maths.free › Trigonometry › 8. Further Applications of Trigonometry › Non-right Triangles: Law of Cosines
Non-right Triangles: Law of Cosines
Use the Law of Cosines to solve oblique triangles.
Using the Law of Cosines to Solve Oblique Triangles
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 \(ABC\) is placed in the coordinate plane with vertex \(A\) at the origin, side \(c\) drawn along the x-axis, and vertex \(C\) located at some point \((x,y)\) in the plane, as illustrated in . Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.
We can drop a perpendicular from \(C\) to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that
\[\cos \ \theta =\frac{x\text{(adjacent)}}{b\text{(hypotenuse)}}\text{ and }\sin \ \theta =\frac{y\text{(opposite)}}{b\text{(hypotenuse)}}\]In terms of \(\theta ,\ x=b\cos \ \theta\) and \(y=b\sin \ \theta .\) The \((x,y)\) point located at \(C\) has coordinates \((b\cos \ \theta ,\) \(b\sin \ \theta ).\) Using the side \((x-c)\) as one leg of a right triangle and \(y\) as the second leg, we can find the length of hypotenuse \(a\) using the Pythagorean Theorem. Thus,
\[\begin{array}{llllll}\ {a}^{2}={(x-c)}^{2}+{y}^{2} & & & & & \\ ={(b\cos \ \theta -c)}^{2}+{(b\sin \ \theta )}^{2} & & & & & \text{Substitute }(b\cos \ \theta )\text{ for}\ x\ \text{and }(b\sin \ \theta )\ \text{for }y. \\ =({b}^{2}{\cos }^{2}\theta -2bc\cos \ \theta +{c}^{2})+{b}^{2}{\sin }^{2}\theta & & & & & \text{Expand the perfect square}. \\ ={b}^{2}{\cos }^{2}\theta +{b}^{2}{\sin }^{2}\theta +{c}^{2}-2bc\cos \ \theta & & & & & \text{Group terms noting that }{\cos }^{2}\theta +{\sin }^{2}\theta =1. \\ ={b}^{2}({\cos }^{2}\theta +{\sin }^{2}\theta )+{c}^{2}-2bc\cos \ \theta & & & & & \text{Factor out }{b}^{2}. \\ {a}^{2}={b}^{2}+{c}^{2}-2bc\cos \ \theta & & & & & \end{array}\]The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion.
Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.
Solving Applied Problems Using the Law of Cosines
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.
Example
Try it.
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 .
Solution
The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, \(180^{\circ}-20^{\circ}=160^{\circ}.\) With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the boat to the port.
\[\begin{array}{l}{x}^{2}={8}^{2}+{10}^{2}-2(8)(10)\cos (160^{\circ}) \\ {x}^{2}=314.35 \\ x=\sqrt{314.35} \\ x\approx 17.7\ \text{miles}\end{array}\]The boat is about 17.7 miles from port.
Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.
Using Heron’s Formula to Find the Area of a Triangle
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.
Example
Try it.
Find the area of the triangle in using Heron’s formula.
Solution
First, we calculate \(s.\)
\[\begin{array}{l}s=\frac{(a+b+c)}{2} \\ s=\frac{(10+15+7)}{2}=16\end{array}\]Then we apply the formula.
\[\begin{array}{l}\text{Area}=\sqrt{s(s-a)(s-b)(s-c)} \\ \text{Area}=\sqrt{16(16-10)(16-15)(16-7)} \\ \text{Area}\approx 29.4\end{array}\]The area is approximately 29.4 square units.
Example
Try it.
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 for a view of the city property.
Solution
Find the measurement for \(s,\) which is one-half of the perimeter.
\[\begin{array}{l}s=\frac{(62.4+43.5+34.1)}{2} \\ s=70\ \text{m}\end{array}\]Apply Heron’s formula.
\[\begin{array}{l}\text{Area}=\sqrt{70(70-62.4)(70-43.5)(70-34.1)} \\ \text{Area}=\sqrt{506,118.2} \\ \text{Area}\approx 711.4\end{array}\]The developer has about 711.4 square meters.
Key Equations
| Law of Cosines | \(\begin{array}{l}{a}^{2}={b}^{2}+{c}^{2}-2bc\cos \ \alpha \\ {b}^{2}={a}^{2}+{c}^{2}-2ac\cos \ \beta \\ {c}^{2}={a}^{2}+{b}^{2}-2abcos\ \gamma \end{array}\) |
| Heron’s formula | \(\begin{array}{l}\text{ Area}=\sqrt{s(s-a)(s-b)(s-c)} \\ \text{where }s=\frac{(a+b+c)}{2}\end{array}\) |
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 and .
- 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 and .
- Heron’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula. See and See .
Using the Law of Cosines to Solve Oblique Triangles
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 \(ABC\) is placed in the coordinate plane with vertex \(A\) at the origin, side \(c\) drawn along the x-axis, and vertex \(C\) located at some point \((x,y)\) in the plane, as illustrated in . Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.
We can drop a perpendicular from \(C\) to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that
\[\cos \ \theta =\frac{x\text{(adjacent)}}{b\text{(hypotenuse)}}\text{ and }\sin \ \theta =\frac{y\text{(opposite)}}{b\text{(hypotenuse)}}\]In terms of \(\theta ,\ x=b\cos \ \theta\) and \(y=b\sin \ \theta .\) The \((x,y)\) point located at \(C\) has coordinates \((b\cos \ \theta ,\) \(b\sin \ \theta ).\) Using the side \((x-c)\) as one leg of a right triangle and \(y\) as the second leg, we can find the length of hypotenuse \(a\) using the Pythagorean Theorem. Thus,
\[\begin{array}{llllll}\ {a}^{2}={(x-c)}^{2}+{y}^{2} & & & & & \\ ={(b\cos \ \theta -c)}^{2}+{(b\sin \ \theta )}^{2} & & & & & \text{Substitute }(b\cos \ \theta )\text{ for}\ x\ \text{and }(b\sin \ \theta )\ \text{for }y. \\ =({b}^{2}{\cos }^{2}\theta -2bc\cos \ \theta +{c}^{2})+{b}^{2}{\sin }^{2}\theta & & & & & \text{Expand the perfect square}. \\ ={b}^{2}{\cos }^{2}\theta +{b}^{2}{\sin }^{2}\theta +{c}^{2}-2bc\cos \ \theta & & & & & \text{Group terms noting that }{\cos }^{2}\theta +{\sin }^{2}\theta =1. \\ ={b}^{2}({\cos }^{2}\theta +{\sin }^{2}\theta )+{c}^{2}-2bc\cos \ \theta & & & & & \text{Factor out }{b}^{2}. \\ {a}^{2}={b}^{2}+{c}^{2}-2bc\cos \ \theta & & & & & \end{array}\]The formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion.
Condensed — the full section is in OpenStax Precalculus 2e.
Solving Applied Problems Using the Law of Cosines
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.
Example
Try it.
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 .
Solution
The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, \(180^{\circ}-20^{\circ}=160^{\circ}.\) With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the boat to the port.
\[\begin{array}{l}{x}^{2}={8}^{2}+{10}^{2}-2(8)(10)\cos (160^{\circ}) \\ {x}^{2}=314.35 \\ x=\sqrt{314.35} \\ x\approx 17.7\ \text{miles}\end{array}\]The boat is about 17.7 miles from port.
Condensed — the full section is in OpenStax Precalculus 2e.
Using Heron’s Formula to Find the Area of a Triangle
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.
Example
Try it.
Find the area of the triangle in using Heron’s formula.
Solution
First, we calculate \(s.\)
\[\begin{array}{l}s=\frac{(a+b+c)}{2} \\ s=\frac{(10+15+7)}{2}=16\end{array}\]Then we apply the formula.
\[\begin{array}{l}\text{Area}=\sqrt{s(s-a)(s-b)(s-c)} \\ \text{Area}=\sqrt{16(16-10)(16-15)(16-7)} \\ \text{Area}\approx 29.4\end{array}\]The area is approximately 29.4 square units.
Example
Try it.
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 for a view of the city property.
Solution
Find the measurement for \(s,\) which is one-half of the perimeter.
\[\begin{array}{l}s=\frac{(62.4+43.5+34.1)}{2} \\ s=70\ \text{m}\end{array}\]Apply Heron’s formula.
\[\begin{array}{l}\text{Area}=\sqrt{70(70-62.4)(70-43.5)(70-34.1)} \\ \text{Area}=\sqrt{506,118.2} \\ \text{Area}\approx 711.4\end{array}\]The developer has about 711.4 square meters.
Key Equations
| Law of Cosines | \(\begin{array}{l}{a}^{2}={b}^{2}+{c}^{2}-2bc\cos \ \alpha \\ {b}^{2}={a}^{2}+{c}^{2}-2ac\cos \ \beta \\ {c}^{2}={a}^{2}+{b}^{2}-2abcos\ \gamma \end{array}\) |
| Heron’s formula | \(\begin{array}{l}\text{ Area}=\sqrt{s(s-a)(s-b)(s-c)} \\ \text{where }s=\frac{(a+b+c)}{2}\end{array}\) |
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 and .
- 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 and .
- Heron’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula. See and See .
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
-
Find the unknown side and angles of the triangle in .
Odhaliť odpoveď
First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines.
Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side \(b,\) as we know the measurement of the opposite angle \(\beta .\)
\[\begin{array}{lllll}{b}^{2}={a}^{2}+{c}^{2}-2ac\cos \ \beta & \\ {b}^{2}={10}^{2}+{12}^{2}-2(10)(12)\cos ({30}^{∘})\begin{array}{llll} & & & \end{array} & \text{Substitute the measurements for the known quantities}. \\ {b}^{2}=100+144-240(\frac{\sqrt{3}}{2}) & \text{Evaluate the cosine and begin to simplify}. \\ {b}^{2}=244-120\sqrt{3} & \\ b=\sqrt{244-120\sqrt{3}} & \text{Use the square root property}. \\ b\approx 6.013 & \end{array}\]Because we are solving for a length, we use only the positive square root. Now that we know the length \(b,\) we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle \(\alpha ,\) we have
\[\begin{array}{lllll}\frac{\sin \ \alpha }{a}=\frac{\sin \ \beta }{b} & \\ \frac{\sin \ \alpha }{10}=\frac{\sin (30^{\circ})}{6.013} & \\ \sin \ \alpha =\frac{10\sin (30^{\circ})}{6.013} & \text{Multiply both sides of the equation by 10}. \\ \alpha ={\sin }^{-1}(\frac{10\sin (30^{\circ})}{6.013})\begin{array}{llll} & & & \end{array} & \text{Find the inverse sine of }\frac{10\sin (30^{\circ})}{6.013}. \\ \alpha \approx 56.3^{\circ} & \end{array}\]The other possibility for \(\alpha\) would be \(\alpha =180^{\circ}-56.3^{\circ}\approx 123.7^{\circ}.\) In the original diagram, \(\alpha\) is adjacent to the longest side, so \(\alpha\) is an acute angle and, therefore, \(123.7^{\circ}\) does not make sense. Notice that if we choose to apply the Law of Cosines, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between \(0^{\circ}\) and \(180^{\circ}.\) Proceeding with \(\alpha \approx 56.3^{\circ},\) we can then find the third angle of the triangle.
\[\gamma =180^{\circ}-30^{\circ}-56.3^{\circ}\approx 93.7^{\circ}\]The complete set of angles and sides is
\[\begin{array}{lllll}\alpha \approx 56.3^{\circ}\begin{array}{llll} & & & \end{array} & a=10 \\ \beta =30^{\circ} & b\approx 6.013 \\ \gamma \approx 93.7^{\circ} & c=12\end{array}\] -
Find the missing side and angles of the given triangle: \(\alpha =30^{\circ},\) \(b=12,\) \(c=24.\)
Odhaliť odpoveď
\(a\approx 14.9,\) \(\beta \approx 23.8^{\circ},\) \(\gamma \approx 126.2^{\circ}.\)
-
Find the angle \(\alpha\) for the given triangle if side \(a=20,\) side \(b=25,\) and side \(c=18.\)
Odhaliť odpoveď
For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle \(\alpha ,\) we have
\[\begin{array}{llll} & & & \\ {a}^{2}={b}^{2}+{c}^{2}-2bc\cos \ \alpha & & & \\ {20}^{2}={25}^{2}+{18}^{2}-2(25)(18)\cos \ \alpha & & & \text{Substitute the appropriate measurements}. \\ 400=625+324-900\cos \ \alpha & & & \text{Simplify in each step}. \\ 400=949-900\cos \ \alpha & & & \\ -549=-900\cos \ \alpha & & & \text{Isolate cos }\alpha . \\ \frac{-549}{-900}=\cos \ \alpha & & & \\ 0.61\approx \cos \ \alpha & & & \\ {\cos }^{-1}(0.61)\approx \alpha & & & \text{Find the inverse cosine}. \\ \alpha \approx 52.4^{\circ} & & & \end{array}\]See .
-
Given \(a=5,b=7,\) and \(c=10,\) find the missing angles.
Odhaliť odpoveď
\(\alpha \approx 27.7^{\circ},\) \(\beta \approx 40.5^{\circ},\) \(\gamma \approx 111.8^{\circ}\)
-
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 5,050 feet from the first tower and 2,420 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.
Odhaliť odpoveď
For simplicity, we start by drawing a diagram similar to and labeling our given information.
Using the Law of Cosines, we can solve for the angle \(\theta .\) Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let \(a=2,420,b=5,050,\) and \(c=6,000.\) Thus, \(\theta\) corresponds to the opposite side \(a=2,420.\)
\[\begin{array}{lllll}{a}^{2} & = & {b}^{2}+{c}^{2}-2bc\cos \theta \\ {(2,420)}^{2} & = & {(5,050)}^{2}+{(6,000)}^{2}-2(5,050)(6,000)\cos \theta \\ \begin{array}{lll}{(2,420)}^{2} & - & {(5,050)}^{2}-{(6,000)}^{2}\end{array} & = & -2(5,050)(6,000)\cos \theta \\ \frac{\begin{array}{lll}{(2,420)}^{2} & - & {(5,050)}^{2}-{(6,000)}^{2}\end{array}}{-2(5,050)(6,000)} & = & \cos \theta \\ \cos \theta & \approx & 0.9183 \\ \theta & \approx & {\cos }^{-1}(0.9183) \\ \theta & \approx & 23.3^{\circ}\end{array}\]To answer the questions about the phone’s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in . This forms two right triangles, although we only need the right triangle that includes the first tower for this problem.
Using the angle \(\theta =23.3^{\circ}\) and the basic trigonometric identities, we can find the solutions. Thus
\[\begin{array}{l}\begin{array}{l} \\ \ \ \ \cos (23.3^{\circ})=\frac{x}{5,050}\end{array} \\ x=5,050\cos (23.3^{\circ}) \\ x\approx 4,638.15\ \text{feet} \\ \ \sin (23.3^{\circ})=\frac{y}{5,050} \\ y=5,050\sin (23.3^{\circ}) \\ y\approx 1,997.5\ \text{feet} \\ \end{array}\]The cell phone is approximately 4,638 feet east and 1998 feet north of the first tower, and 1998 feet 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 .
Odhaliť odpoveď
The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, \(180^{\circ}-20^{\circ}=160^{\circ}.\) With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the boat to the port.
\[\begin{array}{l}{x}^{2}={8}^{2}+{10}^{2}-2(8)(10)\cos (160^{\circ}) \\ {x}^{2}=314.35 \\ x=\sqrt{314.35} \\ x\approx 17.7\ \text{miles}\end{array}\]The boat is about 17.7 miles from port.
-
Find the area of the triangle in using Heron’s formula.
Odhaliť odpoveď
First, we calculate \(s.\)
\[\begin{array}{l}s=\frac{(a+b+c)}{2} \\ s=\frac{(10+15+7)}{2}=16\end{array}\]Then we apply the formula.
\[\begin{array}{l}\text{Area}=\sqrt{s(s-a)(s-b)(s-c)} \\ \text{Area}=\sqrt{16(16-10)(16-15)(16-7)} \\ \text{Area}\approx 29.4\end{array}\]The area is approximately 29.4 square units.
-
Use Heron’s formula to find the area of a triangle with sides of lengths \(a=29.7\ \text{ft},b=42.3\ \text{ft},\) and \(c=38.4\ \text{ft}.\)
Odhaliť odpoveď
Area = 552 square feet
-
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 for a view of the city property.
Odhaliť odpoveď
Find the measurement for \(s,\) which is one-half of the perimeter.
\[\begin{array}{l}s=\frac{(62.4+43.5+34.1)}{2} \\ s=70\ \text{m}\end{array}\]Apply Heron’s formula.
\[\begin{array}{l}\text{Area}=\sqrt{70(70-62.4)(70-43.5)(70-34.1)} \\ \text{Area}=\sqrt{506,118.2} \\ \text{Area}\approx 711.4\end{array}\]The developer has about 711.4 square meters.
-
Find the area of a triangle given \(a=4.38\ \text{ft}\ ,b=3.79\ \text{ft,}\) and \(c=5.22\ \text{ft}\text{.}\)
Odhaliť odpoveď
about 8.15 square feet
-
If you are looking for a missing side of a triangle, what do you need to know when using the Law of Cosines?
Odhaliť odpoveď
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 \(s\) represents in Heron’s formula.
Odhaliť odpoveď
\(s\) is the semi-perimeter, which is half the perimeter of the triangle.
-
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?
Odhaliť odpoveď
The Law of Cosines must be used for any oblique (non-right) triangle.
-
\(\gamma =41.2^{\circ},a=2.49,b=3.13\)
-
\(\alpha =120^{\circ},b=6,c=7\)
Odhaliť odpoveď
11.3
-
\(\beta =58.7^{\circ},a=10.6,c=15.7\)
-
\(\gamma =115^{\circ},a=18,b=23\)
Odhaliť odpoveď
34.7
-
\(\alpha =119^{\circ},a=26,b=14\)
-
\(\gamma =113^{\circ},b=10,c=32\)
Odhaliť odpoveď
26.7
-
\(\beta =67^{\circ},a=49,b=38\)
-
\(\alpha =43.1^{\circ},a=184.2,b=242.8\)
Odhaliť odpoveď
\(c=257.3,96.7\)
-
\(\alpha =36.6^{\circ},a=186.2,b=242.2\)
-
\(\beta =50^{\circ},a=105,b=45\)
Odhaliť odpoveď
not possible
-
\(a=42,b=19,c=30;\) find angle \(A.\)
-
\(a=14,\ b=13,\ c=20;\) find angle \(C.\)
Odhaliť odpoveď
95.5°
-
\(a=16,b=31,c=20;\) find angle \(B.\)
-
\(a=13,\ b=22,\ c=28;\) find angle \(A.\)
Odhaliť odpoveď
26.9°
-
\(a=108,\ b=132,\ c=160;\) find angle \(C.\)
-
\(A=35^{\circ},b=8,c=11\)
Odhaliť odpoveď
\(B\approx 45.9^{\circ},C\approx 99.1^{\circ},a\approx 6.4\)
-
\(B=88^{\circ},a=4.4,c=5.2\)
-
\(C=121^{\circ},a=21,b=37\)
Odhaliť odpoveď
\(A\approx 20.6^{\circ},B\approx 38.4^{\circ},c\approx 51.1\)
-
\(a=13,b=11,c=15\)
-
\(a=3.1,b=3.5,c=5\)
Odhaliť odpoveď
\(A\approx 37.8^{\circ},B\approx 43.8,C\approx 98.4^{\circ}\)
-
\(a=51,b=25,c=29\)
-
Find the area of a triangle with sides of length 18 in, 21 in, and 32 in. Round to the nearest tenth.
Odhaliť odpoveď
177.56 in2
-
Find the area of a triangle with sides of length 20 cm, 26 cm, and 37 cm. Round to the nearest tenth.
-
\(a=\frac{1}{2}\ \text{m},b=\frac{1}{3}\ \text{m},c=\frac{1}{4}\ \text{m}\)
Odhaliť odpoveď
0.04 m2
-
\(a=12.4\text{ ft},\ b=13.7\text{ ft},\ c=20.2\text{ ft}\)
Symbols used here
The non-negative number whose square (n-th power) is x.
The usual name for an angle.
Ratios of sides in a right triangle; coordinates on the unit circle.
1/360 of a full turn. 180° = π radians.
Equal to the precision shown, not exactly.
Ratio of a circle's circumference to its diameter, 3.14159…
The angle whose sine is the given value (and likewise arccos, arctan).
How to: Non-right Triangles: Law of Cosines
- Use the Law of Cosines to solve oblique triangles.
- Solve applied problems using the Law of Cosines.
- Use Heron’s formula to find the area of a triangle.
- Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.
- Apply the Law of Cosines to find the length of the unknown side or angle.
- Apply the
- Compute the measure of the remaining angle.
- The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles.
Questions people ask
Why radians instead of degrees?
A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.
Why does sin x = 1/2 have infinitely many solutions?
Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.
How do I remember the exact values?
Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.
Vyskúšajte si vlastné
Parts of this page are adapted from OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
Viac v kategórii Trigonometry
The unit circleTrigonometric equationsTrigonometric identitiesDegrees and radiansRight-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosinesGraphs of sine, cosine and tangentInverse trigonometric functions