# 10.1 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] .

if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
give me treganamentry question
Solve 2cos x + 3sin x = 0.5
madras university algebra questions papers first year B. SC. maths
Hey
Rightspect
hi
chesky
Give me algebra questions
Rightspect
how to send you
Vandna
What does this mean
cos(x+iy)=cos alpha+isinalpha prove that: sin⁴x=sin²alpha
cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha
rajan
cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha
rajan
is there any case that you can have a polynomials with a degree of four?
victor
***sscc.edu/home/jdavidso/math/catalog/polynomials/fourth/fourth.html
Oliver
can you solve it step b step
give me some important question in tregnamentry
Anshuman
what is linear equation with one unknown 2x+5=3
-4
Joel
x=-4
Joel
x=-1
Joan
I was wrong. I didn't move all constants to the right of the equation.
Joel
x=-1
Cristian
y=x+1
gary
what is the VA Ha D R X int Y int of f(x) =x²+4x+4/x+2 f(x) =x³-1/x-1
can I get help with this?
Wayne
Are they two separate problems or are the two functions a system?
Wilson
Also, is the first x squared in "x+4x+4"
Wilson
x^2+4x+4?
Wilson
thank you
Wilson
Wilson
f(x)=x square-root 2 +2x+1 how to solve this value
Wilson
what is algebra
The product of two is 32. Find a function that represents the sum of their squares.
Paul
if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
hi
John
hi
Grace
what sup friend
John
not much For functions, there are two conditions for a function to be the inverse function:   1--- g(f(x)) = x for all x in the domain of f     2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the  element that you started with, namely, x.
Grace
sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
acha se dhek ke bata sin theta ke value
Ajay
sin theta ke ja gha sin square theta hoga
Ajay