# 8.1 Non-right triangles: law of sines  (Page 4/10)

 Page 4 / 10

Thus,

$\text{Area}=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)=\frac{1}{2}b\left(c\mathrm{sin}\text{\hspace{0.17em}}\alpha \right)$

Similarly,

$\text{Area}=\frac{1}{2}a\left(b\mathrm{sin}\text{\hspace{0.17em}}\gamma \right)=\frac{1}{2}a\left(c\mathrm{sin}\text{\hspace{0.17em}}\beta \right)$

## Area of an oblique triangle

The formula for the area of an oblique triangle is given by

$\begin{array}{l}\text{Area}=\frac{1}{2}bc\mathrm{sin}\text{\hspace{0.17em}}\alpha \hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}ac\mathrm{sin}\text{\hspace{0.17em}}\beta \hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}ab\mathrm{sin}\text{\hspace{0.17em}}\gamma \hfill \end{array}$

This is equivalent to one-half of the product of two sides and the sine of their included angle.

## Finding the area of an oblique triangle

Find the area of a triangle with sides $\text{\hspace{0.17em}}a=90,b=52,\text{\hspace{0.17em}}$ and angle $\text{\hspace{0.17em}}\gamma =102°.\text{\hspace{0.17em}}$ Round the area to the nearest integer.

Using the formula, we have

$\begin{array}{l}\text{Area}=\frac{1}{2}ab\mathrm{sin}\text{\hspace{0.17em}}\gamma \hfill \\ \text{Area}=\frac{1}{2}\left(90\right)\left(52\right)\mathrm{sin}\left(102°\right)\hfill \\ \text{Area}\approx 2289\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{square}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{units}\hfill \end{array}$

Find the area of the triangle given $\text{\hspace{0.17em}}\beta =42°,\text{\hspace{0.17em}}\text{\hspace{0.17em}}a=7.2\text{\hspace{0.17em}}\text{ft},\text{\hspace{0.17em}}\text{\hspace{0.17em}}c=3.4\text{\hspace{0.17em}}\text{ft}.\text{\hspace{0.17em}}$ Round the area to the nearest tenth.

about $\text{\hspace{0.17em}}8.2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{square}\text{\hspace{0.17em}}\text{feet}$

## Solving applied problems using the law of sines

The more we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion.

## Finding an altitude

Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in [link] . Round the altitude to the nearest tenth of a mile.

To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side $\text{\hspace{0.17em}}a,$ and then use right triangle relationships to find the height of the aircraft, $\text{\hspace{0.17em}}h.$

Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship.

The distance from one station to the aircraft is about 14.98 miles.

Now that we know $\text{\hspace{0.17em}}a,\text{\hspace{0.17em}}$ we can use right triangle relationships to solve for $\text{\hspace{0.17em}}h.$

The aircraft is at an altitude of approximately 3.9 miles.

The diagram shown in [link] represents the height of a blimp flying over a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point $\text{\hspace{0.17em}}B,\text{\hspace{0.17em}}$ is 62°, and the distance between the viewing points of the two end zones is 145 yards.

161.9 yd.

Access these online resources for additional instruction and practice with trigonometric applications.

## Key equations

 Law of Sines $\begin{array}{l}\frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{a}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\beta }{b}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\gamma }{c}\text{\hspace{0.17em}}\hfill \\ \frac{a}{\mathrm{sin}\text{\hspace{0.17em}}\alpha }=\frac{b}{\mathrm{sin}\text{\hspace{0.17em}}\beta }=\frac{c}{\mathrm{sin}\text{\hspace{0.17em}}\gamma }\hfill \end{array}$ Area for oblique triangles

## Key concepts

• The Law of Sines can be used to solve oblique triangles, which are non-right triangles.
• According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.
• There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. See [link] .
• The ambiguous case arises when an oblique triangle can have different outcomes.
• There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. See [link] and [link] .
• The Law of Sines can be used to solve triangles with given criteria. See [link] .
• The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. See [link] .
• There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. See [link] .

#### Questions & Answers

what is a function?
CJ Reply
I want to learn about the law of exponent
Quera Reply
explain this
Hinderson Reply
what is functions?
Angel Reply
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
If the plane intersects the cone (either above or below) horizontally, what figure will be created?
Feemark Reply
can you not take the square root of a negative number
Sharon Reply
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
Elaine Reply
can I get some pretty basic questions
Ama Reply
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
Amara Reply
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
Mars Reply
what is the domain of f(x)=x-4/x^2-2x-15 then
Conney Reply
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
jeric Reply
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
jeric Reply
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
Abena Reply
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
Ashley Reply
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

### Read also:

#### Get the best Precalculus course in your pocket!

Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

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

 By By By Rohini Ajay By Brooke Delaney By JavaChamp Team By OpenStax By By Brooke Delaney By Danielrosenberger By Jazzycazz Jackson By OpenStax By Mucho Mizinduko