DJ 3
4JL3 RL2025·VW~·kR{|$dRt △𝐴𝐵𝐶]$∠𝐴𝐶𝐵= 90°$ D e𝐴𝐵B]$𝐶𝐷$
𝐶𝐷= 3$cos𝐵=
2
3$^𝐴𝐶B
eL R
AJ2
BJ4
CJ 5
DJ2 5
5JL3 RL24-25 T·`c·]R{|I4|$𝐶𝐷s4𝐴𝐵$
𝐴𝐷Bte1:1$𝐵𝐶Bte1:2$𝐴𝐷
e4$^𝐶𝐵B
e(}})
AJ4 2
BJ2 10
CJ4 5
DJ4 10
6JL3 RL2025·`·kR{|$d 𝐴𝐵𝐶𝐷]$𝐴𝐶s𝐵𝐷¡¢£𝑂$𝐵𝑀⊥𝐶𝐷$¤¥e M$
𝐵𝑀¢𝐴𝐶£𝑁$𝑂𝐶= 4$𝑂𝐷= 2$^𝑂𝑁B
eL R
AJ
1
2
BJ1
CJ 2
DJ 3
7JL3 RL22-23 STU·`a¦·§R{|$d △𝐴𝐵𝐶]$∠𝐴= 30°$𝐴𝐶= 2 3$tan𝐵=
3
2 $^𝐴𝐵
B
eL R
AJ2 + 2 3
BJ3 +
3
CJ4
DJ5
8JL3 Rd △𝐴𝐵𝐶]$∠𝐴¨∠𝐵©op$ªsin𝐴=
3
2 $cos𝐵=
1
2$^ △𝐴𝐵𝐶L R.
AJ«¬p
BJ«rp
CJpp
DJ®pp
9JL3 R{|$d¯ ABCD ]$𝐴𝐷= 2 3$𝐴𝐵= 5$M CD UBI$° △𝐴𝐷𝑀±² AM .³
´µ △𝐴𝑁𝑀$ AN ∠𝑀𝐴𝐵$^ CN B
eL R
AJ
3 5
2
BJ 5
CJ 7
DJ3
10JL3 R{|$² y=
3
4x+3 ¢ x ¶£ A $°I·«¬pp ¸¹Bpº»£¼ O$½¾
¿º M¨N ÀÁd² y=
3
4x+3 U$ N dHÂ+@$^ tan∠AON B_eL}}R
AJ
1
7
BJ
1
6
CJ
1
5
DJ
1
8
J'("L# 6 O"$ 18 $QO" 3 R
11JL3 RL24-25 STU·`a¦·YZ[\R{|$d △𝐴𝐵𝐶]$∠𝐶= 90°$𝐴𝐶= 3$𝐴𝐵= 6$^
∠𝐴=
J
12JL3 RL2025·`a¦·IkRdRt △𝐴𝐵𝐶]$𝐴𝐶= 3$𝐶𝐷eÅr𝐴𝐵UB]²$𝐶𝐷= 2$^cos𝐴
B_e
J
13JL3 RL24-25 STU·`aÆÇ·YZ[\R{|$dp △𝐴𝐵𝐶]$∠𝐶= 90°$ D d²Z𝐴𝐶U$
ª∠𝐴= 30°$∠𝐵𝐷𝐶= 60°$𝐴𝐷= 2$^𝐵𝐶=
J
14JL3 RL2025·fg`hij·klmnRd △𝐴𝐵𝐶]$𝐷ep @ÈI$𝐴𝐶= 𝐵𝐷$𝐴𝐷⊥𝐶𝐷$
∠𝐵𝐷𝐴+ ∠𝐵𝐴𝐶= 180°$tan∠𝐶𝐴𝐷= 0.5$^tan∠𝐵𝐴𝐷=
J
15JL3 R{|$É]6¿OÊ A$IËÌÍÎWÏÐÑ$d B Òn´OÊ A Ó£ÔBÐÕÖÏ$×
ÌÍsOÊ¡Ø 20 ÉÙ$ÚÛÑÜ D Ò$n´OÊ A dÔBÕÝW 60°ÖÏ$×ÌÍsOÊBØ
Þ𝐴𝐷e
ÉÙJ
16JL3 R△ABC ]$ABß4$ACß5$△ABC B4àe 5 3$áâ∠A B8ã
J
HⅡ
J)*"L# 8 O"$ 72 R
17JL6 RL24-25 STU·æÐçè·YZ[\Réê
(1)2sin60° + 3tan30°
(2)cos245° + tan60°cos30°
18JL6 RL23-24 STU·ëcìí·YZ[\R{|$d4pîïð]$P ∠𝛼Br𝑂𝑄UBI$
ñò P Bóîïe 6$ªsin𝛼=
4
5J
(1)ô P Bõîïö
(2)cos𝛼B_e______J
19JL8 RL2025·`÷c·kR{|$d △𝐴𝐵𝐶]$𝐷𝐵𝐶rUI$ª
𝐴𝐵= 𝐴𝐷,𝐴𝐸⊥𝐵𝐶,𝐴𝐵= 13,𝐴𝐸= 12,𝐶𝐷= 11J
(1)ô𝐵𝐸B
J
(2)ôtan∠𝐴𝐶𝐸B_J
20JL8 RL24-25 STU·æÐøù·YZ[\Rú]ûüýþIÿĀāĂ㥹£ĆĆćĈ$Oĉ
ĊdĂăþ$ċ𝐶Òn´Ăăº𝐴BČpe53°$OčĊdûüýĎþBďYU$ċ𝐷Òn´Ăăº𝐴BČp
e45°$×$¾ĐBØÞ𝐸𝐶e0.45 m$ñòûüýĎþďYÅ𝐶𝐷Bte𝑖= 1:3JLđãĒ
sin53° ≈
4
5$cos53° ≈
3
5$tan53° ≈
4
3R
(1)éêďY𝐷𝐸B18ö
(2)ôĀāĂă𝐴𝐵B18J
21JL10 RL2025·ēĔĕ¼·klmnR|①¨|②¨|③Ė8 × 8BėÖ Ęę$Q¿OėÖ B
r
Ėe1$Q¿OėÖ BºĚeę$ △𝐴𝐵𝐶BºĖdęU$ěĜÆĝ8BĞ$ğdĠġB
Ęę]ĢģôĤ|J
(1)d|①]B𝐵𝐶rUĥI𝐸$Ħtan∠𝐶𝐴𝐸=
1
2ö
(2)d|②]Ĥ △𝐴𝐵𝐶Bp²𝐴𝐷ö
(3)d|③]B𝐴𝐵rUĤI𝑀$Ħ𝐴𝑀:𝐵𝑀= 4:3J
22JL10 RL2025·æWħĨ·kRĩĪīĬBĭĮĪ$įİıÙBIJijĴĵĶķĸTĹĺ$ĜI
ĻļĽľĿŀB<Łł?JdŃńŅņŇU$ŅņOňĦĜÆĐʼnn;ĩĪL| 1RúZBŊ8J{|
2$ŋŌdĪōIŎBŏŐďL¯ 𝐴𝐵𝐶𝐷RUőŒIÆĐʼn$œÆĐʼnŔµĪ4UÖ𝑃Ò$n´ŏŐď
ė.ō𝑄ÒBŕpe63.6°$n´ŏŐďºŖ𝐴ÒBŕpe47.3°$n´ŏŐďºŖ𝐷ÒBŕpe36.7°$ñòŏ
Őď1𝐴𝐵e 10 $ď4Ŋ𝐴𝐷e 16 L|]ŗ6ĖdŘI4@$𝑄,𝐵,𝐶#²R$ô×ĪZBŊ
𝑄𝐶JLřŚśŜIÓOã$đãĒ
tan47.3° ≈1.08$tan36.7° ≈0.75$tan63.6° ≈2.01R
23JL12 RL2022·ŝ·]Ş"RşŠšŢţ
d △𝐴𝐵𝐶]$∠𝐴¨∠𝐵¨∠𝐶ŗ.Brğe𝑎¨𝑏¨𝑐$ôŤ
𝑎
sin𝐴=
𝑏
sin𝐵J
Ťĺ
{| 1$𝐶z𝐶𝐷⊥𝐴𝐵£𝐷$^
dRtΔBCD]$ CD=asinB
dRtΔACD]$𝐶𝐷= 𝑏sin𝐴
∴𝑎sin𝐵= 𝑏sin𝐴
∴
𝑎
sin𝐴=
𝑏
sin𝐵
ťĒU4BŢţ)Ŧšŧ"
(1){| 2$dΔ𝐴𝐵𝐶]$∠𝐴¨∠𝐵¨∠𝐶ŗ.Brğe𝑎¨𝑏¨𝑐$ôŤ
𝑏
sin𝐵=
𝑐
sin𝐶ö
(2)eŨũŪŝūŬŭŮůŰűĬŲ$ųŴŵŶàŷŸŹŮůźŻJ{| 3$żŽ]BIžp ſƀƁƂ
Ź$ñò∠𝐴= 67°$∠𝐵= 53°$𝐴𝐶= 80$ôƃžſƀB4àJLřŚśŜť JđãĒ
sin
53° ≈0.8$sin67° ≈0.9)
24JL12 RL2025·ĪƄc·IkRL1RòƅƆƇƈŌò$pp ]66¿ƉƊ−−¿p$
Ƌr$ÎñòƉƊôƌŗ6ƍòƉƊBƎu)pp $š¿ƋƏ]$ƐƑ)pp B
________J
①ñò¾Ƌrö②ñòIƋrƒI¿opö③ñò¾¿pJ
L2RƓðƔűƕűƖƗ$Ƙp ]66¿ƉƊ−−¿p$Ƌr$ÎñòƉƊôƌŗ6ƍòƉƊ
BƎu)p Jpƙãp rpðBƚƛ$Ɯ9įze)p BbĜƝCJ{| 1$ñò
△𝐴𝐵𝐶]$∠𝐴= 30°$∠𝐵= 45°$𝐴𝐵= 5 + 5 3$)ƃ¿p ö
L3RƞƟƠĜ{| 2$ △𝐴𝐵𝐶]$𝐴𝐶= 2 3$cos𝐴=
3
2 $𝐵𝐶= 𝑚$d)ƃ¿p $ƍòƉƊ
©6¾)B𝑚Bơ_Ƣƣ________J