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Just like regular numbers, angles can be added to obtain a sum, perhaps for the purpose of determining the measure of an unknown angle. Sometimes we can determine a missing angle because we know that the sum must be a certain value. Remember -- the sum of the degree measures of angles in any triangle equals 180 degrees. Below is a picture of triangle ABC, where angle A = 60 degrees, angle B = 50 degrees and angle C = 70 degrees. If we add all three angles in any triangle we get 180 degrees. So, the measure of angle A + angle B + angle C = 180 degrees. This is true for any triangle in the world of geometry. We can use this idea to find the measure of angle(s) where the degree measure is missing or not given. Finding a Missing AngleIn triangle ABC below, angle A = 40 degrees and angle B = 60 degrees. What is the measure of angle C? We know that the sum of the measures of any triangle is 180 degrees. Using the fact that angle A + angle B + angle C = 180 degrees, we can find the measure of angle C.
angle A = 40 angle B = 60 angle C = we don't know. To find angle C, we simply plug into the formula above and solve for C.
A + B + C = 180 C = 180 - A - B C = 180 - 40 - 60 C = 80 To check if 80 degrees is correct, let's add all three angle measures. If we get 180 degrees, then our answer for angle C is right. Here we go:
40 + 60 + 80 = 180 You don't always have to plug in those values to the equation and solve. Once you're comfortable with this sort of problem you'll be able to say "okay, 40 + 60 =100, so the other angle has to be 80!" and it's much quicker. Equilateral TrianglesIf a triangle is equilateral, what is the degree measure of each of its angles? Remember, all sides of an equilateral triangle have equal measure. They also, as you'll learn, have equal angles! Let x = the degree measure of each angle. Triangles have three angles and so we will add x THREE times. We have this:
x + x + x = 180 3x = 180 x = 60 Makes sense, right? If all the angles are equal, and they add up to 180, then it has to be 60 degrees! A Ratio of Angles The degree measures of the angles of a triangle are in the ratio 4 : 5 : 9. Notice that the smallest angle is represented by the smallest number in the ratio given. The smallest number given is 4, right? Since this is a ratio, we have to multiply all those values (4,5,9) by some common factor to get the actual angles. (For example, 60 and 80 are in a 3:4 ratio with a factor of 20) Let 4x = the measure of the smallest angle of the triangle. We can now say that 5x and 9x = the degree measures of the remaining angles of the triangle. We simply add 4x + 5x + 9x, equated the sum to 180 degrees and solve for x. After finding x, we plug the value of x into 4x, and simplify to find the measure of the smallest angle of the triangle at hand.
4x + 5x + 9x = 180 9x + 9x = 180 18x = 180 x = 180/18 x = 10 We found the value of x but it does NOT mean we are done. The answer is 40 degrees. Remember, the sum of the angles of a triangle is 180 degrees. Just take what you are given in a problem and try to determine what will make the final angle add up to 180 degrees. A lesson provided by Mr. Feliz Try the "Triangle Calculator" below:
In this section, you will:
Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in (Figure). How far from port is the boat? Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a SAS (side-angle-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side-side) triangle. In this section, we will investigate another tool for solving oblique triangles described by these last two cases.
The tool we need to solve the problem of the boat’s distance from the port is the Law of Cosines, which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because 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 Law 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. Here 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 point[latex]\,\left(x,y\right)\,[/latex]in the plane, as illustrated in (Figure). Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.
Figure 2. We can drop a perpendicular from[latex]\,C\,[/latex]to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that [latex]\mathrm{cos}\,\theta =\frac{x\text{(adjacent)}}{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(x,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\right)\,[/latex]as one leg of a right triangle and[latex]\,y\,[/latex]as the second leg, we can find the length of hypotenuse[latex]\,a\,[/latex]using the Pythagorean Theorem. Thus, [latex]\begin{array}{llllll} {a}^{2}={\left(x-c\right)}^{2}+{y}^{2}\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \text{ }={\left(b\mathrm{cos}\,\theta -c\right)}^{2}+{\left(b\mathrm{sin}\,\theta \right)}^{2}\hfill & \hfill & \hfill & \hfill & \hfill & \text{Substitute }\left(b\mathrm{cos}\,\theta \right)\text{ for}\,x\,\,\text{and }\left(b\mathrm{sin}\,\theta \right)\,\text{for }y.\hfill \\ \text{ }=\left({b}^{2}{\mathrm{cos}}^{2}\theta -2bc\mathrm{cos}\,\theta +{c}^{2}\right)+{b}^{2}{\mathrm{sin}}^{2}\theta \hfill & \hfill & \hfill & \hfill & \hfill & \text{Expand the perfect square}.\hfill \\ \text{ }={b}^{2}{\mathrm{cos}}^{2}\theta +{b}^{2}{\mathrm{sin}}^{2}\theta +{c}^{2}-2bc\mathrm{cos}\,\theta \hfill & \hfill & \hfill & \hfill & \hfill & \text{Group terms noting that }{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1.\hfill \\ \text{ }={b}^{2}\left({\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \right)+{c}^{2}-2bc\mathrm{cos}\,\theta \hfill & \hfill & \hfill & \hfill & \hfill & \text{Factor out }{b}^{2}.\hfill \\ {a}^{2}={b}^{2}+{c}^{2}-2bc\mathrm{cos}\,\theta \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \end{array}[/latex] The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion. Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve.
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 two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles labeled as in (Figure), with angles[latex]\,\alpha ,\beta ,[/latex] and[latex]\,\gamma ,[/latex] and opposite corresponding sides[latex]\,a,b,[/latex] and[latex]\,c,\,[/latex]respectively, the Law of Cosines is given as three equations. [latex]\begin{array}{l}{a}^{2}={b}^{2}+{c}^{2}-2bc\,\,\mathrm{cos}\,\alpha \\ {b}^{2}={a}^{2}+{c}^{2}-2ac\,\,\mathrm{cos}\,\beta \\ {c}^{2}={a}^{2}+{b}^{2}-2ab\,\,\mathrm{cos}\,\gamma \end{array}[/latex]
Figure 3. To solve for a missing side measurement, the corresponding opposite angle measure is needed. When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle. [latex]\begin{array}{l}\hfill \\ \begin{array}{l}\begin{array}{l}\hfill \\ \mathrm{cos}\text{ }\alpha =\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\hfill \end{array}\hfill \\ \mathrm{cos}\text{ }\beta =\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\hfill \\ \mathrm{cos}\text{ }\gamma =\frac{{a}^{2}+{b}^{2}-{c}^{2}}{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.
Find the unknown side and angles of the triangle in (Figure).
Figure 4.
Find the missing side and angles of the given triangle:[latex]\,\alpha =30°,\,\,b=12,\,\,c=24.[/latex]
Find the angle[latex]\,\alpha \,[/latex]for the given triangle if side[latex]\,a=20,\,[/latex]side[latex]\,b=25,\,[/latex]and side[latex]\,c=18.[/latex]
Because the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method.
Given[latex]\,a=5,b=7,\,[/latex]and[latex]\,c=10,\,[/latex]find the missing angles.
Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few.
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 there are two cell phone towers within range of a cell phone. The two towers are located 6000 feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5050 feet from the first tower and 2420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway.
Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in (Figure).
Figure 8.
We already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use Heron’s formula instead of finding the height. Heron of Alexandria was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known.
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{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}[/latex] where[latex]\,s=\frac{\left(a+b+c\right)}{2}\,[/latex] is one half of the perimeter of the triangle, sometimes called the semi-perimeter.
Find the area of the triangle in (Figure) using Heron’s formula.
Figure 9.
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]
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 (Figure) for a view of the city property.
Figure 10.
Find the area of a triangle given[latex]\,a=4.38\,\text{ft}\,,b=3.79\,\text{ft,}\,[/latex]and[latex]\,c=5.22\,\text{ft}\text{.}[/latex]
If you are looking for a missing side of a triangle, what do you need to know when using the Law of Cosines?
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[latex]\,s\,[/latex]represents in Heron’s formula.
Explain the relationship between the Pythagorean Theorem and the Law of Cosines.
When must you use the Law of Cosines instead of the Pythagorean Theorem?
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=3.13[/latex]
[latex]\alpha =120°,b=6,c=7[/latex]
[latex]\beta =58.7°,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]\alpha =43.1°,a=184.2,b=242.8[/latex]
[latex]\alpha =36.6°,a=186.2,b=242.2[/latex]
[latex]\beta =50°,a=105,b=45{}_{}{}^{}[/latex] For the following exercises, use the Law of Cosines to solve for the missing angle of the oblique triangle. Round to the nearest tenth.
[latex]\,a=42,b=19,c=30;\,[/latex]find angle[latex]\,A.[/latex]
[latex]\,a=14,\text{ }b=13,\text{ }c=20;\,[/latex]find angle[latex]\,C.[/latex]
[latex]\,a=16,b=31,c=20;\,[/latex]find angle[latex]\,B.[/latex]
[latex]\,a=13,\,b=22,\,c=28;\,[/latex]find angle[latex]\,A.[/latex]
[latex]a=108,\,b=132,\,c=160;\,[/latex]find angle[latex]\,C.\,[/latex] For the following exercises, solve the triangle. Round to the nearest tenth.
[latex]A=35°,b=8,c=11[/latex]
[latex]B=88°,a=4.4,c=5.2[/latex]
[latex]C=121°,a=21,b=37[/latex]
[latex]a=13,b=11,c=15[/latex]
[latex]a=3.1,b=3.5,c=5[/latex]
[latex]a=51,b=25,c=29[/latex] For the following exercises, use Heron’s formula to find the area of the triangle. Round to the nearest hundredth.
Find the area of a triangle with sides of length 18 in, 21 in, and 32 in. Round to the nearest tenth.
Find the area of a triangle with sides of length 20 cm, 26 cm, and 37 cm. Round to the nearest tenth.
[latex]a=\frac{1}{2}\,\text{m},b=\frac{1}{3}\,\text{m},c=\frac{1}{4}\,\text{m}[/latex]
[latex]a=12.4\text{ ft},\text{ }b=13.7\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]
For the following exercises, find the length of side [latex]x.[/latex] Round to the nearest tenth.
For the following exercises, find the measurement of angle[latex]\,A.[/latex]
Find the measure of each angle in the triangle shown in (Figure). Round to the nearest tenth.
Figure 11. For the following exercises, solve for the unknown side. Round to the nearest tenth.
For the following exercises, find the area of the triangle. Round to the nearest hundredth.
A parallelogram has sides of length 16 units and 10 units. The shorter diagonal is 12 units. Find the measure of the longer diagonal.
The sides of a parallelogram are 11 feet and 17 feet. The longer diagonal is 22 feet. Find the length of the shorter diagonal.
The sides of a parallelogram are 28 centimeters and 40 centimeters. The measure of the larger angle is 100°. Find the length of the shorter diagonal.
A regular octagon is inscribed in a circle with a radius of 8 inches. (See (Figure).) Find the perimeter of the octagon.
Figure 12.
A regular pentagon is inscribed in a circle of radius 12 cm. (See (Figure).) Find the perimeter of the pentagon. Round to the nearest tenth of a centimeter.
Figure 13. For the following exercises, suppose that[latex]\,{x}^{2}=25+36-60\mathrm{cos}\left(52\right)\,[/latex]represents the relationship of three sides of a triangle and the cosine of an angle.
Find the length of the third side. For the following exercises, find the area of the triangle.
A surveyor has taken the measurements shown in (Figure). Find the distance across the lake. Round answers to the nearest tenth.
Figure 14.
A satellite calculates the distances and angle shown in (Figure) (not 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°, and then 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 be attached to the top of the tower and anchored at a point 98 feet uphill from the base of the tower. Find the length of wire needed.
Figure 16.
Two ships left a port at the same time. One 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°. Find the distance between the two ships after 10 hours of travel.
The graph in (Figure) represents two boats departing at the same time from the same dock. The first boat is traveling at 18 miles per hour at a heading of 327° and the second boat is traveling at 4 miles per hour at a heading of 60°. Find the distance between the two boats after 2 hours.
Figure 17.
A triangular swimming pool measures 40 feet on one side and 65 feet on another side. These sides form an angle that measures 50°. How long is the third side (to the nearest tenth)?
A pilot flies in a straight path for 1 hour 30 min. She then makes a course correction, heading 10° to the right of her original course, and flies 2 hours in the new direction. If she maintains a constant speed of 680 miles per hour, how far is she from her starting position?
Los Angeles is 1,744 miles from Chicago, Chicago is 714 miles from New York, and New York is 2,451 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. Draw a triangle connecting these three cities and find the angles in the triangle.
Two planes leave the same airport at the same time. One flies at 20° east of north at 500 miles per hour. The second flies at 30° east of south at 600 miles per hour. How far apart are the planes after 2 hours?
Two airplanes take off in different directions. One travels 300 mph due west and the other travels 25° north of west at 420 mph. After 90 minutes, how far apart are they, assuming they are flying at the same altitude?
A parallelogram has sides of length 15.4 units and 9.8 units. Its area is 72.9 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 12.9 cm. The angle between the two smallest sides is 117°. What is the area of this quadrilateral?
The four sequential sides of a quadrilateral have lengths 5.7 cm, 7.2 cm, 9.4 cm, and 12.8 cm. The angle between the two smallest sides is 106°. What is the area of this quadrilateral?
Find the area of a triangular piece of land that measures 30 feet on one side and 42 feet on another; the included angle measures 132°. Round to the nearest whole square foot.
Find the area of a triangular piece of land that measures 110 feet on one side and 250 feet on another; the included angle measures 85°. Round to the nearest whole square foot. |
If one angle of a triangle is equal to twice the sum of the other two angles, then the triangle is
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