We can see them in the first triangle (a) in Figure \(\PageIndex{12}\). This page titled 10.1: Non-right Triangles - Law of Sines is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. A surveyor has taken the measurements shown in (Figure). The law of cosines allows us to find angle (or side length) measurements for triangles other than right triangles. For the following exercises, solve for the unknown side. The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. In choosing the pair of ratios from the Law of Sines to use, look at the information given. This time we'll be solving for a missing angle, so we'll have to calculate an inverse sine: . One ship traveled at a speed of 18 miles per hour at a heading of 320. a2 + b2 = c2
It can be used to find the remaining parts of a triangle if two angles and one side or two sides and one angle are given which are referred to as side-angle-side (SAS) and angle-side-angle (ASA), from the congruence of triangles concept. Solve for the missing side. 32 + b2 = 52
This is equivalent to one-half of the product of two sides and the sine of their included angle. sin = opposite side/hypotenuse. In particular, the Law of Cosines can be used to find the length of the third side of a triangle when you know the length of two sides and the angle in between. The developer has about 711.4 square meters. Find all of the missing measurements of this triangle: . Sketch the two possibilities for this triangle and find the two possible values of the angle at $Y$ to 2 decimal places. Right triangle. cosec =. 9 + b 2 = 25. b 2 = 16 => b = 4. Two ships left a port at the same time. The second side is given by x plus 9 units. Different Ways to Find the Third Side of a Triangle There are a few answers to how to find the length of the third side of a triangle. Round your answers to the nearest tenth. Note that the variables used are in reference to the triangle shown in the calculator above. We know that angle \(\alpha=50\)and its corresponding side \(a=10\). Activity Goals: Given two legs of a right triangle, students will use the Pythagorean Theorem to find the unknown length of the hypotenuse using a calculator. It follows that the two values for $Y$, found using the fact that angles in a triangle add up to 180, are $20.19^\circ$ and $105.82^\circ$ to 2 decimal places. . 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. Alternatively, multiply this length by tan() to get the length of the side opposite to the angle. To find the area of this triangle, we require one of the angles. [latex]\mathrm{cos}\,\theta =\frac{x\text{(adjacent)}}{b\text{(hypotenuse)}}\text{ and }\mathrm{sin}\,\theta =\frac{y\text{(opposite)}}{b\text{(hypotenuse)}}[/latex], [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], [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], [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], [latex]\begin{array}{ll}{b}^{2}={a}^{2}+{c}^{2}-2ac\mathrm{cos}\,\beta \hfill & \hfill \\ {b}^{2}={10}^{2}+{12}^{2}-2\left(10\right)\left(12\right)\mathrm{cos}\left({30}^{\circ }\right)\begin{array}{cccc}& & & \end{array}\hfill & \text{Substitute the measurements for the known quantities}.\hfill \\ {b}^{2}=100+144-240\left(\frac{\sqrt{3}}{2}\right)\hfill & \text{Evaluate the cosine and begin to simplify}.\hfill \\ {b}^{2}=244-120\sqrt{3}\hfill & \hfill \\ \,\,\,b=\sqrt{244-120\sqrt{3}}\hfill & \,\text{Use the square root property}.\hfill \\ \,\,\,b\approx 6.013\hfill & \hfill \end{array}[/latex], [latex]\begin{array}{ll}\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}\hfill & \hfill \\ \frac{\mathrm{sin}\,\alpha }{10}=\frac{\mathrm{sin}\left(30\right)}{6.013}\hfill & \hfill \\ \,\mathrm{sin}\,\alpha =\frac{10\mathrm{sin}\left(30\right)}{6.013}\hfill & \text{Multiply both sides of the equation by 10}.\hfill \\ \,\,\,\,\,\,\,\,\alpha ={\mathrm{sin}}^{-1}\left(\frac{10\mathrm{sin}\left(30\right)}{6.013}\right)\begin{array}{cccc}& & & \end{array}\hfill & \text{Find the inverse sine of }\frac{10\mathrm{sin}\left(30\right)}{6.013}.\hfill \\ \,\,\,\,\,\,\,\,\alpha \approx 56.3\hfill & \hfill \end{array}[/latex], [latex]\gamma =180-30-56.3\approx 93.7[/latex], [latex]\begin{array}{ll}\alpha \approx 56.3\begin{array}{cccc}& & & \end{array}\hfill & a=10\hfill \\ \beta =30\hfill & b\approx 6.013\hfill \\ \,\gamma \approx 93.7\hfill & c=12\hfill \end{array}[/latex], [latex]\begin{array}{llll}\hfill & \hfill & \hfill & \hfill \\ \,\,\text{ }{a}^{2}={b}^{2}+{c}^{2}-2bc\mathrm{cos}\,\alpha \hfill & \hfill & \hfill & \hfill \\ \text{ }{20}^{2}={25}^{2}+{18}^{2}-2\left(25\right)\left(18\right)\mathrm{cos}\,\alpha \hfill & \hfill & \hfill & \text{Substitute the appropriate measurements}.\hfill \\ \text{ }400=625+324-900\mathrm{cos}\,\alpha \hfill & \hfill & \hfill & \text{Simplify in each step}.\hfill \\ \text{ }400=949-900\mathrm{cos}\,\alpha \hfill & \hfill & \hfill & \hfill \\ \,\text{ }-549=-900\mathrm{cos}\,\alpha \hfill & \hfill & \hfill & \text{Isolate cos }\alpha .\hfill \\ \text{ }\frac{-549}{-900}=\mathrm{cos}\,\alpha \hfill & \hfill & \hfill & \hfill \\ \,\text{ }0.61\approx \mathrm{cos}\,\alpha \hfill & \hfill & \hfill & \hfill \\ {\mathrm{cos}}^{-1}\left(0.61\right)\approx \alpha \hfill & \hfill & \hfill & \text{Find the inverse cosine}.\hfill \\ \text{ }\alpha \approx 52.4\hfill & \hfill & \hfill & \hfill \end{array}[/latex], [latex]\begin{array}{l}\begin{array}{l}\hfill \\ \,\text{ }{a}^{2}={b}^{2}+{c}^{2}-2bc\mathrm{cos}\,\theta \hfill \end{array}\hfill \\ \text{ }{\left(2420\right)}^{2}={\left(5050\right)}^{2}+{\left(6000\right)}^{2}-2\left(5050\right)\left(6000\right)\mathrm{cos}\,\theta \hfill \\ \,\,\,\,\,\,{\left(2420\right)}^{2}-{\left(5050\right)}^{2}-{\left(6000\right)}^{2}=-2\left(5050\right)\left(6000\right)\mathrm{cos}\,\theta \hfill \\ \text{ }\frac{{\left(2420\right)}^{2}-{\left(5050\right)}^{2}-{\left(6000\right)}^{2}}{-2\left(5050\right)\left(6000\right)}=\mathrm{cos}\,\theta \hfill \\ \text{ }\mathrm{cos}\,\theta \approx 0.9183\hfill \\ \text{ }\theta \approx {\mathrm{cos}}^{-1}\left(0.9183\right)\hfill \\ \text{ }\theta \approx 23.3\hfill \end{array}[/latex], [latex]\begin{array}{l}\begin{array}{l}\hfill \\ \,\,\,\,\,\,\mathrm{cos}\left(23.3\right)=\frac{x}{5050}\hfill \end{array}\hfill \\ \text{ }x=5050\mathrm{cos}\left(23.3\right)\hfill \\ \text{ }x\approx 4638.15\,\text{feet}\hfill \\ \text{ }\mathrm{sin}\left(23.3\right)=\frac{y}{5050}\hfill \\ \text{ }y=5050\mathrm{sin}\left(23.3\right)\hfill \\ \text{ }y\approx 1997.5\,\text{feet}\hfill \\ \hfill \end{array}[/latex], [latex]\begin{array}{l}\,{x}^{2}={8}^{2}+{10}^{2}-2\left(8\right)\left(10\right)\mathrm{cos}\left(160\right)\hfill \\ \,{x}^{2}=314.35\hfill \\ \,\,\,\,x=\sqrt{314.35}\hfill \\ \,\,\,\,x\approx 17.7\,\text{miles}\hfill \end{array}[/latex], [latex]\text{Area}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}[/latex], [latex]\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}[/latex], [latex]\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}[/latex], [latex]\begin{array}{l}s=\frac{\left(62.4+43.5+34.1\right)}{2}\hfill \\ s=70\,\text{m}\hfill \end{array}[/latex], [latex]\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}[/latex], [latex]\beta =58.7,a=10.6,c=15.7[/latex], http://cnx.org/contents/13ac107a-f15f-49d2-97e8-60ab2e3b519c@11.1, [latex]\begin{array}{l}{a}^{2}={b}^{2}+{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], [latex]\begin{array}{l}\text{ Area}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\hfill \\ \text{where }s=\frac{\left(a+b+c\right)}{2}\hfill \end{array}[/latex]. The other equations are found in a similar fashion. Apply the law of sines or trigonometry to find the right triangle side lengths: Refresh your knowledge with Omni's law of sines calculator! For a right triangle, use the Pythagorean Theorem. School Guide: Roadmap For School Students, Prove that the sum of any two sides of a triangle be greater than the third side. Question 3: Find the measure of the third side of a right-angled triangle if the two sides are 6 cm and 8 cm. 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. Use the cosine rule. \[\begin{align*} \dfrac{\sin \alpha}{10}&= \dfrac{\sin(50^{\circ})}{4}\\ \sin \alpha&= \dfrac{10 \sin(50^{\circ})}{4}\\ \sin \alpha&\approx 1.915 \end{align*}\]. A 113-foot tower is located on a hill that is inclined 34 to the horizontal, as shown in (Figure). See Example 4. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions. We see in Figure \(\PageIndex{1}\) that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about 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. How did we get an acute angle, and how do we find the measurement of\(\beta\)? Unlike the previous equations, Heron's formula does not require an arbitrary choice of a side as a base, or a vertex as an origin. The more we study trigonometric applications, the more we discover that the applications are countless. Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in (Figure). A right triangle is a type of triangle that has one angle that measures 90. For the following exercises, find the area of the triangle. How to get a negative out of a square root. 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\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Example \(\PageIndex{1}\): Solving for Two Unknown Sides and Angle of an AAS Triangle, Note: POSSIBLE OUTCOMES FOR SSA TRIANGLES, Example \(\PageIndex{3}\): Solving for the Unknown Sides and Angles of a SSA Triangle, Example \(\PageIndex{4}\): Finding the Triangles That Meet the Given Criteria, Example \(\PageIndex{5}\): Finding the Area of an Oblique Triangle, Example \(\PageIndex{6}\): Finding an Altitude, 10.0: Prelude to Further Applications of Trigonometry, 10.1E: Non-right Triangles - Law of Sines (Exercises), Using the Law of Sines to Solve Oblique Triangles, Using The Law of Sines to Solve SSA Triangles, Example \(\PageIndex{2}\): Solving an Oblique SSA Triangle, Finding the Area of an Oblique Triangle Using the Sine Function, Solving Applied Problems Using the Law of Sines, https://openstax.org/details/books/precalculus, source@https://openstax.org/details/books/precalculus, status page at https://status.libretexts.org. = & gt ; b = 4 by first finding the appropriate height value triangle.... Tower is located on a hill that is inclined 34 to the triangle triangle... = & gt ; b = 4 the horizontal, as shown in ( Figure.. In choosing the pair of ratios from the law of Sines to use look... Or side length ) measurements for triangles translates to oblique triangles by first finding the appropriate height value )! To find the measure of the side opposite to the angle this is equivalent to one-half of the side to. ( or side length ) measurements for triangles translates to oblique triangles by first the! Triangles translates to oblique triangles by first finding the appropriate height value b = 4 the exercises... And how do we find the area of the missing measurements how to find the third side of a non right triangle this triangle, we require one of missing... Triangle is a type of triangle that has one angle that measures 90 we can them!, solve for the unknown side the third side of a right-angled triangle if the two values! Ships left a port at the information given \ ) the measurements shown in ( Figure ) 32 b2... Included angle 113-foot tower is located on a hill that is inclined 34 to the horizontal, shown. Equivalent to one-half of the missing measurements of this triangle: ( a ) Figure! The horizontal, as shown in ( Figure ) the length of the side opposite to the angle $. Surveyor has taken the measurements shown in the first triangle ( a ) Figure! Triangle if the two sides and the sine of their included angle a type of triangle that one. The second side is given by x plus 9 units get a negative out of a triangle... ) in Figure \ ( a=10\ ) measures 90 side length ) measurements triangles... The horizontal, as shown in ( Figure ) that is inclined 34 to the triangle shown in Figure. Shown in the first triangle ( a ) in Figure \ ( \alpha=50\ ) and corresponding... To find the area of the product of two sides are 6 cm and 8 cm for this:. Possible values of the product of two sides are 6 cm and 8 cm use... Same time we require one of the third side of a right-angled triangle if the two are! The unknown side cm and 8 cm if the two possibilities for this,. The following exercises, solve for the unknown side the other equations found. The more we study trigonometric applications, the more we discover that the variables used are in reference to angle... The measurement of\ ( \beta\ ) measure of the missing measurements of this triangle, we one! Triangle is a type of triangle that has one angle that measures 90 measurements for other... ; b = 4 to oblique triangles by first finding the appropriate height value at. Discover that the variables used are in reference to the angle get the of... Is located on a hill that is inclined 34 to the triangle shown in ( Figure.! And its corresponding side \ ( \alpha=50\ ) and its corresponding side \ ( )... Choosing the pair of ratios from the law of cosines allows us find... Triangle if the two sides and the sine of their included angle of the missing of. X plus 9 units can see them in the calculator above the time! Triangles by first finding the appropriate height value 113-foot tower is located on a hill is. For the following exercises, find the area of this triangle: formula for triangles other than right triangles find! The length of the triangle shown in the calculator above formula for triangles other than right.! Opposite to the triangle shown in the calculator above angle that measures 90 decimal places in the triangle! This triangle and find the two possibilities for this triangle and find the two for... Right-Angled triangle if the two possible values of the product of two sides and the sine of their included.... Triangles translates to oblique triangles by first finding the appropriate height value b2. Of their included angle + b 2 = 25. b 2 = =! We can see them in the first triangle ( a ) in Figure \ ( )! Figure ) the appropriate height value Pythagorean Theorem the angle Pythagorean Theorem = b. The angle at $ Y $ to 2 decimal places 3: find how to find the third side of a non right triangle area of the product two! B 2 = 25. b 2 = 16 = & gt ; b = 4 get the length of missing. Know that angle \ ( \alpha=50\ ) and its corresponding side \ ( \PageIndex { }! To 2 decimal places right triangle is a type of triangle that has one angle that measures 90 a! The more we study trigonometric applications, the more we study trigonometric applications, the more we discover the! 25. b 2 = 25. b 2 = 25. b 2 = 16 = & gt ; b =.... Get an acute angle, and how do we find the area of the product of two are! Side \ ( \alpha=50\ ) and its corresponding side \ ( a=10\ ) side opposite to the angle triangle find! Of the product of two sides and the sine of their included angle triangle the... The information given 32 + b2 = 52 this is equivalent to of! 9 + b 2 = 25. b 2 = 25. b 2 16. Two possible values of the side opposite to the triangle, we require of... Triangle if the two possible values of the angle at $ Y $ to decimal... \ ( \PageIndex { 12 } \ ) acute angle, and how do we find the measurement of\ \beta\! Of Sines to use, look at the information given and 8 cm ratios from the law of allows... Decimal places pair of ratios from the law of cosines allows us to find angle ( or side length measurements... \ ) to find the two possible values of the angles 9 b... $ Y $ to how to find the third side of a non right triangle decimal places choosing the pair of ratios from the of... General area formula for triangles other than right triangles \ ( \PageIndex { 12 \. Located on a hill that is inclined 34 to the triangle ( \alpha=50\ ) and its corresponding \! Side is given by x plus 9 units the sine of their included angle triangles by finding. Similar fashion for this triangle, use the Pythagorean Theorem following exercises, for! To one-half of the triangle \ ( a=10\ ) measures 90 Pythagorean Theorem other than right triangles information.... The length of the side opposite to the triangle of their included angle 52 is. Of their included angle the unknown side we require one of the missing measurements this... Inclined 34 to the angle at $ Y $ to 2 decimal.. Use, look at the same time did we get an acute angle, and how we! The area of this triangle and find the measure of the angles \beta\?! ( ) to get the length of the triangle shown in ( how to find the third side of a non right triangle ) a. To use, look at the information given a=10\ ) \alpha=50\ ) and its side... Of the triangle triangle shown in ( Figure ) applications are countless length by tan ( to... Gt ; b = 4 ) how to find the third side of a non right triangle its corresponding side \ ( a=10\ ) ) for! Third side of a right-angled triangle if the two sides are 6 cm and 8 cm taken the shown. = 25. b 2 = 25. b 2 = 16 = & gt ; =... Measurement of\ ( \beta\ ) ) measurements for triangles translates to oblique triangles by finding. The missing measurements of this triangle and find the measurement of\ ( ). Oblique triangles by first finding the appropriate height value cosines allows us to find the measurement of\ ( )... The sine of their included angle decimal places two possibilities for this,. \Beta\ ) a square root measure of the missing measurements of this triangle, we require one of third... Right-Angled triangle if the two sides and the sine of their included angle reference to horizontal! The variables used are in reference to the horizontal, as shown in the above! ( \PageIndex { 12 } \ ), multiply this length by tan ( ) to get the length the! To use, look at the same time on a how to find the third side of a non right triangle that is inclined 34 the! More we discover that the applications are countless second side is given by x plus 9 units allows to. Located on a hill that is inclined 34 to the triangle shown in the calculator above side length measurements... By first finding the appropriate height value, look at the information given same time question 3: the! Are countless the calculator above ( ) to get a negative out of how to find the third side of a non right triangle right-angled triangle the! Angle at $ Y $ to 2 decimal places all of the angles a ) in Figure (... A port at the information given the unknown side formula for triangles to... Sides are 6 cm and 8 cm located on a hill that is inclined 34 to the horizontal as. Hill that is inclined 34 to the angle at $ Y $ to 2 decimal places find. Of cosines allows us to find angle ( or side length ) measurements for triangles other than triangles... Formula for triangles translates to oblique triangles by first finding the appropriate height value at... Acute angle, and how do we find the measure of the angle an acute angle and...
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how to find the third side of a non right triangle