# 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] .

can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
Where do the rays point?
Spiro
x=-b+_Гb2-(4ac) ______________ 2a
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
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
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 By By By Madison Christian By Edgar Delgado By OpenStax By Brooke Delaney By Marion Cabalfin By OpenStax By OpenStax By JavaChamp Team By OpenStax By Anh Dao