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

 Page 2 / 10

## 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 the sum of $\text{\hspace{0.17em}}{\alpha }^{\prime }\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\beta }^{\prime },$ 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] .

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.
The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
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?
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
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
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions