23.4.1解直角三角形及应用 课件 -2026-2027学年华东师大版数学九年级上册

#华东师大版九年级上册\(\boldsymbol{2}\)3.4《解直角三角形及应用》专项练习##一、知识点梳理###1.什么是解直角三角形在$\boldsymbol{Rt\triangleABC\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)\angleC=90^\circ}$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)已知**\(\boldsymbol{2}\)个元素（至少一条边）**\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)求出其余所有未知边、角。共有5个元素：3条边$\(a,b,c\)$；\(\boldsymbol{2}\)个锐角$\angleA,\angleB$。用到4类工具：1.两锐角互余：$\boldsymbol{\angleA+\angleB=90^\circ}$\(\boldsymbol{2}\).勾股定理：$\boldsymbol{a^\(\boldsymbol{2}\)+b^\(\boldsymbol{2}\)=c^\(\boldsymbol{2}\)}$3.三角函数：$\sinA=\dfrac{a}{c},\cosA=\dfrac{b}{c},\\(\boldsymbol{\tan}\)A=\dfrac{a}{b}$4.特殊角三角函数（30°、\(\boldsymbol{45}\)°、60°）>原则：**尽量用已知条件\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)少用中间求出的数\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)减少计算误差**###\(\boldsymbol{2}\).实际应用名词（必考）-**仰角**：视线向上\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)水平线与视线夹角-**俯角**：视线向下\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)水平线与视线夹角-**坡度（坡比）$i$**：坡面的垂直高度$h$和水平宽度$l$之比$$\boldsymbol{\(i=\)\dfrac{h}{l}=\\(\boldsymbol{\tan}\)\alpha}$$$\alpha$叫坡角-**方位角**：北偏东、北偏西、南偏东、南偏西###解题通用步骤1.画图\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)抽象出直角三角形；\(\boldsymbol{2}\).标出已知边、已知角；3.选择合适公式求未知量；4.作答。⚠易错提醒1.坡度$i$是$\boldsymbol{\\(\boldsymbol{\tan}\)}$坡角\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)**不是斜边比**；\(\boldsymbol{2}\).仰角俯角都是**和水平线**的夹角\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)不是和竖直线；3.不是直角三角形\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)要**作高分割成两个直角三角形**（双直角模型\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)大题高频）。##二、选择题1.在$Rt\triangleABC\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)\angleC=90^\circ$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)已知$a,\angleA$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)求斜边$c$用（）A.$c=a\cdot\sinA$B.$c=\dfrac{a}{\sinA}$C.$c=a\cdot\cosA$D.$c=\dfrac{a}{\cosA}$\(\boldsymbol{2}\).坡比$\(i=\)1:\sqrt{3}$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)则坡角为（）A.$30^\circ$B.$\(\boldsymbol{45}\)^\circ$C.$60^\circ$D.$90^\circ$3.从楼上看地面的点\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)视线与水平线夹角叫（）A.仰角B.俯角C.坡角D.方位角4.\(Rt\triangleABC\)\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)$\angleC=90^\circ$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)$BC=\(\boldsymbol{2}\)\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)\angleA=30^\circ$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)则$\(AB=\)$（）A.$\(\boldsymbol{2}\)$B.$4$C.$\(\boldsymbol{2}\)\sqrt{3}$D.$\sqrt{3}$5.某斜坡坡角$\alpha$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)水平宽$\sqrt{3}$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)高1\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)则坡比$\(i=\)$（）A.$1:\sqrt{3}$B.$\sqrt{3}:1$C.$\(1:\(\boldsymbol{2}\)\)$D.$\(\(\boldsymbol{2}\):1\)$##三、填空题1.解直角三角形\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)除直角外\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)至少要知道____个条件\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)且至少有一个是____。\(\boldsymbol{2}\).仰角和俯角\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)都是视线与____之间的夹角。3.坡比$\(i=\)\dfrac{\text{垂直高度}}{\text{________}}=\\(\boldsymbol{\tan}\)\alpha$。4.$Rt\triangleABC,\angleC=90^\circ$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)$\(a=3,b=3\)$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)则$\angleA=\boldsymbol{\_\_\_^\circ}$。5.观测目标\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)仰角$\(\boldsymbol{2}\)5^\circ$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)水平线以上\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)那么俯角看回来也是____°。##四、解答题1.在$Rt\triangleABC$中\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)$\angleC=90^\circ$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)$\(a=4\)\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)\angleA=30^\circ$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)解这个直角三角形。\(\boldsymbol{2}\).如图\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)小明在地面$A$点测旗杆顶端$C$仰角$30^\circ$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)$A$离旗杆底部$B$水平距离$\(AB=\)1\(\boldsymbol{2}\)\mathrm{m}$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)求旗杆$BC$高度。（保留根号）3.一段斜坡\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)坡比$\(i=\)1:\sqrt{3}$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)斜坡长$\(\boldsymbol{2}\)0\mathrm{m}$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)求斜坡的垂直高度。4.拓展（双直角\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)经典大题）：大楼$AB$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)地面一点$C$测楼顶$A$仰角$\(\boldsymbol{45}\)^\circ$；向前走\(\(\boldsymbol{2}\)0\mathrm{m}\)到$D$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)测楼顶仰角$60^\circ$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)求大楼高度$AB$。（保留根号）##五、参考答案与解析###选择题1.B解析：$\sinA=\dfrac{a}{c}\Rightarrowc=\dfrac{a}{\sinA}$\(\boldsymbol{2}\).A解析：$\\(\boldsymbol{\tan}\)\alpha=\dfrac1{\sqrt{3}}=\dfrac{\sqrt{3}}{3},\alpha=30^\circ$3.B4.B解析：$\angleA=30^\circ$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)对边$BC=\dfrac1\(\boldsymbol{2}\)AB\Rightarrow\(AB=\)4$5.A###填空题1.$\boldsymbol{\(\boldsymbol{2}\)}$；**边**\(\boldsymbol{2}\).水平线3.水平宽度4.$\boldsymbol{\(\boldsymbol{45}\)}$5.$\boldsymbol{\(\boldsymbol{2}\)5}$###解答题1.解：$\angleC=90^\circ,\angleA=30^\circ$$\angleB=90^\circ-30^\circ=60^\circ$$\(a=4\)$是$\angleA$对边\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)$\(AB=\)c=\(\boldsymbol{2}\)a=8$$b=\sqrt{c^\(\boldsymbol{2}\)-a^\(\boldsymbol{2}\)}=\sqrt{8^\(\boldsymbol{2}\)-4^\(\boldsymbol{2}\)}=4\sqrt{3}$$\therefore\angleB=60^\circ\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)b=4\sqrt{3}\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)c=8$\(\boldsymbol{2}\).解：$Rt\triangleABC\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)\angleB=90^\circ\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)\angleA=30^\circ\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)\(AB=\)1\(\boldsymbol{2}\)$$\\(\boldsymbol{\tan}\)30^\circ=\dfrac{BC}{AB}$$BC=AB\cdot\\(\boldsymbol{\tan}\)30^\circ=1\(\boldsymbol{2}\)\times\dfrac{\sqrt{3}}{3}=4\sqrt{3}$答：旗杆高$4\sqrt{3}\mathrm{m}$。3.解：设垂直高$h$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)水平宽$l$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)$\(i=\)h:l=1:\sqrt{3}$则$l=\sqrt{3}h$由勾股：$h^\(\boldsymbol{2}\)+(\sqrt{3}h)^\(\boldsymbol{2}\)=\(\boldsymbol{2}\)0^\(\boldsymbol{2}\)$$h^\(\boldsymbol{2}\)+3h^\(\boldsymbol{2}\)=400,4h^\(\boldsymbol{2}\)=400,h^\(\boldsymbol{2}\)=100$$\(h=10\)$（长度正数）答：垂直高度10m。4.解：设$\(AB=\)x$$Rt\triangleABC,\angleC=\(\boldsymbol{45}\)^\circ$\(\boldsymbol{Rt\triangleABC\(Rt\triangleABC\(BC=\(\boldsymbol{2}\)\(\(a=4\)，\angleA=30^\circ\)\angleA=30^\circ\)\angleC=90^\circ\)\angleC=90^\circ}\)$\thereforeBC=\(AB=\)x$$BD=BC-CD=x-\(\boldsymbol{2}\)0$$Rt\triangleABD,\angleADB=60^\circ$$\\(\boldsymbol{\tan}\)60^\circ=\dfrac{AB}{BD}$$\sqrt{3}=\dfrac{x}{x-\(\boldsymbol{2}\)0}$$\sqrt{3}(x-\(\boldsymbol{2}\)0)=x$$\sqrt{3}x-\(\boldsymbol{2}\)0\sqrt{3}=x$$x(\sqrt{3}-1)=\(\boldsymbol{2}\)0\sqrt{3}$$x=\dfrac{\(\boldsymbol{2}\)0\sqrt{3}}{\sqrt{3}-1}=\dfrac{\(\boldsymb