4.6 锡奈法

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  Sines法律

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  When given a right triangle, you can use basic trigonometry to solve for missing information. When given SSS or SAS, you can use the to solve for the missing information. But what happens when you are given two sides of a triangle and an angle that is not included? There are many ways to show two triangles are congruent, but SSA is not one of them. Why not? 
  ::当给定一个右三角时, 您可以使用基本的三角测量来解析缺失的信息。 当给定 SSS 或 SAS 时, 您可以用此来解析缺失的信息 。 但是, 如果给定三角的两侧和一个不包含角的角, 那么会发生什么 ? 有许多方法可以显示两个三角是相容的, 但是 SSA 不是其中之一 。 为什么不呢 ?

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  The Law of Sines
  ::Sines法

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  When given two sides and an angle that is not included between the two sides, you can use the Law of Sines. The Law of Sines states that in every triangle the ratio of each side to the sine of its corresponding angle is always the same. Essentially, it clarifies the general concept that opposite the largest angle is always the longest side. 
  ::当给定两边和两边之间没有包括的角度时, 您可以使用“ 辛尼定律 ” 。 《 辛尼定律 》 规定, 在每一个三角形中, 每一边与相应角的正弦之比总是一样。 基本上, 它澄清了对齐最大角的对面总是最长的一面这个一般概念。

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  $\begin{align*}\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\end{align*}$
  ::asinA=bsinB=csinC A=bsinB=csinC

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  Here is a proof of the Law of Sines:
  ::以下是《辛尼定律》的证明:

  

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  Looking at the right triangle formed on the left:
  ::查看左侧形成的右三角形 :

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  $$\begin{align*}\sin A & = \frac{h}{b}\\ h & = b \sin A\end{align*}$$
  ::A=hbh=bsin=A=sin=A=sin=sin=A=sin=A=sin=A=a=hbh=bsin=bsin=A=sin=A=a=a=hbh=bsin=bsin=A=a=a=bsin=a=a=a=a=bsin=a=bsin=a=a=a=a=bsin=a=bsin=A=a=a=a=a=bhhhh=bhh=bsin=bsin_A=A=a=a=a=a=a=a=a=A=a=a=a=bsin=A=A=a=a=a=a=a=a=a=a=a=a=a=a=a=a=a=a=a=a=a=bs=a=a=a=a=a=a=a=a=a=a=a=a=a=a

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  Looking at the right triangle formed on the right:
  ::查看右侧的三角形 :

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  $$\begin{align*}\sin B & = \frac{h}{a}\\ h & = a \sin B\end{align*}$$
  ::B=Hah=Asin=B=B=B=Asin=B=B=Asin=B=B=B=B=B=B=Asin=B=B=B=B=B=B=B=Asin=B=B=B=B=B=B=B=Asin=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B=B

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  Equating the heights which must be identical:
  ::平衡必须相同的高度 :

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  $$\begin{align*}a \sin B & = b \sin A\\ \frac{a}{\sin A} & = \frac{b}{\sin B}\end{align*}$$
  ::asin_B=bsin_Aasin_A=bsin_B =bsin_B asin_A=bsin_B =bsin_B =bsin_B =bsin_B =bsin_B =bsin_B asin_Aasin_A=A=bsin_B

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  The best way to use the Law of Sines is to draw an extremely consistent picture each and every time even if that means redrawing and relabeling a picture. The reason why the consistency is important is because sometimes given SSA information defines zero, one or even two possible triangles. 
  ::使用Sines法则的最佳方法是每次绘制非常一致的图片,即使这意味着重新绘制和重新标注图片。 一致性之所以重要,是因为有时给SSA信息定义了零,一个甚至两个可能的三角。

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  Always draw the given angle in the bottom left with the two given sides above. 
  ::总是在左下方与上方两侧绘制给定角度 。

  

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  In this image side  $\begin{align*}a\end{align*}$ is deliberately too short, but in most problems you will not know this. You will need to compare $\begin{align*}a\end{align*}$ to the height.
  ::在这个图像侧面,一个刻意的太短,但在大多数问题中,你不会知道这一点。你需要比较一个和高度。

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  $$\begin{align*}\sin A & = \frac{h}{c}\\ h & = c \ \sin A\end{align*}$$
  ::一九九九年一月一日 一月一日 一月一日 一月一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日一日

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  This is commonly referred to testing the ambiguous case. There are four different tests to determine the number of triangles that exist given the measurements.
  ::这通常是指测试模糊性案例,根据测量结果,有四种不同的测试来确定存在三角形的数量。

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  Case 1: $\begin{align*}a < h\end{align*}$
  ::案例1:a<h

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  Simply put, side $\begin{align*}a\end{align*}$  is not long enough to reach the opposite side and constructing the triangle is impossible. Zero triangles exist.
  ::简言之,A侧距离不够长,无法到达对面,不可能建造三角形。存在零三角形。

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  Case 2: $\begin{align*}a = h\end{align*}$
  ::案例2:a=h

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  Side $\begin{align*}a\end{align*}$  just barely reaches the opposite side forming a $\begin{align*}90^\circ\end{align*}$  angle. 
  ::侧面只勉强到达对面的一面 形成90角角

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  Case 3: $\begin{align*}h < a < c\end{align*}$
  ::判例3: h<a<c

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  In this case side  $\begin{align*}a\end{align*}$ can swing toward the interior of the triangle or the exterior of the triangle- there are two possible triangles. This is called the ambiguous case because the given information does not uniquely identify one triangle. To solve for both triangles, use the Law of Sines to solve for angle $\begin{align*}C_1\end{align*}$  first and then use the supplement to determine $\begin{align*}C_2\end{align*}$ .
  ::在此情况下, 一个可以向三角形内部或三角形外部摇摆, 可能有两个可能的三角形。 因为给定的信息并不独有指定一个三角形, 所以这被称为模糊的立体 。 要解决这两个三角形, 请先使用 Sines 法来解决角 C1 或三角形外部, 然后使用 补充 来确定 C2 。

  

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  Case 4:  $\begin{align*}c \le a\end{align*}$
  ::判例4:ca

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  In this case, side $\begin{align*}a\end{align*}$  can only swing towards the exterior of the triangle, only producing $\begin{align*}C_1\end{align*}$ .
  ::在此情况下, 一方只能向三角形外侧摇摆, 只生产C1。

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  For the case of SSA, you should always check how many triangles there are before starting to find measures. Take the following triangle: 
  ::对于特别服务协定,在开始找到措施之前,您应该总是先检查有多少三角形。请使用以下三角形:

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  $\begin{align*}\angle A = 40^\circ, c = 13\end{align*}$ , and $\begin{align*}a = 2\end{align*}$ . 
  ::A=40,c=13,和a=2。

  

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  Before trying find  $\begin{align*}\angle C\end{align*}$ , you need to check that a triangle is possible and if there is more than one solution. Use the equation from above,
  ::在尝试找到 QC 之前, 您需要检查三角形是否可行, 如果存在多个解决方案, 则检查是否可行 。 使用上面的方程,

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  $$\begin{array}{rl}\sin {40}^{\circ }& =\frac{h}{13}\\ h& =13\sin {40}^{\circ }\approx 8.356\end{array}$$

  $$\begin{align*}\sin 40^\circ & = \frac{h}{13}\\ h & = 13 \sin 40^\circ \approx 8.356\end{align*}$$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}什么? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}什么? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}什么? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}什么?

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  Because $\begin{align*}a < h \ (2 < 8.356)\end{align*}$ , this information does not form a proper triangle.
  ::因为 <h (2<8.356) , 此信息不构成合适的三角形 。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked why SSA is not a method to show that two triangles are congruent. SSA is not a method from Geometry that shows two triangles are congruent because it does not always define a unique triangle. Sometimes there is no triangle, one triangle, or two triangles.
  ::早些时候,有人问为什么SSA不是显示两个三角一致的方法。 SSA不是几何中显示两个三角一致的方法,因为它并不总是定义一个独特的三角。 有时没有三角、一个三角或两个三角。

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  Example 2
  ::例2

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  $\begin{align*}\angle A = 17^\circ, c = 14\end{align*}$ , and $\begin{align*}a = 4.0932 \ldots\end{align*}$   If possible, find $\begin{align*}\angle C\end{align*}$ .
  ::A=17,c=14,和a=4.0932...

  

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  Check that a triangle is possible: 

  $$\begin{align*}\sin 17^\circ & = \frac{h}{14}\\ h & = 14 \ \sin 17^\circ \approx 4.0932\end{align*}$$
  ::检查三角形是否可行: sin @ 17 @ h14h=14 sin @ 17 @ 4. 0932

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  Since $\begin{align*}a = h\end{align*}$ , this information forms exactly one triangle and angle  $\begin{align*}C\end{align*}$ must be $\begin{align*}90^\circ\end{align*}$ .
  ::由于 a=h, 此信息完全代表一个三角形, 角度 C 必须是 90 。

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  Example 3
  ::例3

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  $\begin{align*}\angle A = 22^\circ, c = 11\end{align*}$  and $\begin{align*}a = 9\end{align*}$ . If possible, find $\begin{align*}\angle C\end{align*}$ .
  ::A=22,c=11和a=9,如果可能,请找到C。

  

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  Check that a triangle is possible:
  ::检查三角形是否可行 :

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  $$\begin{array}{rl}\sin {22}^{\circ }& =\frac{h}{11}\\ h& =11\sin {22}^{\circ }\approx 4.12\end{array}$$

  $$\begin{align*}\sin 22^\circ & = \frac{h}{11}\\ h & = 11 \ \sin 22^\circ \approx 4.12\end{align*}$$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}为什么? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}为什么? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}为什么?

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  Since $\begin{align*}h < a < c\end{align*}$ , there must be two possible angles for angle $\begin{align*}C\end{align*}$ . 
  ::由于 h<a<c, 角C必须有两个可能的角 。

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  Apply the Law of Sines:
  ::适用Sines法则:

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  $$\begin{align*}\frac{9}{\sin 22^\circ} & = \frac{11}{\sin C_1}\\ 9 \sin C_1 & = 11 \sin 22^\circ\\ \sin C_1 & = \frac{11 \sin 22^\circ}{9}\\ C_1 & = \sin^{-1} \left( \frac{11 \sin 22^\circ}{9} \right) \approx 27.24 ^\circ\\ C_2 & = 180 - C_1 \approx 152.75^\circ\end{align*}$$
  ::9sin2211sinC19sinC1=11sin2229C1=11sin229C1=1sin_1{(11sin229)\27.24C2=180-C1152.75

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  Example 4
  ::例4

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  Given  $\begin{align*}\Delta ABC\end{align*}$ where $\begin{align*}A = 12^\circ, B = 50^\circ, a = 14\end{align*}$  find $\begin{align*}b\end{align*}$ .
  ::根据A=12的ABC,B=50,a=14 找到b。

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  $\begin{align*}\frac{14}{\sin 12^\circ} = \frac{b}{\sin 50^\circ}\end{align*}$
  ::14sin12bsin5014sin12bsin5014sin12bsin}14sin12bsin12bsin50

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  $\begin{align*}b = \frac{14 \sin 50^\circ}{\sin 12^\circ} \approx 51.58\end{align*}$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}为什么? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}为什么? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

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  Example 5
  ::例5

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  Given  $\begin{align*}\Delta ABC\end{align*}$ where $\begin{align*}A = 70^\circ, b = 8, a = 3\end{align*}$ , find  $\begin{align*}\angle B\end{align*}$ if possible. 
  ::根据ABC的A=70,b=8,a=3,如果可能,请找到B。

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  $\begin{align*}\sin 70^\circ = \frac{h}{8}\end{align*}$
  ::70h8的罪 70\\\\\\\\\\\\\\\\h8\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  $\begin{align*}h = 8 \sin 70^\circ \approx 7.51 \ldots\end{align*}$
  ::h=8sin707.51...

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  Because $\begin{align*}a < h\end{align*}$ , this triangle is impossible. 
  ::因为这个三角形是不可能的

  Summary

  播放段落 The Law of Sines states that in every triangle, the ratio of each side to the sine of its corresponding angle is always the same. $$\begin{align*}\frac{a}{\sin A} = \frac{b}{\sin B} =\frac{c}{\sin C}\end{align*}$$ ::Sines法则指出,在每一个三角形中,每一边与相应角的正弦之比总是相同的。asin_A=bsin_B=csin_C

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  Review
  ::回顾

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  For 1-3, draw a picture of the triangle and state how many triangles could be formed with the given values.
  ::对于 1-3, 绘制三角形的图片, 并显示以给定值可以组成多少三角形 。

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  1.  $\begin{align*}A = 30^\circ, a = 13, b = 15\end{align*}$
  ::1. A=30,a=13,b=15

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  2.  $\begin{align*}A = 22^\circ, a = 21, b = 12\end{align*}$
  ::2. A=22,a=21,b=12

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  3.  $\begin{align*}A = 42^\circ, a = 36, b = 37\end{align*}$
  ::3. A=42a=36,b=37

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  For 4-7, find all possible measures of $\begin{align*}\angle B\end{align*}$  (if any exist) for each of the following triangle values.
  ::对于 4-7, 找到以下三角值中每个值的所有可能的 +B (如果存在的话) 。

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  4. $\begin{align*}A = 86^\circ, a =15, b = 11\end{align*}$
  ::4. A=86,a=15,b=11

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  5.  $\begin{align*}A = 30^\circ, a = 24, b = 43\end{align*}$
  ::5. A=30,a=24,b=43

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  6.  $\begin{align*}A = 48^\circ, a = 34, b = 39\end{align*}$
  ::6. A=48,a=34,b=39

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  7.  $\begin{align*}A = 80^\circ, a = 22, b = 20\end{align*}$
  ::7. A=80,a=22,b=20

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  For 8-12, find the length of  $\begin{align*}b\end{align*}$ for each of the following triangle values.
  ::对于 8 - 12, 查找以下三角值中的每个值的 b 长度 。

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  8. $\begin{align*}A = 94^\circ, a = 31, B = 34^\circ\end{align*}$
  ::8. A=94,a=31,B=34

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  9.  $\begin{align*}A = 112^\circ, a = 12, B = 15^\circ\end{align*}$
  ::9. A=112,a=12,B=15

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  10.  $\begin{align*}A = 78^\circ, a = 20, B = 16^\circ\end{align*}$
  ::10. A=78,a=20,B=16

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  11. $\begin{align*}A = 54^\circ, a = 15, B = 112^\circ\end{align*}$
  ::11. A=54a=15,B=112

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  12. $\begin{align*}A = 39^\circ, a = 9, B = 98^\circ\end{align*}$
  ::12. A=39a=9,B=98

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  13. In  $\begin{align*}\Delta ABC, b=10\end{align*}$ and $\begin{align*}\angle A = 39^\circ\end{align*}$ . What's a possible value for $\begin{align*}a\end{align*}$  that would produce two triangles?
  ::13. 在 QABC,b=10 和 QA=39\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\A=39\\\\\\\\\\\\\\\\\\\\\\\\\\\A\\\\\\\\\\\\\\\A\\\\\\\\\\\\\\\\\\\\A\\\\\\\\\\\\\A\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以可以可以可以可以可以可以

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  14. In  $\begin{align*}\Delta ABC, b=10\end{align*}$ and $\begin{align*}\angle A = 39^\circ\end{align*}$ . What's a possible value for $\begin{align*}a\end{align*}$  that would produce no triangles?
  ::14. 在 QABC,b=10 和 QA=39\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\A=39\\\\\\\\\\\\\\\\\\\\\\\\\\\A\\\\\\\\\\\\\\\A\\\\\\\\\\\\\\\\\\\\\A\\\\\\\\\\\\A\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以可以可以可以可以可以可以可以可以可以可以可以

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  15. In  $\begin{align*}\Delta ABC, b=10\end{align*}$ and $\begin{align*}\angle A = 39^\circ\end{align*}$ . What's a possible value for $\begin{align*}a\end{align*}$  that would produce one triangle?
  ::15. 在 QABC,b=10 和 QA=39\\\\\\\\\\\\\\"A=39\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"ABC\\\\\\\\"ABCB=10\\\\"A=39\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"A=39\\\\\\\\\\\\\\\\\\\\\\\\\\"A\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  Review (Answers)
  ::回顾(答复)

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  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。