|概念|定义|图形|
|----|----|----|
|仰角、俯角|在视线与水平线所成的锐角中，视线在水平线上方的角叫做仰角，视线在水平线下方的角叫做俯角|![img id=9]|
|坡度(坡比)、坡角|坡面的垂直高度h与水平宽度l的比叫做坡度（坡比），用字母i表示；坡面与水平面的夹角α叫做坡角，i = tan α = $\frac{h}{l}$|![img id=10]|
|方向角|一般指以观测者的位置为中心，将正北或正南方向作为起始方向旋转到目标方向线所成的角（一般指锐角），通常表示成北(南)偏东(西)多少度，方向角的角度在0°～90°之间. 如图，点A，B，C关于点O的方向角分别是北偏东30°，南偏东60°，北偏西45°(也称西北方向)||

典例1(2023·宿迁)如图,在网格中,每个小正方形的边长均为1,A,B,C三点都在格点(网格线的交点)上,则sin∠ABC = ________.

如图，连接AC. 由勾股定理，得$AB^{2}=2^{2}+4^{2}=20$，$BC^{2}=1^{2}+3^{2}=10$，$AC^{2}=1^{2}+3^{2}=10$，$\therefore BC^{2}+AC^{2}=AB^{2}$.
$\therefore \triangle ACB$是直角三角形，且$\angle ACB = 90^{\circ}$. $\therefore \sin \angle ABC = \frac{AC}{AB}=\frac{\sqrt{10}}{2\sqrt{5}}=\frac{\sqrt{2}}{2}$.

典例2(2024·扬州高邮模拟)在△ABC中，若cosA = $\frac{\sqrt{3}}{2}$，tanB = 1，则∠C = ________.

$\because \cos A=\frac{\sqrt{3}}{2}$，$\tan B = 1$，$\therefore \angle A = 30^{\circ}$，$\angle B = 45^{\circ}$.
$\therefore \angle C=180^{\circ}-\angle A-\angle B=180^{\circ}-30^{\circ}-45^{\circ}=105^{\circ}$.

典例3(2023·连云港)如图，矩形OABC的顶点A在函数y = $\frac{k}{x}$(x ＜ 0)的图像上，顶点B,C在第一象限，对角线AC//x轴，交y轴于点D.若矩形OABC的面积是6，cos∠OAC = $\frac{2}{3}$，则k = ________.

过点A作$AE\perp x$轴于点E. $\because$矩形OABC的面积是6，
$\therefore$易得$\triangle AOC$的面积是3. 由题意，得$\angle AOC = 90^{\circ}$，$\cos \angle OAC=\frac{2}{3}$，$\therefore \frac{OA}{AC}=\frac{2}{3}$. $\because AC// x$轴，$\therefore \angle AOE = \angle CAO$. $\because \angle OEA=\angle AOC = 90^{\circ}$，$\therefore \triangle OEA\sim \triangle AOC$.
$\therefore \frac{S_{\triangle OEA}}{S_{\triangle AOC}}=(\frac{OA}{AC})^{2}$，即$\frac{S_{\triangle OEA}}{3}=\frac{4}{9}$. $\therefore S_{\triangle OEA}=\frac{4}{3}$. $\because S_{\triangle OEA}=\frac{1}{2}|k|$，$k＜0$，$\therefore k = -\frac{8}{3}$.

典例4(2024·苏州)如图,在△ABC中,AB = 4$\sqrt{2}$,D为AB的中点，∠BAC = ∠BCD，cos∠ADC = $\frac{\sqrt{2}}{4}$,⊙O是△ACD的外接圆.求:
(1)BC的长;
(2)⊙O的半径.

(1) $\because \angle BAC=\angle BCD$，$\angle B=\angle B$，$\therefore \triangle BAC\sim \triangle BCD$. $\therefore \frac{BC}{BD}=\frac{BA}{BC}$. $\because AB = 4\sqrt{2}$，D为AB的中点，$\therefore BD = AD = 2\sqrt{2}$. $\therefore BC^{2}=BA\cdot BD = 4\sqrt{2}\times 2\sqrt{2}=16$. $\therefore BC = 4$.
(2) 如图，过点A作$AE\perp CD$于点E，连接CO并延长，交$\odot O$于点F，连接AF. $\because$在$Rt\triangle AED$中，$\cos \angle ADC=\frac{DE}{AD}=\frac{\sqrt{2}}{4}$，$AD = 2\sqrt{2}$，$\therefore DE = 1$. $\therefore AE=\sqrt{AD^{2}-DE^{2}}=\sqrt{(2\sqrt{2})^{2}-1^{2}}=\sqrt{7}$. 由(1)知，$\triangle BAC\sim \triangle BCD$，$\therefore \frac{AC}{CD}=\frac{AB}{CB}=\frac{4\sqrt{2}}{4}=\sqrt{2}$. 设$CD = x$，则$AC=\sqrt{2}x$，$CE = x - 1$. $\because$在$Rt\triangle ACE$中，$AC^{2}=CE^{2}+AE^{2}$，$\therefore (\sqrt{2}x)^{2}=(x - 1)^{2}+(\sqrt{7})^{2}$，即$x^{2}+2x - 8 = 0$，解得$x = 2$或$x = - 4$(舍去). $\therefore CD = 2$，$AC = 2\sqrt{2}$.
$\because \angle AFC$与$\angle ADC$都是$\overset{\frown}{AC}$所对的圆周角，$\therefore \angle AFC = \angle ADC$. $\because CF$为$\odot O$的直径，$\therefore \angle CAF = 90^{\circ}$. $\therefore \sin \angle AFC=\frac{AC}{CF}=\sin \angle ADC=\frac{AE}{AD}=\frac{\sqrt{7}}{2\sqrt{2}}$，即$\frac{2\sqrt{2}}{CF}=\frac{\sqrt{7}}{2\sqrt{2}}$. $\therefore CF=\frac{8\sqrt{7}}{7}$.
$\therefore \frac{1}{2}CF=\frac{4\sqrt{7}}{7}$，即$\odot O$的半径为$\frac{4\sqrt{7}}{7}$.