10.2 Non-right triangles: law of cosines  (Page 4/8)

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Heron’s formula

Heron’s formula finds the area of oblique triangles in which sides $\text{\hspace{0.17em}}a,b\text{,}$ and $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ are known.

$\text{Area}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

where $\text{\hspace{0.17em}}s=\frac{\left(a+b+c\right)}{2}\text{\hspace{0.17em}}$ is one half of the perimeter of the triangle, sometimes called the semi-perimeter.

Using heron’s formula to find the area of a given triangle

Find the area of the triangle in [link] using Heron’s formula.

First, we calculate $\text{\hspace{0.17em}}s.$

$\begin{array}{l}\begin{array}{l}\\ s=\frac{\left(a+b+c\right)}{2}\end{array}\hfill \\ s=\frac{\left(10+15+7\right)}{2}=16\hfill \end{array}$

Then we apply the formula.

$\begin{array}{l}\begin{array}{l}\\ \text{Area}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\end{array}\hfill \\ \text{Area}=\sqrt{16\left(16-10\right)\left(16-15\right)\left(16-7\right)}\hfill \\ \text{Area}\approx 29.4\hfill \end{array}$

The area is approximately 29.4 square units.

Use Heron’s formula to find the area of a triangle with sides of lengths $\text{\hspace{0.17em}}a=29.7\text{\hspace{0.17em}}\text{ft},b=42.3\text{\hspace{0.17em}}\text{ft},\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}c=38.4\text{\hspace{0.17em}}\text{ft}.$

Area = 552 square feet

Applying heron’s formula to a real-world problem

A Chicago city developer wants to construct a building consisting of artist’s lofts on a triangular lot bordered by Rush Street, Wabash Avenue, and Pearson Street. The frontage along Rush Street is approximately 62.4 meters, along Wabash Avenue it is approximately 43.5 meters, and along Pearson Street it is approximately 34.1 meters. How many square meters are available to the developer? See [link] for a view of the city property.

Find the measurement for $\text{\hspace{0.17em}}s,\text{\hspace{0.17em}}$ which is one-half of the perimeter.

$\begin{array}{l}s=\frac{\left(62.4+43.5+34.1\right)}{2}\hfill \\ s=70\text{\hspace{0.17em}}\text{m}\hfill \end{array}$

Apply Heron’s formula.

$\begin{array}{l}\text{Area}=\sqrt{70\left(70-62.4\right)\left(70-43.5\right)\left(70-34.1\right)}\hfill \\ \text{Area}=\sqrt{506,118.2}\hfill \\ \text{Area}\approx 711.4\hfill \end{array}$

The developer has about 711.4 square meters.

Find the area of a triangle given $\text{\hspace{0.17em}}a=4.38\text{\hspace{0.17em}}\text{ft}\text{\hspace{0.17em}},b=3.79\text{\hspace{0.17em}}\text{ft,}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}c=5.22\text{\hspace{0.17em}}\text{ft}\text{.}$

Access these online resources for additional instruction and practice with the Law of Cosines.

Key equations

 Law of Cosines $\begin{array}{l}{a}^{2}={b}^{2}+{c}^{2}-2bc\mathrm{cos}\text{\hspace{0.17em}}\alpha \hfill \\ {b}^{2}={a}^{2}+{c}^{2}-2ac\mathrm{cos}\text{\hspace{0.17em}}\beta \hfill \\ {c}^{2}={a}^{2}+{b}^{2}-2abcos\text{\hspace{0.17em}}\gamma \hfill \end{array}$ Heron’s formula

Key concepts

• The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles.
• The Generalized Pythagorean Theorem is the Law of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into two right triangles or forms one right triangle, which allows sides to be related and measurements to be calculated. See [link] and [link] .
• The Law of Cosines is useful for many types of applied problems. The first step in solving such problems is generally to draw a sketch of the problem presented. If the information given fits one of the three models (the three equations), then apply the Law of Cosines to find a solution. See [link] and [link] .
• Heron’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula. See [link] and See [link] .

Verbal

If you are looking for a missing side of a triangle, what do you need to know when using the Law of Cosines?

two sides and the angle opposite the missing side.

If you are looking for a missing angle of a triangle, what do you need to know when using the Law of Cosines?

Explain what $\text{\hspace{0.17em}}s\text{\hspace{0.17em}}$ represents in Heron’s formula.

$\text{\hspace{0.17em}}s\text{\hspace{0.17em}}$ is the semi-perimeter, which is half the perimeter of the triangle.

f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
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Amit
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Dorbor
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Biswajit
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Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
which of these functions is not uniformly continuous on 0,1
solve this equation by completing the square 3x-4x-7=0
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
what's the formula
Modress
-x=7
Modress
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siame
what is trigonometry
deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
Thomas
solve for me this equational y=2-x
what are you solving for
Alex
solve x
Rubben
you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
Nikki
then I got x=-2
Rubben
it will b -y+2=x
Alex
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Nikki
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Rubben
how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m
More example of algebra and trigo
What is Indices
If one side only of a triangle is given is it possible to solve for the unkown two sides?
cool
Rubben
kya
Khushnama
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Okey tell me, what's your problem is?
Navin