How To Find Area Of A Non Right Triangle

Learning Objectives

In this section, you will:

• Utilise the Constabulary of Cosines to solve oblique triangles.
• Solve applied problems using the Police of Cosines.
• Apply Heron's formula to find the area of a triangle.

Suppose a boat leaves port, travels x miles, turns xx degrees, and travels some other 8 miles as shown in (Figure). How far from port is the gunkhole?

Effigy 1.

Unfortunately, while the Law of Sines enables us to address many non-correct triangle cases, information technology does not help u.s.a. with triangles where the known angle is between two known sides, a SAS (side-bending-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side-side) triangle. In this department, nosotros will investigate some other tool for solving oblique triangles described by these concluding two cases.

Using the Law of Cosines to Solve Oblique Triangles

The tool we need to solve the problem of the boat'due south distance from the port is the Constabulary of Cosines, which defines the human relationship amidst bending measurements and side lengths in oblique triangles. Iii formulas make up the Constabulary of Cosines. At first glance, the formulas may announced complicated considering they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level.

Understanding how the Police of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the Pythagorean Theorem to non-right triangles. Hither is how it works: An arbitrary non-right triangle[latex]\,ABC\,[/latex]is placed in the coordinate plane with vertex[latex]\,A\,[/latex]at the origin, side[latex]\,c\,[/latex]drawn along the x-axis, and vertex[latex]\,C\,[/latex]located at some signal[latex]\,\left(ten,y\right)\,[/latex]in the aeroplane, as illustrated in (Figure). Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle every bit noted.

Effigy 2.

Nosotros can driblet a perpendicular from[latex]\,C\,[/latex]to the x-axis (this is the altitude or superlative). Recalling the bones trigonometric identities, we know that

[latex]\mathrm{cos}\,\theta =\frac{ten\text{(side by side)}}{b\text{(hypotenuse)}}\text{ and }\mathrm{sin}\,\theta =\frac{y\text{(opposite)}}{b\text{(hypotenuse)}}[/latex]

In terms of[latex]\,\theta ,\text{ }x=b\mathrm{cos}\,\theta \,[/latex]and[latex]y=b\mathrm{sin}\,\theta .\text{ }[/latex]The[latex]\,\left(ten,y\right)\,[/latex]point located at[latex]\,C\,[/latex]has coordinates[latex]\,\left(b\mathrm{cos}\,\theta ,\,\,b\mathrm{sin}\,\theta \right).\,[/latex]Using the side[latex]\,\left(x-c\correct)\,[/latex]as i leg of a right triangle and[latex]\,y\,[/latex]as the second leg, we can discover the length of hypotenuse[latex]\,a\,[/latex]using the Pythagorean Theorem. Thus,

[latex]\begin{array}{llllll} {a}^{two}={\left(ten-c\right)}^{2}+{y}^{2}\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \text{ }={\left(b\mathrm{cos}\,\theta -c\right)}^{ii}+{\left(b\mathrm{sin}\,\theta \right)}^{2}\hfill & \hfill & \hfill & \hfill & \hfill & \text{Substitute }\left(b\mathrm{cos}\,\theta \right)\text{ for}\,10\,\,\text{and }\left(b\mathrm{sin}\,\theta \right)\,\text{for }y.\hfill \\ \text{ }=\left({b}^{2}{\mathrm{cos}}^{ii}\theta -2bc\mathrm{cos}\,\theta +{c}^{ii}\correct)+{b}^{two}{\mathrm{sin}}^{2}\theta \hfill & \hfill & \hfill & \hfill & \hfill & \text{Expand the perfect square}.\hfill \\ \text{ }={b}^{2}{\mathrm{cos}}^{two}\theta +{b}^{two}{\mathrm{sin}}^{2}\theta +{c}^{two}-2bc\mathrm{cos}\,\theta \hfill & \hfill & \hfill & \hfill & \hfill & \text{Group terms noting that }{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =i.\hfill \\ \text{ }={b}^{two}\left({\mathrm{cos}}^{two}\theta +{\mathrm{sin}}^{2}\theta \correct)+{c}^{2}-2bc\mathrm{cos}\,\theta \hfill & \hfill & \hfill & \hfill & \hfill & \text{Factor out }{b}^{2}.\hfill \\ {a}^{two}={b}^{ii}+{c}^{2}-2bc\mathrm{cos}\,\theta \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \end{array}[/latex]

The formula derived is one of the 3 equations of the Law of Cosines. The other equations are institute in a similar fashion.

Proceed in heed that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, effort to draw a diagram of the state of affairs. Every bit more than information emerges, the diagram may take to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve.

Law of Cosines

The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other 2 sides minus twice the production of the other ii sides and the cosine of the included angle. For triangles labeled as in (Figure), with angles[latex]\,\blastoff ,\beta ,[/latex] and[latex]\,\gamma ,[/latex] and reverse corresponding sides[latex]\,a,b,[/latex] and[latex]\,c,\,[/latex]respectively, the Law of Cosines is given equally three equations.

[latex]\brainstorm{array}{50}{a}^{ii}={b}^{2}+{c}^{2}-2bc\,\,\mathrm{cos}\,\alpha \\ {b}^{2}={a}^{2}+{c}^{2}-2ac\,\,\mathrm{cos}\,\beta \\ {c}^{2}={a}^{two}+{b}^{2}-2ab\,\,\mathrm{cos}\,\gamma \end{assortment}[/latex]

Figure three.

To solve for a missing side measurement, the corresponding opposite angle measure is needed.

When solving for an angle, the corresponding opposite side measure out is needed. We tin can use another version of the Law of Cosines to solve for an bending.

[latex]\brainstorm{array}{l}\hfill \\ \begin{array}{50}\begin{array}{l}\hfill \\ \mathrm{cos}\text{ }\blastoff =\frac{{b}^{ii}+{c}^{ii}-{a}^{2}}{2bc}\hfill \cease{array}\hfill \\ \mathrm{cos}\text{ }\beta =\frac{{a}^{two}+{c}^{ii}-{b}^{2}}{2ac}\hfill \\ \mathrm{cos}\text{ }\gamma =\frac{{a}^{ii}+{b}^{2}-{c}^{ii}}{2ab}\hfill \end{array}\hfill \end{array}[/latex]

Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.

1. Sketch the triangle. Identify the measures of the known sides and angles. Utilize variables to represent the measures of the unknown sides and angles.
2. Employ the Law of Cosines to find the length of the unknown side or angle.
3. Utilise the Law of Sines or Cosines to find the measure of a second angle.
4. Compute the mensurate of the remaining bending.

Finding the Unknown Side and Angles of a SAS Triangle

Observe the unknown side and angles of the triangle in (Effigy).

Figure 4.

Endeavor Information technology

Find the missing side and angles of the given triangle:[latex]\,\blastoff =thirty°,\,\,b=12,\,\,c=24.[/latex]

Testify Solution

[latex]a\approx 14.nine,\,\,\beta \approx 23.eight°,\,\,\gamma \approx 126.two°.[/latex]

Solving for an Angle of a SSS Triangle

Discover the angle[latex]\,\alpha \,[/latex]for the given triangle if side[latex]\,a=xx,\,[/latex]side[latex]\,b=25,\,[/latex]and side[latex]\,c=18.[/latex]

Analysis

Because the changed cosine can render any angle between 0 and 180 degrees, at that place will not be any cryptic cases using this method.

Try Information technology

Given[latex]\,a=five,b=7,\,[/latex]and[latex]\,c=10,\,[/latex]discover the missing angles.

Evidence Solution

[latex]\alpha \approx 27.7°,\,\,\beta \approx forty.5°,\,\,\gamma \approx 111.8°[/latex]

Solving Practical Issues Using the Police of Cosines

Merely as the Police of Sines provided the advisable equations to solve a number of applications, the Constabulary of Cosines is applicable to situations in which the given data fits the cosine models. Nosotros may see these in the fields of navigation, surveying, astronomy, and geometry, merely to name a few.

Using the Law of Cosines to Solve a Advice Problem

On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose at that place are two cell phone towers within range of a cell telephone. The ii towers are located 6000 feet apart forth a straight highway, running eastward to west, and the prison cell phone is northward of the highway. Based on the betoken filibuster, it tin can exist determined that the signal is 5050 anxiety from the start tower and 2420 feet from the second belfry. Decide the position of the jail cell telephone due north and east of the first tower, and determine how far information technology is from the highway.

Calculating Altitude Traveled Using a SAS Triangle

Returning to our problem at the starting time of this department, suppose a boat leaves port, travels 10 miles, turns twenty degrees, and travels another viii miles. How far from port is the gunkhole? The diagram is repeated here in (Effigy).

Figure viii.

Using Heron's Formula to Observe the Expanse of a Triangle

We already learned how to find the expanse of an oblique triangle when nosotros know two sides and an bending. Nosotros also know the formula to find the surface area of a triangle using the base and the summit. When nosotros know the three sides, withal, nosotros tin apply Heron's formula instead of finding the height. Heron of Alexandria was a geometer who lived during the starting time century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known.

Heron's Formula

Heron's formula finds the area of oblique triangles in which sides[latex]\,a,b\text{,}[/latex]and[latex]\,c\,[/latex]are known.

[latex]\text{Area}=\sqrt{due south\left(s-a\right)\left(s-b\right)\left(s-c\right)}[/latex]

where[latex]\,south=\frac{\left(a+b+c\correct)}{two}\,[/latex] is one one-half of the perimeter of the triangle, sometimes called the semi-perimeter.

Using Heron'southward Formula to Find the Surface area of a Given Triangle

Find the area of the triangle in (Figure) using Heron'south formula.

Figure 9.

Try It

Use Heron's formula to find the area of a triangle with sides of lengths[latex]\,a=29.7\,\text{ft},b=42.3\,\text{ft},\,[/latex]and[latex]\,c=38.4\,\text{ft}.[/latex]

Show Solution

Surface area = 552 foursquare feet

Applying Heron'southward Formula to a Existent-Earth Trouble

A Chicago city developer wants to construct a building consisting of artist'due south lofts on a triangular lot bordered past Rush Street, Wabash Avenue, and Pearson Street. The frontage forth Rush Street is approximately 62.iv meters, along Wabash Avenue information technology is approximately 43.five meters, and along Pearson Street it is approximately 34.ane meters. How many foursquare meters are available to the developer? Encounter (Effigy) for a view of the city property.

Figure x.

Endeavor It

Find the surface area of a triangle given[latex]\,a=iv.38\,\text{ft}\,,b=3.79\,\text{ft,}\,[/latex]and[latex]\,c=5.22\,\text{ft}\text{.}[/latex]

Show Solution

about 8.xv square feet

Key Equations

Law of Cosines	[latex]\brainstorm{array}{fifty}{a}^{ii}={b}^{ii}+{c}^{2}-2bc\mathrm{cos}\,\alpha \hfill \\ {b}^{2}={a}^{2}+{c}^{2}-2ac\mathrm{cos}\,\beta \hfill \\ {c}^{2}={a}^{2}+{b}^{2}-2abcos\,\gamma \hfill \end{array}[/latex]
Heron's formula	[latex]\begin{assortment}{l}\text{ Expanse}=\sqrt{s\left(southward-a\correct)\left(due south-b\correct)\left(due south-c\right)}\hfill \\ \text{where }s=\frac{\left(a+b+c\correct)}{2}\hfill \cease{assortment}[/latex]

Central Concepts

• The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles.
• The Generalized Pythagorean Theorem is the Police of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into ii correct triangles or forms i correct triangle, which allows sides to exist related and measurements to be calculated. See (Figure) and (Figure).
• The Police force of Cosines is useful for many types of applied issues. The beginning step in solving such issues is by and large to depict a sketch of the trouble presented. If the information given fits 1 of the three models (the iii equations), then apply the Police force of Cosines to notice a solution. Encounter (Effigy) and (Figure).
• Heron'due south formula allows the calculation of surface area in oblique triangles. All three sides must be known to apply Heron'due south formula. See (Figure) and See (Effigy).

Section Exercises

Verbal

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

Show Solution

two sides and the angle opposite the missing side.

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

Explain what[latex]\,s\,[/latex]represents in Heron'southward formula.

Evidence Solution

[latex]\,s\,[/latex]is the semi-perimeter, which is half the perimeter of the triangle.

Explicate the relationship between the Pythagorean Theorem and the Constabulary of Cosines.

When must you use the Law of Cosines instead of the Pythagorean Theorem?

Testify Solution

The Law of Cosines must be used for any oblique (not-right) triangle.

Algebraic

For the following exercises, assume[latex]\,\alpha \,[/latex]is opposite side[latex]\,a,\beta \,[/latex] is opposite side[latex]\,b,\,[/latex]and[latex]\,\gamma \,[/latex] is opposite side[latex]\,c.\,[/latex]If possible, solve each triangle for the unknown side. Round to the nearest tenth.

[latex]\gamma =41.2°,a=2.49,b=3.thirteen[/latex]

[latex]\alpha =120°,b=6,c=7[/latex]

[latex]\beta =58.vii°,a=10.6,c=15.7[/latex]

[latex]\gamma =115°,a=18,b=23[/latex]

[latex]\alpha =119°,a=26,b=14[/latex]

[latex]\gamma =113°,b=10,c=32[/latex]

[latex]\beta =67°,a=49,b=38[/latex]

[latex]\blastoff =43.1°,a=184.2,b=242.eight[/latex]

[latex]\alpha =36.half-dozen°,a=186.2,b=242.2[/latex]

[latex]\beta =50°,a=105,b=45{}_{}{}^{}[/latex]

Show Solution

not possible

For the following exercises, use the Law of Cosines to solve for the missing angle of the oblique triangle. Round to the nearest 10th.

[latex]\,a=42,b=19,c=xxx;\,[/latex]detect angle[latex]\,A.[/latex]

[latex]\,a=xiv,\text{ }b=13,\text{ }c=twenty;\,[/latex]discover angle[latex]\,C.[/latex]

[latex]\,a=16,b=31,c=20;\,[/latex]find bending[latex]\,B.[/latex]

[latex]\,a=13,\,b=22,\,c=28;\,[/latex]discover angle[latex]\,A.[/latex]

[latex]a=108,\,b=132,\,c=160;\,[/latex]observe bending[latex]\,C.\,[/latex]

For the following exercises, solve the triangle. Circular to the nearest tenth.

[latex]A=35°,b=8,c=11[/latex]

Show Solution

[latex]B\approx 45.9°,C\approx 99.1°,a\approx 6.4[/latex]

[latex]B=88°,a=4.4,c=v.2[/latex]

[latex]C=121°,a=21,b=37[/latex]

Show Solution

[latex]A\approx 20.6°,B\approx 38.4°,c\approx 51.1[/latex]

[latex]a=13,b=11,c=15[/latex]

[latex]a=3.1,b=three.5,c=5[/latex]

Show Solution

[latex]A\approx 37.8°,B\approx 43.8,C\approx 98.4°[/latex]

[latex]a=51,b=25,c=29[/latex]

For the following exercises, employ Heron's formula to find the area of the triangle. Circular to the nearest hundredth.

Find the area of a triangle with sides of length 18 in, 21 in, and 32 in. Circular to the nearest tenth.

Observe the area of a triangle with sides of length 20 cm, 26 cm, and 37 cm. Round to the nearest tenth.

[latex]a=\frac{i}{2}\,\text{g},b=\frac{one}{three}\,\text{thousand},c=\frac{1}{4}\,\text{m}[/latex]

[latex]a=12.four\text{ ft},\text{ }b=xiii.seven\text{ ft},\text{ }c=20.2\text{ ft}[/latex]

[latex]a=1.6\text{ yd},\text{ }b=2.6\text{ yd},\text{ }c=4.1\text{ yd}[/latex]

Graphical

For the following exercises, find the length of side [latex]ten.[/latex] Round to the nearest tenth.

For the following exercises, observe the measurement of angle[latex]\,A.[/latex]

Find the measure of each angle in the triangle shown in (Effigy). Round to the nearest tenth.

Figure 11.

For the post-obit exercises, solve for the unknown side. Round to the nearest tenth.

For the post-obit exercises, notice the area of the triangle. Round to the nearest hundredth.

Extensions

A parallelogram has sides of length 16 units and x units. The shorter diagonal is 12 units. Find the measure out of the longer diagonal.

The sides of a parallelogram are eleven feet and 17 feet. The longer diagonal is 22 anxiety. Find the length of the shorter diagonal.

The sides of a parallelogram are 28 centimeters and 40 centimeters. The measure out of the larger angle is 100°. Observe the length of the shorter diagonal.

A regular octagon is inscribed in a circle with a radius of viii inches. (See (Effigy).) Notice the perimeter of the octagon.

Figure 12.

A regular pentagon is inscribed in a circle of radius 12 cm. (Come across (Effigy).) Find the perimeter of the pentagon. Round to the nearest tenth of a centimeter.

Figure thirteen.

For the following exercises, suppose that[latex]\,{x}^{ii}=25+36-60\mathrm{cos}\left(52\right)\,[/latex]represents the relationship of three sides of a triangle and the cosine of an bending.

Draw the triangle.

Evidence Solution

Discover the length of the third side.

For the following exercises, find the expanse of the triangle.

Existent-World Applications

A surveyor has taken the measurements shown in (Figure). Notice the distance across the lake. Circular answers to the nearest tenth.

Effigy fourteen.

A satellite calculates the distances and angle shown in (Figure) (non to scale). Find the distance between the two cities. Round answers to the nearest tenth.

Figure 15.

An airplane flies 220 miles with a heading of 40°, so flies 180 miles with a heading of 170°. How far is the plane from its starting point, and at what heading? Round answers to the nearest tenth.

A 113-foot tower is located on a hill that is inclined 34° to the horizontal, as shown in (Figure). A guy-wire is to exist attached to the acme of the tower and anchored at a betoken 98 feet uphill from the base of operations of the tower. Observe the length of wire needed.

Figure xvi.

Two ships left a port at the same time. Ane ship traveled at a speed of 18 miles per hour at a heading of 320°. The other ship traveled at a speed of 22 miles per hour at a heading of 194°. Notice the altitude between the ii ships after x hours of travel.

The graph in (Figure) represents 2 boats departing at the same time from the same dock. The first boat is traveling at 18 miles per 60 minutes at a heading of 327° and the second gunkhole is traveling at four miles per hr at a heading of 60°. Find the distance between the two boats later 2 hours.

Effigy 17.

A triangular pond pool measures 40 feet on one side and 65 feet on another side. These sides class an bending that measures l°. How long is the third side (to the nearest tenth)?

A pilot flies in a straight path for ane hour xxx min. She then makes a course correction, heading 10° to the right of her original grade, and flies 2 hours in the new direction. If she maintains a constant speed of 680 miles per hr, how far is she from her starting position?

Los Angeles is i,744 miles from Chicago, Chicago is 714 miles from New York, and New York is ii,451 miles from Los Angeles. Draw a triangle connecting these three cities, and find the angles in the triangle.

Philadelphia is 140 miles from Washington, D.C., Washington, D.C. is 442 miles from Boston, and Boston is 315 miles from Philadelphia. Describe a triangle connecting these three cities and find the angles in the triangle.

Show Solution

Two planes leave the aforementioned airport at the aforementioned time. One flies at twenty° east of north at 500 miles per hr. The second flies at 30° east of south at 600 miles per hour. How far apart are the planes after ii hours?

Two airplanes have off in dissimilar directions. 1 travels 300 mph due west and the other travels 25° north of due west at 420 mph. After ninety minutes, how far apart are they, assuming they are flight at the same altitude?

A parallelogram has sides of length 15.four units and 9.8 units. Its expanse is 72.nine square units. Find the measure of the longer diagonal.

The four sequential sides of a quadrilateral have lengths 4.5 cm, 7.9 cm, 9.4 cm, and 12.ix cm. The angle between the 2 smallest sides is 117°. What is the expanse of this quadrilateral?

The four sequential sides of a quadrilateral take lengths v.seven cm, 7.2 cm, 9.4 cm, and 12.8 cm. The bending between the 2 smallest sides is 106°. What is the expanse of this quadrilateral?

Notice the area of a triangular piece of state that measures 30 anxiety on one side and 42 feet on some other; the included bending measures 132°. Round to the nearest whole square foot.

Find the area of a triangular slice of country that measures 110 feet on one side and 250 anxiety on another; the included bending measures 85°. Round to the nearest whole square pes.

Glossary

Law of Cosines

states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the production of the other two sides and the cosine of the included angle

Generalized Pythagorean Theorem

an extension of the Law of Cosines; relates the sides of an oblique triangle and is used for SAS and SSS triangles

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