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miss Akunina [59]
2 years ago
11

Find the approximate values of the trigonometric functions of θ given the following information. Enter the values correct to 2 d

ecimal places. θ is in standard position the terminal side of θ is in quadrant III the terminal side is parallel to the line 2y - 5x + 16 = 0
sin θ =
cos θ =
tan θ =
cot θ =
sec θ =
csc θ =

Mathematics
1 answer:
kati45 [8]2 years ago
7 0

Answer:

Step-by-step explanation:

slope of any line is same as the tan θ . so we first try to find the slope of the given line and then using that we can find remaining trigonometric functions .

To find the slope of a line we need to change the equation of line to slope intercept form .

2y - 5x +16 =0

move all terms to right

2y = 5x - 16

divide all by 2

y = 5/2 x - 8

compare this with y =mx+b

slope = m = 5/2

It means

tan θ =  5/2 = 2.5

tan θ =  2.50

now use the trigonometric ratios (see the image attached )

sin θ = \frac{y}{z} = \frac{5}{\sqrt{29} }  = 0.93

cos θ =  \frac{x}{z} = \frac{2}{\sqrt{29} }  = 0.37

tan θ =  2.50

cot θ =  \frac{x}{y} = \frac{2}{5 }  = 0.40

sec θ =  \frac{z}{y} = \frac{\sqrt{29}{5} }  = 1.08

csc θ =  \frac{z}{y} = \frac{\sqrt{29}{2} }  = 2.69

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Vikki [24]

Answer:

if you mean find OS, then:

OS = 42

Step-by-step explanation:

if you mean find OS, then:

8x-51 = 3x-6

5x = 45

x = 9

OS = 2(3x-6)

OS = 6x-12

substitute for x

OS = 6(9)-12 =42

8 0
2 years ago
Read 2 more answers
What is 5.316 - 1.942 btw (show ur work) :)
erastovalidia [21]

Answer:

3.374

Step-by-step explanation:

\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}

\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}

\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}

\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4\\

\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}

\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}\\

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{7}&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&11&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&\linethrough{1}&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{3}&7&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}

=3.374

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1 year ago
Determine if the segment lengths form a triangle. If so, would the triangle be acute, right, or obtuse? 4.1, 8.2, 12.2
goldfiish [28.3K]
Three line segments can form a triangle if the length of the longest segment is greater than the sum of the lengths of the shorter segments.

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12.3 \ \textgreater \  12.2 \\
true
They can form a triangle.

Now if c is the length of the longest side and a and b are the lengths of the shorter sides, then:
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- if c^2\ \textless \ a^2+b^2, the triangle is acute
- if c^2\ \textgreater \ a^2+b^2, the triangle is obtuse

4.1^2+8.2^2=16.81+67.24=84.05 \\
12.2^2=148.84 \\ \\
148.84\ \textgreater \ 84.05
The triangle is obtuse.

The answer is D.
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