Course: 13.16 使用特别保安系统(以寻找角度)使用 " 共信法 " | The Way To Learn

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Select section 使用有SSS的科辛斯定律(以寻找角度)

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使用有SSS的科辛斯定律(以寻找角度)
Sarine draws a triangle. She measures the length of the sides and records her measurements as follows. What is the measure of angle C of the triangle?
::沙林绘制三角形。她测量两侧的长度, 记录她的测量结果如下。三角形角 C 的度量是多少 ?

$\begin{aligned} a = 3 \\ b = 4 \\ c = 5 \end{aligned}$
$\begin{align*}a=3\\ b=4\\ c=5\end{align*}$
::a=3b=4c=5

Law of Cosines with SSS
::具有SSSS的Casines定律

The , $\begin{align*}a^2+b^2-2ab \cos C\end{align*}$ , can be rearranged to facilitate the calculation of the measure of angle $\begin{align*}C\end{align*}$ when $\begin{align*}a, b\end{align*}$ and $\begin{align*}c\end{align*}$ are all known lengths.
::a2+b2-2abcosC可重新排列,以便在a、b和c均为已知长度时,便于计算角C的度量。

$\begin{aligned} a^{2} + b^{2} - 2 a b \cos C & = c^{2} \\ a^{2} + b^{2} - c^{2} & = 2 a b \cos C \\ \frac{a^{2} + b^{2} - c^{2}}{2 a b} & = \cos C \end{aligned}$
$\begin{align*}a^2+b^2-2ab \cos C &=c^2 \\ a^2+b^2-c^2 &=2ab \cos C \\ \frac{a^2+b^2-c^2}{2ab} &=\cos C\end{align*}$
::a2+b2-2abcos*C=c2a2+b2-c2=2abcos*Ca2+b2-c22ab=cos*C

which can be further manipulated to $\begin{align*}C=\cos^{-1} \left(\frac{a^2+b^2-c^2}{2ab} \right)\end{align*}$ .
::C=cos-1(a2+b2-c22ab)可被进一步操纵到 C=cos-1(a2+b2-c22ab)。

Let's find the measure of the largest angle in the triangle with side lengths 12, 18 and 21.
::让我们找出三角形中最大角的大小角度, 边长12、18和21。

First, we must determine which angle will be the largest. Recall from Geometry that the longest side is opposite the largest angle. The longest side is 21 so we will let $\begin{align*}c = 21\end{align*}$ since $\begin{align*}C\end{align*}$ is the angle we are trying to find. Let $\begin{align*}a =12\end{align*}$ and $\begin{align*}b = 18\end{align*}$ and use the formula to solve for $\begin{align*}C\end{align*}$ as shown. It doesn’t matter which sides we assign to $\begin{align*}a\end{align*}$ and $\begin{align*}b\end{align*}$ . They are interchangeable in the formula.
::首先,我们必须确定哪个角度最大。从几何中回顾, 最长的一边是最大角度对面的。最长的一面是 21 , 因此我们会让 C= 21 , 因为 C 是我们试图找到的角。请用 a= 12 和 b= 18 来解析 C 的公式。我们给 a 和 b 指定哪边并不重要。它们可以在公式中互换。

$\begin{aligned} m ∠ C = \cos^{- 1} (\frac{12^{2} + 18^{2} - 21^{2}}{2 (12) (18)}) \approx 86^{\circ} \end{aligned}$
$\begin{align*}m \angle C=\cos^{-1} \left(\frac{12^2+18^2-21^2}{2(12)(18)} \right) \approx 86^\circ\end{align*}$
::mC=cos-1(122+182-2122(12)(18))86

Note: Be careful to put parenthesis around the entire numerator and entire denominator on the calculator to ensure the proper order of operations. Your calculator screen should look like this:
::注意: 注意将整个分子和整个分母的括号放在计算器上,以确保操作的正确顺序。您的计算器屏幕应该这样看 :

$\begin{aligned} \cos^{- 1} ((12^{2} + 18^{2} - 21^{2}) / (2 (12) (18))) \end{aligned}$
$\begin{align*}\cos^{-1}((12^2+18^2-21^2)/(2(12)(18)))\end{align*}$
:(122+182-212)/(2(12)(18)))

Now let's find the value of $\begin{align*}x\end{align*}$ , to the nearest degree .
::现在让我们找到 x 的值, 至最近的度。

The angle with measure $\begin{align*}x^\circ\end{align*}$ will be angle $\begin{align*}C\end{align*}$ so $\begin{align*}c = 16, a = 22\end{align*}$ and $\begin{align*}b = 8\end{align*}$ . Remember, $\begin{align*}a\end{align*}$ and $\begin{align*}b\end{align*}$ are interchangeable in the formula. Now we can replace the variables with the known measures and solve.
::使用度量 x 的角度将是角度 C , c= 16, a=22 和 b=8. 记住, a 和 b 在公式中可以互换。现在我们可以用已知的度量和解析替换变量。

$\begin{aligned} \cos^{- 1} (\frac{22^{2} + 8^{2} - 16^{2}}{2 (22) (8)}) \approx 34^{\circ} \end{aligned}$
$\begin{align*}\cos^{-1} \left(\frac{22^2+8^2-16^2}{2(22)(8)} \right) \approx 34^\circ\end{align*}$
:222+82 - 1622(22)(8))_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Finally, let's find the $\begin{align*}m \angle A\end{align*}$ , if $\begin{align*}a = 10, b = 15\end{align*}$ and $\begin{align*}c = 21\end{align*}$ .
::最后,让我们找到MA,如果a=10,b=15,c=21。

First, let’s rearrange the formula to reflect the sides given and requested angle:
::首先,让我们重新排列公式,以反映给定和要求的角度:

$\begin{align*}\cos A=\left(\frac{b^2+c^2-a^2}{2(b)(c)} \right)\end{align*}$ , now plug in our values $\begin{align*}m \angle A=\cos^{-1} \left(\frac{15^2+21^2-10^2}{2(15)(21)} \right) \approx 26^\circ\end{align*}$
:b2+c2-a22(b(c))),现在插入我们的值 mA=cos-1(152+212-1022(15)(21)26)

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the measure of angle C of the triangle that has sides a = 3, b = 4, and c = 5.
::早些时候,有人要求你找到三角形角C的测量尺度,三角形的角C的边角a = 3,b = 4,c = 5。

We can use the manipulated Law of Cosines to solve for C.
::我们可以用被操纵的科辛斯定律解决C

$\begin{aligned} C = \cos^{- 1} \frac{3^{2} + 4^{2} - 5^{2}}{2 (3) (4)} \\ C = \cos^{- 1} \frac{9 + 16 - 25}{24} \\ C = \cos^{- 1} \frac{0}{24} = \cos^{- 1} 0 \\ C = 90^{\circ} \end{aligned}$
$\begin{align*}C=\cos^{-1} \frac{3^2+4^2-5^2}{2(3)(4)} \\ C=\cos^{-1} \frac{9 + 16 - 25}{24}\\ C=\cos^{-1} \frac{0}{24} = \cos^{-1} 0\\ C = 90^\circ\end{align*}$
::C=cos-132+42-522(3)(4)C=cos-19+16-22524C=cos-1024=cos-10C=90

Therefore, the triangle is a right triangle.
::因此,三角形是一个右三角形。

Example 2
::例2

Find the measure of $\begin{align*}x\end{align*}$ in the diagram:
::在图表中查找 x 的度量 :

$\begin{align*}\cos^{-1} \left(\frac{14^2+8^2-19^2}{2(14)(8)} \right) \approx 117^\circ\end{align*}$
::-1(142+82-1922(14)(8))117(142+82-1922(14)

Example 3
::例3

Find the measure of the smallest angle in the triangle with side lengths 47, 54 and 72.
::查找三角形中最小角的度量,边长47、54和72。

The smallest angle will be opposite the side with length 47, so this will be our $\begin{align*}c\end{align*}$ in the equation.
::最小的角是侧面的47长的对面, 所以这是我们的方程中的C。

$\begin{aligned} \cos^{- 1} (\frac{54^{2} + 72^{2} - 47^{2}}{2 (54) (72)}) \approx 41^{\circ} \end{aligned}$
$\begin{align*}\cos^{-1} \left(\frac{54^2+72^2-47^2}{2(54)(72)} \right) \approx 41^\circ\end{align*}$
:542+722-4722(54)(72)41)

Example 4
::例4

Find $\begin{align*}m \angle B\end{align*}$ , if $\begin{align*}a = 68, b = 56\end{align*}$ and $\begin{align*}c = 25\end{align*}$ .
::找到 mB,如果a=68,b=56和c=25。

Rearrange the formula to solve for $\begin{align*}m \angle B, \cos B=\left(\frac{a^2+c^2-b^2}{2(a)(c)} \right); \cos^{-1} \left(\frac{68^2+25^2-56^2}{2(68)(25)} \right) \approx 52^\circ\end{align*}$
::mB,cosB=(a2+c2-b22(a)(c));cos-1(682+252-5622(68)(25))52=(a2+c2-b22(a)(c));cos-1(682+252-5622(68)(25))

Review
::回顾

Use the Law of Cosines to find the value of $\begin{align*}x\end{align*}$ , to the nearest degree, in problems 1 through 6.
::使用《科辛斯定律》在问题1至6中找到x的值,最接近于x的值。

Find the measure of the smallest angle in the triangle with side lengths 150, 165 and 200 meters.
::在三角形中找到最小角度的测量值,边长150、165和200米。

Find the measure of the largest angle in the triangle with side length 59, 83 and 100 yards.
::查找三角形中最大角的测量值,边长59、83和100码。

Find the $\begin{align*}m \angle C\end{align*}$ if $\begin{align*}a = 6, b = 9\end{align*}$ and $\begin{align*}c=13\end{align*}$ .
::如果a=6,b=9和c=13,请查找 mC。

Find the $\begin{align*}m \angle B\end{align*}$ if $\begin{align*}a = 15, b = 8\end{align*}$ and $\begin{align*}c = 9\end{align*}$ .
::如果a=15,b=8和c=9,请查找 mB。

Find the $\begin{align*}m \angle A\end{align*}$ if $\begin{align*}a = 24, b = 20\end{align*}$ and $\begin{align*}c = 14\end{align*}$ .
::如果 a=24,b=20 和 c=14, 则查找 mA 。

A triangular plot of land is bordered by a road, a fence and a creek. If the stretch along the road is 100 meters, the length of the fence is 115 meters and the side along the creek is 90 meters, at what angle do the fence and road meet?
::一条三角地带的陆地与一条道路、一道栅栏和一条小溪相邻,如果沿这条道路的长度为100米,那么围栏的长度为115米,沿这条小溪的一侧为90米,围栏和道路的交汇点是什么?

Review (Answers)
::回顾(答复)

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。

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