Practice – Non Right-Angled Triangle Trigonometry 117 June 12, 2020 1. 2. At the corner, a park is being built in the shape of a triangle. How did we get an acute angle, and how do we find the measurement of[latex]\,\beta ?\,[/latex]Let’s investigate further. Assuming that the street is level, estimate the height of the building to the nearest foot. Find the distance of the plane from point[latex]\,A\,[/latex]to the nearest tenth of a kilometer. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. A triangle with two given sides and a non-included angle. To determine how far a boat is from shore, two radar stations 500 feet apart find the angles out to the boat, as shown in (Figure). As the GCSE mathematics curriculum increasingly challenges students to solve multiple step problems it is important for students to understand how to prove, apply and link together the various formulae associated to non-righ… There are three possible cases: ASA, AAS, SSA. The cosine rule can be used to find a missing side when all sides and an angle are involved in the question. Find[latex]\,AD\,[/latex]in (Figure). A yield sign measures 30 inches on all three sides. Round each answer to the nearest tenth. There are three possible cases: ASA, AAS, SSA. To do so, we need to start with at least three of these values, including at least one of the sides. A pilot is flying over a straight highway. Calculate the area of the triangle ABC. A pole leans away from the sun at an angle of[latex]\,7°\,[/latex]to the vertical, as shown in (Figure). We will investigate three possible oblique triangle problem situations: 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. The Law of Sines can be used to solve oblique triangles, which are non-right triangles. Trigonometry in Non-Right Angled Triangles Area of a Triangle You need to know 2 sides and the angle in between. The Law of Sines can be used to solve oblique triangles, which are non-right triangles. Roll over or tap the triangle to see what that means … Students learn how to derive the Sine, Cosine and Area formulae for non-right-angled triangles. It may also be used to find a missing angle if all the sides of a non-right angled triangle are known. These formulae represent the cosine rule. Read more. Since a must be positive, the value of c in the original question is 4.54 cm. How long is the pole? See, The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. Solve applied problems using the Law of Sines. This formula represents the sine rule. (Hint: Draw a perpendicular from[latex]\,N\,[/latex]to[latex]\,LM).\,[/latex]Round each answer to the nearest tenth. (Figure) shows a satellite orbiting Earth. This is a good indicator to use the sine rule in a question rather than the cosine rule. In this case, we know the angle[latex]\,\gamma =85°,\,[/latex]and its corresponding side[latex]\,c=12,\,[/latex]and we know side[latex]\,b=9.\,[/latex]We will use this proportion to solve for[latex]\,\beta .[/latex]. Note that the angle of elevation is the angle up from the ground; for example, if you look up at something, this angle is the angle between the ground and your line of site.. Round to the nearest tenth. See, The Law of Sines can be used to solve triangles with given criteria. Find the angle marked x in the following triangle to 3 decimal places: Note how much accuracy is retained throughout this calculation. How can we determine the altitude of the aircraft? What is the altitude of the climber? Find the area of the triangle given[latex]\,\beta =42°,\,\,a=7.2\,\text{ft},\,\,c=3.4\,\text{ft}.\,[/latex]Round the area to the nearest tenth. • Detailed solution with non-right-angled triangle trigonometry formulas. MS-M6 - Non-right-angled trigonometry Measurement It is the responsibility of individual teachers to ensure their students are adequately prepared for the HSC examinations, identifying the suitability of resources, and adapting resources to the students’ context when required. For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. The inverse sine will produce a single result, but keep in mind that there may be two values for[latex]\,\beta .\,[/latex]It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions. Find the diameter of the circle in (Figure). Solve both triangles in (Figure). A street light is mounted on a pole. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship. Observing the two triangles in (Figure), one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property[latex]\,\mathrm{sin}\,\alpha =\frac{\text{opposite}}{\text{hypotenuse}}\,[/latex]to write an equation for area in oblique triangles. The angle of elevation from the tip of the man’s shadow to the top of his head of 28°. Use the Law of Sines to find angle[latex]\,\beta \,[/latex]and angle[latex]\,\gamma ,\,[/latex]and then side[latex]\,c.\,[/latex]Solving for[latex]\,\beta ,\,[/latex]we have the proportion. We see in (Figure) 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. Points[latex]\,A\,[/latex]and[latex]\,B\,[/latex]are on opposite sides of a lake. The angle of inclination of the hill is[latex]\,67°.\,[/latex]A guy wire is to be attached to the top of the tower and to the ground, 165 meters downhill from the base of the tower. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Students tend to memorise the bottom one as it is the one that looks most like Pythagoras. Given[latex]\,\alpha =80°,a=100,\,\,b=10,\,[/latex]find the missing side and angles. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution. When the known values are the side opposite the missing angle and another side and its opposite angle. To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side[latex]\,a,[/latex] and then use right triangle relationships to find the height of the aircraft,[latex]\,h.[/latex]. Again, it is not necessary to memorise them all – one will suffice (see Example 2 for relabelling). Trigonometry The three trigonometric ratios; sine, cosine and tangent are used to calculate angles and lengths in right-angled triangles. The interior angles of a triangle always add up to 180° while the exterior angles of a triangle are equal to the sum of the two interior angles that are not adjacent to it. In this case, if we subtract[latex]\,\beta \,[/latex]from 180°, we find that there may be a second possible solution. Solve the triangle shown in (Figure) to the nearest tenth. Using the sine and cosine rules in non right angled triangles to find the missing sides and angles, and a brief look at the ambiguity in the Sine rule. To find the area of this triangle, we require one of the angles. Find[latex]\,AB\,[/latex]in the parallelogram shown in (Figure). Read about Non-right Triangle Trigonometry (Trigonometry Reference) in our free Electronics Textbook 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. Non-right angled triangles - cosine and sine rule - StudyWell For right-angled triangles, we have Pythagoras’ Theorem and SOHCAHTOA. The rule also stands if you write the entire thing the other way up. To find the remaining missing values, we calculate[latex]\,\alpha =180°-85°-48.3°\approx 46.7°.\,[/latex]Now, only side[latex]\,a\,[/latex]is needed. [/latex], The formula for the area of an oblique triangle is given by. Using trigonometry: tan=35=tan−135=30.96° Labelling Sides of Non-Right Angle Triangles. Solving for a side in right … Entering sides of values 1.00, 2.00, and 2.00 will yield much more acurate results of 75.5, 75.5, and 29.0. Trigonometry – Non-Right-Angled Triangles Lessons Powerpoint comes with two assessments, a homework and revision questions. Round the answer to the nearest tenth. They then move 300 feet closer to the building and find the angle of elevation to be 50°. For the following exercises, find the area of the triangle with the given measurements. Two search teams spot a stranded climber on a mountain. If there is more than one possible solution, show both. Are you ready to test your Pure Maths knowledge? Non - Right angled Trigonometry. Here are some types of word problems (applications) that you might see when studying right angle trigonometry.. In order to estimate the height of a building, two students stand at a certain distance from the building at street level. We can stop here without finding the value of[latex]\,\alpha .\,[/latex]Because the range of the sine function is[latex]\,\left[-1,1\right],\,[/latex]it is impossible for the sine value to be 1.915. This gives, which is impossible, and so[latex]\,\beta \approx 48.3°.[/latex]. one triangle,[latex]\,\alpha \approx 50.3°,\beta \approx 16.7°,a\approx 26.7[/latex], [latex]b=3.5,\,\,c=5.3,\,\,\gamma =\,80°[/latex], [latex]a=12,\,\,c=17,\,\,\alpha =\,35°[/latex], two triangles,[latex] \,\gamma \approx 54.3°,\beta \approx 90.7°,b\approx 20.9[/latex]or[latex] {\gamma }^{\prime }\approx 125.7°,{\beta }^{\prime }\approx 19.3°,{b}^{\prime }\approx 6.9[/latex], [latex]a=20.5,\,\,b=35.0,\,\,\beta =25°[/latex], [latex]a=7,\,c=9,\,\,\alpha =\,43°[/latex], two triangles,[latex] \beta \approx 75.7°, \gamma \approx 61.3°,b\approx 9.9[/latex]or[latex] {\beta }^{\prime }\approx 18.3°,{\gamma }^{\prime }\approx 118.7°,{b}^{\prime }\approx 3.2[/latex], two triangles,[latex]\,\alpha \approx 143.2°,\beta \approx 26.8°,a\approx 17.3\,[/latex]or[latex]\,{\alpha }^{\prime }\approx 16.8°,{\beta }^{\prime }\approx 153.2°,{a}^{\prime }\approx 8.3[/latex]. [latex]\beta \approx 5.7°,\gamma \approx 94.3°,c\approx 101.3[/latex]. For the following exercises, find the area of each triangle. The Greeks focused on the calculation of chords, while mathematicians in India … A man and a woman standing[latex]\,3\frac{1}{2}\,[/latex]miles apart spot a hot air balloon at the same time. Round each answer to the nearest tenth. In fact, inputting[latex]\,{\mathrm{sin}}^{-1}\left(1.915\right)\,[/latex]in a graphing calculator generates an ERROR DOMAIN. There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. The Bermuda triangle is a region of the Atlantic Ocean that connects Bermuda, Florida, and Puerto Rico. 1. PRO Features : 1) View calculation steps 2) View formulas 3) No ads • Giving solution based on your input. Access these online resources for additional instruction and practice with trigonometric applications. Solving for a side in a right triangle using the trigonometric ratios. Brian’s house is on a corner lot. This is equivalent to one-half of the product of two sides and the sine of their included angle. Round each answer to the nearest hundredth. It is simply half of b times h. Area = 12 bh (The Triangles page explains more). Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in (Figure). There are several ways to find the area of a triangle. • Support Ambiguous Case. Preview. What is the distance from[latex]\,A\,[/latex]to[latex]\,B,\,[/latex]rounded to the nearest whole meter? Solve the triangle in (Figure). Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion. To find[latex]\,\beta ,\,[/latex]apply the inverse sine function. Find the area of the park if, along one road, the park measures 180 feet, and along the other road, the park measures 215 feet. Solving problems with non-right-angled triangles involves multiple areas of mathematics ranging from complex formulae to angles in a triangle and on a straight line. Similar to an angle of elevation, an angle of depression is the acute angle formed by a horizontal line and an observer’s line of sight to an object below the horizontal. Measurement involves the application of knowledge, skills and understanding of numbers and geometry to quantify and solve problems in practical situations. Use the Law of Sines to solve for[latex]\,a\,[/latex]by one of the proportions. When the elevation of the sun is[latex]\,55°,\,[/latex]the pole casts a shadow 42 feet long on the level ground. Note that it is not necessary to memorise all of them – one will suffice, since a relabelling of the angles and sides will give you the others. Example:- Calculate the area of this triangle. about[latex]\,8.2\,\,\text{square}\,\text{feet}[/latex]. This is different to the cosine rule since two angles are involved. Round each answer to the nearest tenth. Trigonometry: Non Right-Angled Triangles MichaelExamSolutionsKid 2020-03-11T23:34:40+00:00 Trigonometry Non Right Angled Triangles When finding the area of a segment you will often need to find the area of a triangle given two sides and an included angle when the angle is given in degrees or radians. In choosing the pair of ratios from the Law of Sines to use, look at the information given. In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. What type of triangle results in an ambiguous case? The angle of elevation from the second search team to the climber is 22°. Sketch the two possibilities for this triangle and find the two possible values of the angle at Y to 2 decimal places. It follows that the area is given by. The distance from the satellite to station[latex]\,A\,[/latex]is approximately 1716 miles. ), it is very obvious that most triangles that could be constructed for navigational or surveying reasons would not contain a right angle. Naomi bought a modern dining table whose top is in the shape of a triangle. If we rounded earlier and used 4.699 in the calculations, the final result would have been x=26.545 to 3 decimal places and this is incorrect. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. Dropping a perpendicular from[latex]\,\gamma \,[/latex]and viewing the triangle from a right angle perspective, we have (Figure). Find the radius of the circle in (Figure). The most important thing is that the base and height are at right angles. This formula works for a right triangle as well, since the since of 90 is one. However, these methods do not work for non-right angled triangles. Designed to solve triangle trigonometry problem with well explanation. A communications tower is located at the top of a steep hill, as shown in (Figure). See. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°. The sides of a triangle are in arithmetic sequence and the greatest angle is double the smallest angle. The roof of a house is at a[latex]\,20°\,[/latex]angle. Assuming that the street is level, estimate the height of the building to the nearest foot. Find[latex]\,m\angle ADC\,[/latex]in (Figure). A 6-foot-tall man is standing on the street a short distance from the pole, casting a shadow. The diagram shown in (Figure) represents the height of a blimp flying over a football stadium. The angle used in calculation is[latex]\,{\alpha }^{\prime },\,[/latex]or[latex]\,180-\alpha . The angle of elevation from the tip of her shadow to the top of her head is 28°. We will work on three key rules. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. To solve an oblique triangle, use any pair of applicable ratios. MS-M6 Non-right-angled trigonometry. Trigonometry and Non-Right-Angled Triangles. Note that when using the sine rule, it is sometimes possible to get two answers for a given angle\side length, both of which are valid. [/latex], [latex]A\approx 47.8°\,[/latex]or[latex]\,{A}^{\prime }\approx 132.2°[/latex], Find angle[latex]\,B\,[/latex]when[latex]\,A=12°,a=2,b=9.[/latex]. Knowing Base and Height. Trigonometry: Right and Non-Right Triangles Area of a Triangle Using Sine We can use sine to determine the area of non-right triangles. Our mission is to provide a free, world-class education to anyone, anywhere. 3. Answering the question given amounts to finding side a in this new triangle. Round to the nearest tenth. The sine and cosine rules calculate lengths and angles … Need to know one pair (angle and side) plus Round each answer to the nearest tenth. 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. In the acute triangle, we have[latex]\,\mathrm{sin}\,\alpha =\frac{h}{c}\,[/latex]or[latex]c\mathrm{sin}\,\alpha =h.\,[/latex]However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base[latex]\,b\,[/latex]to form a right triangle. (Hint: Draw a perpendicular from[latex]\,H\,[/latex]to[latex]\,JK).\,[/latex]Round each answer to the nearest tenth. Here we take trigonometry to the next level by working with triangles that do not have a right angle. The angle of depression is the angle that comes down from a … When the satellite is on one side of the two stations, the angles of elevation at[latex]\,A\,[/latex]and[latex]\,B\,[/latex]are measured to be[latex]\,86.2°\,[/latex]and[latex]\,83.9°,\,[/latex]respectively. See (Figure). In the triangle shown in (Figure), solve for the unknown side and angles. Round to the nearest tenth of a mile. [latex]\,\angle m\,[/latex]is obtuse. See. Use the Law of Sines to solve oblique triangles. For non-right angled triangles, we have the cosine rule, the sine rule and a new expression for finding area. Compare right triangles and oblique triangles. The Law of Sines is based on proportions and is presented symbolically two ways. The angle supplementary to[latex]\,\beta \,[/latex]is approximately equal to 49.9°, which means that[latex]\,\beta =180°-49.9°=130.1°.\,[/latex](Remember that the sine function is positive in both the first and second quadrants.) © Copyright of StudyWell Publications Ltd. 2020. Similarly, to solve for[latex]\,b,\,[/latex]we set up another proportion. Visit our Practice Papers page and take StudyWell’s own Pure Maths tests. The sine rule can be used to find a missing angle or a missing side when two corresponding pairs of angles and sides are involved in the question. The ambiguous case arises when an oblique triangle can have different outcomes. Therefore, no triangles can be drawn with the provided dimensions. The more we study trigonometric applications, the more we discover that the applications are countless. However, in the diagram, angle[latex]\,\beta \,[/latex]appears to be an obtuse angle and may be greater than 90°. You can round when jotting down working but you should retain accuracy throughout calculations. 4.3 4 customer reviews. If there is more than one possible solution, show both. They then move 250 feet closer to the building and find the angle of elevation to be 53°. Round each answer to the nearest tenth. Right-Angled Triangles: h Non-Right-Angled Triangles: Note that to maintain accuracy, store values on your calculator and leave rounding until the end of the question. In the Law of Sines, what is the relationship between the angle in the numerator and the side in the denominator? Note the standard way of labeling triangles: angle[latex]\,\alpha \,[/latex](alpha) is opposite side[latex]\,a;\,[/latex]angle[latex]\,\beta \,[/latex](beta) is opposite side[latex]\,b;\,[/latex]and angle[latex]\,\gamma \,[/latex](gamma) is opposite side[latex]\,c.\,[/latex]See (Figure). As is the case with the sine rule and the cosine rule, the sides and angles are not fixed. Round the distance to the nearest tenth of a foot. They’re really not significantly different, though the derivation of the formula for a non-right triangle is a little different. It is not possible for a triangle to have more than one vertex with internal angle greater than or equal to 90°, or it would no longer be a triangle. Preview and details Files included (6) pdf, 136 KB. There are three possible cases: ASA, AAS, SSA. Round the altitude to the nearest tenth of a mile. The three trigonometric ratios; sine, cosine and tangent are used to calculate angles and lengths in right-angled triangles. The sine rule will give us the two possibilities for the angle at Z, this time using the second equation for the sine rule above: Solving gives or . The Sine rule, the Cosine rule and the formula for th area of a triangle. For oblique triangles, we must find[latex]\,h\,[/latex]before we can use the area formula. Round answers to the nearest whole mile. The distance from one station to the aircraft is about 14.98 miles. It's a PRO app and easy to use with eye-catching User Interface. However, these methods do not work for non-right angled triangles. Find the length of the side marked x in the following triangle: The triangle PQR has sides PQ=6.5cm, QR=9.7cm and PR = c cm. Round your answers to the nearest tenth. Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180°−15°−35°=130°. Area of Triangles. We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. 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]Solve each triangle, if possible. \hfill \\ \text{ }\,\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}\hfill & \hfill \end{array}[/latex], [latex]\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\gamma }{c}\text{ and }\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\gamma }{c}[/latex], [latex]\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\lambda }{c}[/latex], [latex]\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\gamma }{c}[/latex], [latex]\frac{a}{\mathrm{sin}\,\alpha }=\frac{b}{\mathrm{sin}\,\beta }=\frac{c}{\mathrm{sin}\,\gamma }[/latex], [latex]\begin{array}{l}\begin{array}{l}\hfill \\ \beta =180°-50°-30°\hfill \end{array}\hfill \\ \,\,\,\,=100°\hfill \end{array}[/latex], [latex]\begin{array}{llllll}\,\,\frac{\mathrm{sin}\left(50°\right)}{10}=\frac{\mathrm{sin}\left(30°\right)}{c}\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ c\frac{\mathrm{sin}\left(50°\right)}{10}=\mathrm{sin}\left(30°\right)\hfill & \hfill & \hfill & \hfill & \hfill & \text{Multiply both sides by }c.\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c=\mathrm{sin}\left(30°\right)\frac{10}{\mathrm{sin}\left(50°\right)}\hfill & \hfill & \hfill & \hfill & \hfill & \text{Multiply by the reciprocal to isolate }c.\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\approx 6.5\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \end{array}[/latex], [latex]\begin{array}{ll}\begin{array}{l}\hfill \\ \,\text{ }\frac{\mathrm{sin}\left(50°\right)}{10}=\frac{\mathrm{sin}\left(100°\right)}{b}\hfill \end{array}\hfill & \hfill \\ \text{ }b\mathrm{sin}\left(50°\right)=10\mathrm{sin}\left(100°\right)\hfill & \text{Multiply both sides by }b.\hfill \\ \text{ }b=\frac{10\mathrm{sin}\left(100°\right)}{\mathrm{sin}\left(50°\right)}\begin{array}{cccc}& & & \end{array}\hfill & \text{Multiply by the reciprocal to isolate }b.\hfill \\ \text{ }b\approx 12.9\hfill & \hfill \end{array}[/latex], [latex]\begin{array}{l}\begin{array}{l}\hfill \\ \alpha =50°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a=10\hfill \end{array}\hfill \\ \beta =100°\,\,\,\,\,\,\,\,\,\,\,\,b\approx 12.9\hfill \\ \gamma =30°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\approx 6.5\hfill \end{array}[/latex], [latex]\begin{array}{r}\hfill \frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}\\ \hfill \frac{\mathrm{sin}\left(35°\right)}{6}=\frac{\mathrm{sin}\,\beta }{8}\\ \hfill \frac{8\mathrm{sin}\left(35°\right)}{6}=\mathrm{sin}\,\beta \,\\ \hfill 0.7648\approx \mathrm{sin}\,\beta \,\\ \hfill {\mathrm{sin}}^{-1}\left(0.7648\right)\approx 49.9°\\ \hfill \beta \approx 49.9°\end{array}[/latex], [latex]\gamma =180°-35°-130.1°\approx 14.9°[/latex], [latex]{\gamma }^{\prime }=180°-35°-49.9°\approx 95.1°[/latex], [latex]\begin{array}{l}\frac{c}{\mathrm{sin}\left(14.9°\right)}=\frac{6}{\mathrm{sin}\left(35°\right)}\hfill \\ \text{ }c=\frac{6\mathrm{sin}\left(14.9°\right)}{\mathrm{sin}\left(35°\right)}\approx 2.7\hfill \end{array}[/latex], [latex]\begin{array}{l}\frac{{c}^{\prime }}{\mathrm{sin}\left(95.1°\right)}=\frac{6}{\mathrm{sin}\left(35°\right)}\hfill \\ \text{ }{c}^{\prime }=\frac{6\mathrm{sin}\left(95.1°\right)}{\mathrm{sin}\left(35°\right)}\approx 10.4\hfill \end{array}[/latex], [latex]\begin{array}{ll}\alpha =80°\hfill & a=120\hfill \\ \beta \approx 83.2°\hfill & b=121\hfill \\ \gamma \approx 16.8°\hfill & c\approx 35.2\hfill \end{array}[/latex], [latex]\begin{array}{l}{\alpha }^{\prime }=80°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a}^{\prime }=120\hfill \\ {\beta }^{\prime }\approx 96.8°\,\,\,\,\,\,\,\,\,\,\,\,\,{b}^{\prime }=121\hfill \\ {\gamma }^{\prime }\approx 3.2°\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{c}^{\prime }\approx 6.8\hfill \end{array}[/latex], [latex]\begin{array}{ll}\,\,\,\frac{\mathrm{sin}\left(85°\right)}{12}=\frac{\mathrm{sin}\,\beta }{9}\begin{array}{cccc}& & & \end{array}\hfill & \text{Isolate the unknown}.\hfill \\ \,\frac{9\mathrm{sin}\left(85°\right)}{12}=\mathrm{sin}\,\beta \hfill & \hfill \end{array}[/latex], [latex]\begin{array}{l}\beta ={\mathrm{sin}}^{-1}\left(\frac{9\mathrm{sin}\left(85°\right)}{12}\right)\hfill \\ \beta \approx {\mathrm{sin}}^{-1}\left(0.7471\right)\hfill \\ \beta \approx 48.3°\hfill \end{array}[/latex], [latex]\alpha =180°-85°-131.7°\approx -36.7°,[/latex], [latex]\begin{array}{l}\begin{array}{l}\hfill \\ \begin{array}{l}\hfill \\ \frac{\mathrm{sin}\left(85°\right)}{12}=\frac{\mathrm{sin}\left(46.7°\right)}{a}\hfill \end{array}\hfill \end{array}\hfill \\ \,a\frac{\mathrm{sin}\left(85°\right)}{12}=\mathrm{sin}\left(46.7°\right)\hfill \\ \text{ }\,\,\,\,\,\,a=\frac{12\mathrm{sin}\left(46.7°\right)}{\mathrm{sin}\left(85°\right)}\approx 8.8\hfill \end{array}[/latex], [latex]\begin{array}{l}\begin{array}{l}\hfill \\ \alpha \approx 46.7°\text{ }a\approx 8.8\hfill \end{array}\hfill \\ \beta \approx 48.3°\text{ }b=9\hfill \\ \gamma =85°\text{ }c=12\hfill \end{array}[/latex], [latex]\begin{array}{l}\,\frac{\mathrm{sin}\,\alpha }{10}=\frac{\mathrm{sin}\left(50°\right)}{4}\hfill \\ \,\,\mathrm{sin}\,\alpha =\frac{10\mathrm{sin}\left(50°\right)}{4}\hfill \\ \,\,\mathrm{sin}\,\alpha \approx 1.915\hfill \end{array}[/latex], [latex]\text{Area}=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)=\frac{1}{2}b\left(c\mathrm{sin}\,\alpha \right)[/latex], [latex]\text{Area}=\frac{1}{2}a\left(b\mathrm{sin}\,\gamma \right)=\frac{1}{2}a\left(c\mathrm{sin}\,\beta \right)[/latex], [latex]\begin{array}{l}\text{Area}=\frac{1}{2}bc\mathrm{sin}\,\alpha \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}ac\mathrm{sin}\,\beta \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}ab\mathrm{sin}\,\gamma \hfill \end{array}[/latex], [latex]\begin{array}{l}\text{Area}=\frac{1}{2}ab\mathrm{sin}\,\gamma \hfill \\ \text{Area}=\frac{1}{2}\left(90\right)\left(52\right)\mathrm{sin}\left(102°\right)\hfill \\ \text{Area}\approx 2289\,\,\text{square}\,\,\text{units}\hfill \end{array}[/latex], [latex]\begin{array}{l}\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \end{array}\hfill \\ \text{ }\frac{\mathrm{sin}\left(130°\right)}{20}=\frac{\mathrm{sin}\left(35°\right)}{a}\hfill \end{array}\hfill \\ a\mathrm{sin}\left(130°\right)=20\mathrm{sin}\left(35°\right)\hfill \\ \text{ }a=\frac{20\mathrm{sin}\left(35°\right)}{\mathrm{sin}\left(130°\right)}\hfill \\ \text{ }a\approx 14.98\hfill \end{array}[/latex], [latex]\begin{array}{l}\mathrm{sin}\left(15°\right)=\frac{\text{opposite}}{\text{hypotenuse}}\hfill \\ \mathrm{sin}\left(15°\right)=\frac{h}{a}\hfill \\ \mathrm{sin}\left(15°\right)=\frac{h}{14.98}\hfill \\ \text{ }\,\text{ }h=14.98\mathrm{sin}\left(15°\right)\hfill \\ \text{ }\,h\approx 3.88\hfill \end{array}[/latex],[email protected], [latex]\begin{array}{l}\frac{\mathrm{sin}\,\alpha }{a}=\frac{\mathrm{sin}\,\beta }{b}=\frac{\mathrm{sin}\,\gamma }{c}\,\hfill \\ \frac{a}{\mathrm{sin}\,\alpha }=\frac{b}{\mathrm{sin}\,\beta }=\frac{c}{\mathrm{sin}\,\gamma }\hfill \end{array}[/latex], [latex]\begin{array}{r}\hfill \text{Area}=\frac{1}{2}bc\mathrm{sin}\,\alpha \\ \hfill \text{ }=\frac{1}{2}ac\mathrm{sin}\,\beta \\ \hfill \text{ }=\frac{1}{2}ab\mathrm{sin}\,\gamma \end{array}[/latex]. The satellite is approximately 1706 miles above the ground. [/latex], Find side[latex]\,a[/latex] when[latex]\,A=132°,C=23°,b=10. The satellite passes directly over two tracking stations[latex]\,A\,[/latex]and[latex]\,B,\,[/latex]which are 69 miles apart. In this section, we will find out how to solve problems involving non-right triangles. In triangle XYZ, length XY=6.14m, length YZ=3.8m and the angle at X is 27 degrees. The altitude extends from any vertex to the opposite side or to the line containing the opposite side at a 90° angle. (See (Figure)). Therefore, the complete set of angles and sides is, [latex]\begin{array}{l}\alpha ={98}^{\circ }\,\,\,\,\,\,\,\,\,\,\,\,a=34.6\\ \beta ={39}^{\circ }\,\,\,\,\,\,\,\,\,\,\,\,b=22\\ \gamma ={43}^{\circ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,c=23.8\end{array}[/latex].