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

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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] .

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20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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