# 8.1 Non-right triangles: law of sines  (Page 2/10)

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## Solving for two unknown sides and angle of an aas triangle

Solve the triangle shown in [link] to the nearest tenth.

The three angles must add up to 180 degrees. From this, we can determine that

$\begin{array}{l}\begin{array}{l}\hfill \\ \beta =180°-50°-30°\hfill \end{array}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=100°\hfill \end{array}$

To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle $\alpha =50°$ and its corresponding side $a=10.\text{\hspace{0.17em}}$ We can use the following proportion from the Law of Sines to find the length of $\text{\hspace{0.17em}}c.\text{\hspace{0.17em}}$

Similarly, to solve for $\text{\hspace{0.17em}}b,\text{\hspace{0.17em}}$ we set up another proportion.

Therefore, the complete set of angles and sides is

$\begin{array}{l}\begin{array}{l}\hfill \\ \alpha =50°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a=10\hfill \end{array}\hfill \\ \beta =100°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}b\approx 12.9\hfill \\ \gamma =30°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c\approx 6.5\hfill \end{array}$

Solve the triangle shown in [link] to the nearest tenth.

$\begin{array}{l}\alpha ={98}^{\circ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a=34.6\\ \beta ={39}^{\circ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=22\\ \gamma ={43}^{\circ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=23.8\end{array}$

## Using the law of sines to solve ssa triangles

We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case    . Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution.

## Possible outcomes for ssa triangles

Oblique triangles in the category SSA may have four different outcomes. [link] illustrates the solutions with the known sides $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ and known angle $\text{\hspace{0.17em}}\alpha .$

## Solving an oblique ssa triangle

Solve the triangle in [link] for the missing side and find the missing angle measures to the nearest tenth.

Use the Law of Sines to find angle $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ and angle $\text{\hspace{0.17em}}\gamma ,\text{\hspace{0.17em}}$ and then side $\text{\hspace{0.17em}}c.\text{\hspace{0.17em}}$ Solving for $\text{\hspace{0.17em}}\beta ,\text{\hspace{0.17em}}$ we have the proportion

$\begin{array}{r}\hfill \frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{a}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\beta }{b}\\ \hfill \frac{\mathrm{sin}\left(35°\right)}{6}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\beta }{8}\\ \hfill \frac{8\mathrm{sin}\left(35°\right)}{6}=\mathrm{sin}\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}\\ \hfill 0.7648\approx \mathrm{sin}\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}\\ \hfill {\mathrm{sin}}^{-1}\left(0.7648\right)\approx 49.9°\\ \hfill \beta \approx 49.9°\end{array}$

However, in the diagram, angle $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of $\text{\hspace{0.17em}}\beta ?\text{\hspace{0.17em}}$ Let’s investigate further. Dropping a perpendicular from $\text{\hspace{0.17em}}\gamma \text{\hspace{0.17em}}$ and viewing the triangle from a right angle perspective, we have [link] . It appears that there may be a second triangle that will fit the given criteria.

The angle supplementary to $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ is approximately equal to 49.9°, which means that $\text{\hspace{0.17em}}\beta =180°-49.9°=130.1°.\text{\hspace{0.17em}}$ (Remember that the sine function is positive in both the first and second quadrants.) Solving for $\text{\hspace{0.17em}}\gamma ,$ we have

$\gamma =180°-35°-130.1°\approx 14.9°$

We can then use these measurements to solve the other triangle. Since $\text{\hspace{0.17em}}{\gamma }^{\prime }\text{\hspace{0.17em}}$ is supplementary to $\text{\hspace{0.17em}}\gamma ,$ we have

${\gamma }^{\prime }=180°-35°-49.9°\approx 95.1°$

Now we need to find $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{c}^{\prime }.$

We have

Finally,

To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in [link] .

However, we were looking for the values for the triangle with an obtuse angle $\text{\hspace{0.17em}}\beta .\text{\hspace{0.17em}}$ We can see them in the first triangle (a) in [link] .

I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
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Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
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ok
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f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
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pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
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Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
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how can are find the domain and range of a relations
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Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
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...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert