Find the missing side of a obtuse triangle calculator

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Find the missing side of a obtuse triangle calculator

In the module Further trigonometry (Year 10), we introduced and proved the sine rule, which is used to find sides and angles in non-right-angled triangles. In the triangle \(ABC\), labelled as shown, we have \[ \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}. \] Clearly, we may also write this as \[ \dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}. \] In general, one of the three angles may be obtuse. The formula still holds true, although the geometric proof is slightly different. Exercise 4 Find two expressions for \(h\) in the diagram below, and hence deduce the sine rule.Detailed description of diagram Repeat the method of part (a), using the following diagram, to show that the sine rule holds in obtuse-angled triangles. The triangle \(ABC\) has \(AB = 9\text{ cm}\), \(\angle ABC = 76^\circ\) and \(\angle ACB = 58^\circ\). Find, correct to two decimal places, Solution Applying the sine rule gives \[ \dfrac{AC}{\sin 76^\circ} = \dfrac{9}{\sin 58^\circ} \] and so \begin{align*} AC = \dfrac{9 \sin 76^\circ}{\sin 58^\circ}\\ \approx 10.30\text{ cm} \qquad \text{(to two decimal places).} \end{align*} To find \(BC\), we first find the angle \(\angle CAB\) opposite it. \begin{align*} \angle CAB = 180^\circ - 58^\circ - 76^\circ\\ = 46^\circ. \end{align*} Thus, by the sine rule, \[ \dfrac{BC}{\sin 46^\circ} = \dfrac{9}{\sin58^\circ} \] and so \[ BC \approx 7.63\text{ cm} \qquad \text{(to two decimal places).} \] The ambiguous case In the module Congruence (Year 8), it was emphasised that, when applying the SAS congruence test, the angle in question has to be the angle included between the two sides. For example, the following diagram shows two non-congruent triangles \(ABC\) and \(ABC'\) having two pairs of matching sides and sharing a common (non-included) angle. Detailed description of diagram Suppose we are told that a triangle \(PQR\) has \(PQ = 9\), \(\angle PQR = 45^\circ\) and \(PR = 7\). Then the angle opposite \(PQ\) is not uniquely determined. There are two non-congruent triangles that satisfy the given data. Detailed description of diagram Applying the sine rule to the triangle, we have \[ \dfrac{\sin \theta}{9} = \dfrac{\sin 45^\circ}{7} \] and so \begin{align*} \sin\theta = \dfrac{9\sin 45^\circ}{7}\\ \approx 0.9091. \end{align*} Thus \(\theta \approx 65^\circ\), assuming that \(\theta\) is acute. But the supplementary angle is \(\theta' \approx 115^\circ\). The triangle \(PQR'\) also satisfies the given data. This situation is sometimes referred to as the ambiguous case. Since the angle sum of a triangle is \(180^\circ\), in some circumstances only one of the two angles calculated is geometrically valid. The cosine rule We know from the SAS congruence test that a triangle is completely determined if we are given two sides and the included angle. However, if we know two sides and the included angle in a triangle, the sine rule does not help us determine the remaining side or the remaining angles. The second important formula for general triangles is the cosine rule. Suppose \(ABC\) is a triangle and that the angles \(A\) and \(C\) are acute. Drop a perpendicular from \(B\) to the line interval \(AC\) and mark the lengths as shown in the following diagram. Detailed description of diagram In the triangle \(ABD\), applying Pythagoras' theorem gives \[ c^2 = h^2 + (b - x)^2. \] Also, in the triangle \(BCD\), another application of Pythagoras' theorem gives \[ h^2 = a^2 - x^2. \] Substituting this expression for \(h^2\) into the first equation and expanding, \begin{align*} c^2 = a^2 - x^2 + (b - x)^2\\ = a^2 - x^2 + b^2 - 2bx + x^2\\ = a^2 + b^2 - 2bx. \end{align*} Finally, from triangle \(BCD\), we have \(x = a\,\cos C\) and so \[ c^2 = a^2 + b^2 - 2ab\,\cos C. \] This last formula is known as the cosine rule. Notice that, if \(C = 90^\circ\), then since \(\cos C = 0\) we obtain Pythagoras' theorem, and so we can regard the cosine rule as Pythagoras' theorem with a correction term. The cosine rule is also true when the angle \(C\) is obtuse. But note that, in this case, the final term in the formula will produce a positive number, because the cosine of an obtuse angle is negative. Some care must be taken in this instance. By relabelling the sides and angles of the triangle, we can also write the cosine rule as \(a^2 = b^2 + c^2 - 2bc\,\cos A\) and \(b^2 = a^2 + c^2 - 2ac\,\cos B\). Find the value of \(x\) to one decimal place. Solution Applying the cosine rule gives \begin{align*} x^2 = 7^2 + 8^2 - 2 \times 7 \times 8 \times \cos 110^\circ\\ = 113 + 112\cos70^\circ\\ \approx 151.306, \end{align*} so \(x \approx 12.3\) (to one decimal place). Finding angles If the three sides of a triangle are known, then the three angles are uniquely determined. (This is the SSS congruence test.) Again, the sine rule is of no help in finding the three angles, since it requires the knowledge of (at least) one angle, but we can use the cosine rule instead. We can substitute the three side lengths \(a\), \(b\), \(c\) into the formula \(c^2 = a^2 + b^2 - 2ab\,\cos C\), where \(C\) is the angle opposite the side \(c\), and then rearrange to find \(\cos C\) and hence \(C\). Alternatively, we can rearrange the formula to obtain \[ \cos C = \dfrac{a^2+b^2-c^2}{2ab} \] and then substitute. Students may choose to rearrange the cosine rule or to learn this further formula. Using this form of the cosine rule often reduces arithmetical errors. Recall that, in any triangle \(ABC\) labelled as shown, if \(a < b\), then \(\text{angle } A < \text{angle } B\). A triangle has side lengths 6 cm, 8 cm and 11 cm. Find the smallest angle in the triangle. Solution The smallest angle in the triangle is opposite the smallest side. Applying the cosine rule: \begin{align*} 6^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos \theta \\ \cos\theta = \dfrac{8^2+11^2-6^2}{2\times 8 \times 11}\\ =\dfrac{149}{176}. \end{align*} So \(\theta \approx 32.2^\circ\) (correct to one decimal place). The area of a triangle We saw in the module Introductory trigonometry (Years 9--10) that, if we take any triangle with two given sides \(a\) and \(b\) about a given (acute) angle \(\theta\), then the area of the triangle is \[ \text{Area} = \dfrac{1}{2} ab\,\sin\theta. \] This formula also holds when \(\theta\) is obtuse. Exercise 5 A triangle has two sides of length 5 cm and 4 cm containing an angle \(\theta\). Its area is 5 cm\(^2\). Find the two possible (exact) values of \(\theta\) and draw the two triangles that satisfy the given information. Exercise 6 Write down two different expressions for the area of a triangle \(ABC\) and derive the sine rule from them. Next page - Content - Trigonometric identities An obtuse triangle is a triangle in which one of the angles is an obtuse angle. (Obviously, only a single angle in a triangle can be obtuse or it wouldn't be a triangle.) A triangle must be either obtuse, acute, or right. From the law of cosines, for a triangle with side lengths , , and , (1) with the angle opposite side . For an angle to be obtuse, . Therefore, an obtuse triangle satisfies one of , , or . An obtuse triangle can be dissected into no fewer than seven acute triangles (Wells 1986, p. 71). A famous problem is to find the chance that three points picked randomly in a plane are the polygon vertices of an obtuse triangle (Eisenberg and Sullivan 1996). Unfortunately, the solution of the problem depends on the procedure used to pick the "random" points (Portnoy 1994). In fact, it is impossible to pick random variables which are uniformly distributed in the plane (Eisenberg and Sullivan 1996). Guy (1993) gives a variety of solutions to the problem. Woolhouse (1886) solved the problem by picking uniformly distributed points in the unit disk, and obtained (2) The problem was generalized by Hall (1982) to -dimensional ball triangle picking, and Buchta (1986) gave closed form evaluations for Hall's integrals. In 1893, Lewis Carroll (1976) posed and gave another solution to the problem as follows. Call the longest side of a triangle , and call the diameter . Draw arcs from and of radius . Because the longest side of the triangle is defined to be , the third polygon vertex of the triangle must lie within the region . If the third polygon vertex lies within the semicircle, the triangle is an obtuse triangle. If the polygon vertex lies on the semicircle (which will happen with probability 0), the triangle is a right triangle. Otherwise, it is an acute triangle. The chance of obtaining an obtuse triangle is then the ratio of the area of the semicircle to that of . The area of is then twice the area of a circular sector minus the area of the triangle. (3) Therefore, (4) Mathematica ? The #1 tool for creating Demonstrations and anything technical. Wolfram|Alpha ? Explore anything with the first computational knowledge engine. Wolfram Demonstrations Project ? Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. ? Join the initiative for modernizing math education. Online Integral Calculator ? Solve integrals with Wolfram|Alpha. Step-by-step Solutions ? Walk through homework problems step-by-step from beginning to end. Hints help you try the next step on your own. Wolfram Problem Generator ? Unlimited random practice problems and answers with built-in Step-by-step solutions. Practice online or make a printable study sheet. Wolfram Education Portal ? Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. Wolfram Language ? Knowledge-based programming for everyone. home / math / triangle calculator Please provide 3 values including at least one side to the following 6 fields, and click the "Calculate" button. When radians are selected as the angle unit, it can take values such as pi/2, pi/4, etc. A triangle is a polygon that has three vertices. A vertex is a point where two or more curves, lines, or edges meet; in the case of a triangle, the three vertices are joined by three line segments called edges. A triangle is usually referred to by its vertices. Hence, a triangle with vertices a, b, and c is typically denoted as abc. Furthermore, triangles tend to be described based on the length of their sides, as well as their internal angles. For example, a triangle in which all three sides have equal lengths is called an equilateral triangle while a triangle in which two sides have equal lengths is called isosceles. When none of the sides of a triangle have equal lengths, it is referred to as scalene, as depicted below. Tick marks on an edge of a triangle are a common notation that reflects the length of the side, where the same number of ticks means equal length. Similar notation exists for the internal angles of a triangle, denoted by differing numbers of concentric arcs located at the triangle's vertices. As can be seen from the triangles above, the length and internal angles of a triangle are directly related, so it makes sense that an equilateral triangle has three equal internal angles, and three equal length sides. Note that the triangle provided in the calculator is not shown to scale; while it looks equilateral (and has angle markings that typically would be read as equal), it is not necessarily equilateral and is simply a representation of a triangle. When actual values are entered, the calculator output will reflect what the shape of the input triangle should look like. Triangles classified based on their internal angles fall into two categories: right or oblique. A right triangle is a triangle in which one of the angles is 90?, and is denoted by two line segments forming a square at the vertex constituting the right angle. The longest edge of a right triangle, which is the edge opposite the right angle, is called the hypotenuse. Any triangle that is not a right triangle is classified as an oblique triangle and can either be obtuse or acute. In an obtuse triangle, one of the angles of the triangle is greater than 90?, while in an acute triangle, all of the angles are less than 90?, as shown below. Triangle facts, theorems, and laws 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. 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. Another way to calculate the exterior angle of a triangle is to subtract the angle of the vertex of interest from 180?. The sum of the lengths of any two sides of a triangle is always larger than the length of the third side Pythagorean theorem: The Pythagorean theorem is a theorem specific to right triangles. For any right triangle, the square of the length of the hypotenuse equals the sum of the squares of the lengths of the two other sides. It follows that any triangle in which the sides satisfy this condition is a right triangle. There are also special cases of right triangles, such as the 30? 60? 90, 45? 45? 90?, and 3 4 5 right triangles that facilitate calculations. Where a and b are two sides of a triangle, and c is the hypotenuse, the Pythagorean theorem can be written as: a2 + b2 = c2 EX: Given a = 3, c = 5, find b: 32 + b2 = 52 9 + b2 = 25 b2 = 16 => b = 4 Law of sines: the ratio of the length of a side of a triangle to the sine of its opposite angle is constant. Using the law of sines makes it possible to find unknown angles and sides of a triangle given enough information. Where sides a, b, c, and angles A, B, C are as depicted in the above calculator, the law of sines can be written as shown below. Thus, if b, B and C are known, it is possible to find c by relating b/sin(B) and c/sin(C). Note that there exist cases when a triangle meets certain conditions, where two different triangle configurations are possible given the same set of data. Given the lengths of all three sides of any triangle, each angle can be calculated using the following equation. Refer to the triangle above, assuming that a, b, and c are known values. Area of a Triangle There are multiple different equations for calculating the area of a triangle, dependent on what information is known. Likely the most commonly known equation for calculating the area of a triangle involves its base, b, and height, h. The "base" refers to any side of the triangle where the height is represented by the length of the line segment drawn from the vertex opposite the base, to a point on the base that forms a perpendicular. Given the length of two sides and the angle between them, the following formula can be used to determine the area of the triangle. Note that the variables used are in reference to the triangle shown in the calculator above. Given a = 9, b = 7, and C = 30?: Another method for calculating the area of a triangle uses Heron's formula. 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. However, it does require that the lengths of the three sides are known. Again, in reference to the triangle provided in the calculator, if a = 3, b = 4, and c = 5: Median, inradius, and circumradius Median The median of a triangle is defined as the length of a line segment that extends from a vertex of the triangle to the midpoint of the opposing side. A triangle can have three medians, all of which will intersect at the centroid (the arithmetic mean position of all the points in the triangle) of the triangle. Refer to the figure provided below for clarification. The medians of the triangle are represented by the line segments ma, mb, and mc. The length of each median can be calculated as follows: Where a, b, and c represent the length of the side of the triangle as shown in the figure above. As an example, given that a=2, b=3, and c=4, the median ma can be calculated as follows: Inradius The inradius is the radius of the largest circle that will fit inside the given polygon, in this case, a triangle. The inradius is perpendicular to each side of the polygon. In a triangle, the inradius can be determined by constructing two angle bisectors to determine the incenter of the triangle. The inradius is the perpendicular distance between the incenter and one of the sides of the triangle. Any side of the triangle can be used as long as the perpendicular distance between the side and the incenter is determined, since the incenter, by definition, is equidistant from each side of the triangle. For the purposes of this calculator, the inradius is calculated using the area (Area) and semiperimeter (s) of the triangle along with the following formulas: where a, b, and c are the sides of the triangle Circumradius The circumradius is defined as the radius of a circle that passes through all the vertices of a polygon, in this case, a triangle. The center of this circle, where all the perpendicular bisectors of each side of the triangle meet, is the circumcenter of the triangle, and is the point from which the circumradius is measured. The circumcenter of the triangle does not necessarily have to be within the triangle. It is worth noting that all triangles have a circumcircle (circle that passes through each vertex), and therefore a circumradius. For the purposes of this calculator, the circumradius is calculated using the following formula: Where a is a side of the triangle, and A is the angle opposite of side a Although side a and angle A are being used, any of the sides and their respective opposite angles can be used in the formula.

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