# 10.1 Non-right triangles: law of sines  (Page 3/10)

 Page 3 / 10

Given $\text{\hspace{0.17em}}\alpha =80°,a=120,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b=121,\text{\hspace{0.17em}}$ find the missing side and angles. If there is more than one possible solution, show both.

Solution 1

$\begin{array}{ll}\alpha =80°\hfill & a=120\hfill \\ \beta \approx 83.2°\hfill & b=121\hfill \\ \gamma \approx 16.8°\hfill & c\approx 35.2\hfill \end{array}$

Solution 2

$\begin{array}{l}{\alpha }^{\prime }=80°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{a}^{\prime }=120\hfill \\ {\beta }^{\prime }\approx 96.8°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{b}^{\prime }=121\hfill \\ {\gamma }^{\prime }\approx 3.2°\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{c}^{\prime }\approx 6.8\hfill \end{array}$

## Solving for the unknown sides and angles of a ssa triangle

In the triangle shown in [link] , solve for the unknown side and angles. Round your answers to the nearest tenth.

In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle $\text{\hspace{0.17em}}\gamma =85°,\text{\hspace{0.17em}}$ and its corresponding side $\text{\hspace{0.17em}}c=12,\text{\hspace{0.17em}}$ and we know side $\text{\hspace{0.17em}}b=9.\text{\hspace{0.17em}}$ We will use this proportion to solve for $\text{\hspace{0.17em}}\beta .$

To find $\text{\hspace{0.17em}}\beta ,\text{\hspace{0.17em}}$ apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for $\text{\hspace{0.17em}}\beta .\text{\hspace{0.17em}}$ It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions.

$\begin{array}{l}\beta ={\mathrm{sin}}^{-1}\left(\frac{9\mathrm{sin}\left(85°\right)}{12}\right)\hfill \\ \beta \approx {\mathrm{sin}}^{-1}\left(0.7471\right)\hfill \\ \beta \approx 48.3°\hfill \end{array}$

In this case, if we subtract $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ from 180°, we find that there may be a second possible solution. Thus, $\text{\hspace{0.17em}}\beta =180°-48.3°\approx 131.7°.\text{\hspace{0.17em}}$ To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives

$\alpha =180°-85°-131.7°\approx -36.7°,$

which is impossible, and so $\text{\hspace{0.17em}}\beta \approx 48.3°.$

To find the remaining missing values, we calculate $\text{\hspace{0.17em}}\alpha =180°-85°-48.3°\approx 46.7°.\text{\hspace{0.17em}}$ Now, only side $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is needed. Use the Law of Sines to solve for $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ by one of the proportions.

The complete set of solutions for the given triangle is

Given $\text{\hspace{0.17em}}\alpha =80°,a=100,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=10,\text{\hspace{0.17em}}$ find the missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth.

$\beta \approx 5.7°,\gamma \approx 94.3°,c\approx 101.3$

## Finding the triangles that meet the given criteria

Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10.

Using the given information, we can solve for the angle opposite the side of length 10. See [link] .

$\begin{array}{l}\text{\hspace{0.17em}}\frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{10}=\frac{\mathrm{sin}\left(50°\right)}{4}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{10\mathrm{sin}\left(50°\right)}{4}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha \approx 1.915\hfill \end{array}$

We can stop here without finding the value of $\text{\hspace{0.17em}}\alpha .\text{\hspace{0.17em}}$ Because the range of the sine function is $\text{\hspace{0.17em}}\left[-1,1\right],\text{\hspace{0.17em}}$ it is impossible for the sine value to be 1.915. In fact, inputting $\text{\hspace{0.17em}}{\mathrm{sin}}^{-1}\left(1.915\right)\text{\hspace{0.17em}}$ in a graphing calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the provided dimensions.

Determine the number of triangles possible given $\text{\hspace{0.17em}}a=31,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=26,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\beta =48°.\text{\hspace{0.17em}}\text{\hspace{0.17em}}$

two

## Finding the area of an oblique triangle using the sine function

Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as $\text{\hspace{0.17em}}\text{Area}=\frac{1}{2}bh,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is base and $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ is height. For oblique triangles, we must find $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ before we can use the area formula. Observing the two triangles in [link] , one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{\text{opposite}}{\text{hypotenuse}}\text{\hspace{0.17em}}$ to write an equation for area in oblique triangles. In the acute triangle, we have $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{h}{c}\text{\hspace{0.17em}}$ or $c\mathrm{sin}\text{\hspace{0.17em}}\alpha =h.\text{\hspace{0.17em}}$ However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ to form a right triangle. The angle used in calculation is $\text{\hspace{0.17em}}{\alpha }^{\prime },\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}180-\alpha .$

#### Questions & Answers

A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
Kaitlyn Reply
The sequence is {1,-1,1-1.....} has
amit Reply
circular region of radious
Kainat Reply
how can we solve this problem
Joel Reply
Sin(A+B) = sinBcosA+cosBsinA
Eseka Reply
Prove it
Eseka
Please prove it
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
Arleathia Reply
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
Prince Reply
I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
SANDESH Reply
write down the polynomial function with root 1/3,2,-3 with solution
Gift Reply
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
Pream Reply
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Oroke Reply
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
kiruba Reply
what is the answer to dividing negative index
Morosi Reply
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
Shivam Reply
give me the waec 2019 questions
Aaron Reply

### Read also:

#### Get the best Algebra and trigonometry course in your pocket!

Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Algebra and trigonometry' conversation and receive update notifications?

 By By By Lakeima Roberts By Anindyo Mukhopadhyay By