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This Lesson (BASIC TRIG FUNCTIONS) was created by by Theo(13342)
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This lesson provides an overview of basic trigonometric functions used to solve problems involving right triangle, acute triangle, and obtuse triangles.
REFERENCES Dave's Short Course in Trigonometry http://www.sosmath.com/trig/trig.html http://www.themathpage.com/atrig/trigonometry.htm http://www.intmath.com/Trigonometric-functions/Trig-functions-intro.php http://www.syvum.com/math/trigo/ http://jwilson.coe.uga.edu/emt668/EMAT6680.2003.fall/Bismarck/6690%20folder/trigproofs/triproofs.html http://www.themathpage.com/atrig/proof.htm The term "select" is used in place of "click on" or "left click" in many places. They mean the same thing. You use your mouse, or mouse pad, or any other method available to you, to select the option or link referenced. CALCULATOR If you don't have a calculator, or you want to find cosecant, secant, or cotangent directly, the Online Trig Function Calculator can help you. It's very easy to use. You enter your angle and select the right function key and you've got it. You can set the accuracy (significant decimal digits) as well. Just scroll down further until you see the area where you can input the number of decimal places accuracy you require. It even contains a short tutorial on Trigonometry which you might find useful. For those unfamiliar with the terminology, arc in front of the function means "the angle who function is". This is also sometimes written as Example: you enter 30 (for 30 degrees and then select the sin button. you get .5 you enter .5 and then select the arcsin button. you get 30 The calculator gives you results in degrees and in radians. radians = degrees * (pi/180) degrees = radians * (180/pi) pi = Example: 360 degrees = 2*pi radians 1 degree = pi/180 radians = .017453293 radians. 1 radian = 180/pi degrees = 57.29577951 degrees. If you enter degrees, then look for the answer in degrees. If you enter radians, then look for the answer in radians. For this lesson, we will be working in degrees only. The default for rounding is 5 decimal places. If you need more or less, then scroll down and enter the number of decimal places you need. You will have to do this every time you exit and reenter the calculator because the default of 5 decimal places is set each time. Don't try to use the enter key. You will think your change took, but it didn't. Just enter the value in the box and select any area outside the box with your mouse in order to save the entry. The calculator opens in a separate window so you should be able to keep it open and access it any time you need it until you specifically close that window. ABBREVIATIONS OF THE TRIGONOMETRIC FUNCTIONS Let x be the angle. Sin(x) = Sine of x Cos(x) = Cosine of x Tan(x) = Tangent of x Csc(x) = Cosecant of x Sec(x) = Secant of x Cot(x) = Cotangent of x RIGHT TRIANGLE This triangle has one angle equal to 90 degrees. The sum of the other 2 angles of this triangle equals 90 degrees. The other 2 angles of this triangle are complements of each other. select the following link to see a picture of a Basic Right Triangle The triangle is labeled ABC. Side a is opposite angle A. Side b is opposite angle B. Side c is opposite angle C. The hypotenuse of the right triangle is side c which is opposite right angle C. Sides a and b are legs of the right triangle. They are opposite acute angles A and B. BASIC TRIG FUNCTIONS OF A RIGHT TRIANGLE. The basic trig functions are: sine cosine tangent Sine of an angle is equal to the opposite side divided by the hypotenuse. Cosine of an angle is equal to the adjacent side divided by the hypotenuse. Tangent of an angle is equal to the opposite side divided by the adjacent side. The reciprocals of the basic trig functions are: cosecant secant cotangent Cosecant of an angle is the reciprocal of the sine of an angle. It is equal to the hypotenuse divided by the opposite side. This is the same as 1 divided by the sine of the angle. Secant of an angle is the reciprocal of the cosine of an angle. It is equal to the hypotenuse divided by the adjacent side. This is the same as 1 divided by the cosine of the angle. Cotangent of an angle is the reciprocal of the tangent of an angle. It is equal to the adjacent side divided by the opposite side. This is the same as 1 divided by the tangent of the angle. To see how this works, we'll show you the cosecant of an angle. Csc (B) = hyp/opp Sin (B) = opp/hyp 1 / Sin (B) = 1 / (opp/hyp) = hyp/opp = Csc (B) Select the following link to see a picture of a right triangle and its associated Basic Trigonometric Functions TYPES OF TRIANGLES There are 2 basic types of triangles. They are: ACUTE TRIANGLE All angles of this triangle are less than 90 degrees. OBTUSE TRIANGLE One of the angles of this triangle is greater than 90 degrees and less than 180 degrees. The other two angles are acute. Select this link to see a picture of an Acute Triangle and an Obtuse Triangle ALTITUDE OF A TRIANGLE An altitude of a triangle is a perpendicular line that is drawn from an angle of the triangle to the side opposite that angle. The line that is drawn can intersect with the original base of the triangle or it can intersect with an extension of the original base of the triangle. Select the following link to see a picture of an altitude dropped from an acute triangle and an altitude dropped from an obtuse triangle. You can see that the altitude dropped from the acute triangle intersects with the existing base of the triangle, while the altitude dropped from an acute angle of the obtuse triangle intersects with an extension of the existing base of the triangle. Dropping an altitude is one of the basic ways to use trigonometry to solve problems involving triangles that are not right triangles. A PROBLEM INVOLVING A RIGHT TRIANGLE You are given Triangle ABC which is a right triangle. You are given side BC = 5 units in length. You are given side AC = 7 units in length. You need to find the angles of the triangle and the third side. You have forgotten the pythagorean formula which states that c^2 = a^2 + b^2, but you have not forgotten your basic trig formulas of sine, cosine, and tangent. Since this is a right triangle, you do not need to drop an altitude because you already have one. That would be side BC. Select the following link to see a picture of your right triangle. Picture of your Right Triangle You find the Tangent of Angle A by using the formula: Tan(A) = opposite / hypotenuse = BC/AC = 5/7 = .714285714 You find the Angle whose Tangent is .714285714 by using the formula: Angle A = Arctangent (.714285714) = 35.53767779 degrees. You find Angle B by using the formula: Angle A + Angle B = 90 degrees, which becomes: Angle B = 90 degrees - Angle A = 90 - 35.53767779 = 54.46232221 degrees. You find AB by using the formula: Sin(A) = opposite / hypotenuse. This formula becomes: Sin(35.53767779) = 5/AB Divide both sides of this equation by Sin(35.53767779) and multiply both sides of this equation by AB to get: AB = 5/Sin(35.53767779) = 5/.581238194 = 8.602325267 You solved this problem using the trigonometric functions of a right triangle. Since this was a right triangle to start with, you only had to use the trigonometric functions that applied to the given criteria. In this case, it was the Tangent function that got you started. Once you were able to obtain one angle, you could then use the Sine function to find your hypotenuse. You did not have to drop any altitudes to form a right triangle because you were given a right triangle to start with. A PROBLEM INVOLVING AN ACUTE TRIANGLE You are given triangle ABC which is an acute triangle. You want to find the height of the triangle. You are given that Angle A = 70 degrees, Angle B = 80 degrees, Angle C = 30 degrees. You are also given that side AC = 7 units in length. You drop an altitude from B intersecting with, and perpendicular to, side AC at Point D. The length of this altitude is the height that you are looking to find. Select the following link to see a picture of your triangle with it's new altitude. Acute Triangle with New Altitude We identified the length of the altitude as x. We identified the length of line segment AD as y, and the length of line segment DC as (7-y). We know that Tan(70) = x/y (equation 1) We know that Tan(30) = x/(7-y) (equation 2) In order to find x, we will need to know y. Using equation 1 and 2, we solve for x. We get: x = y*Tan(70) and we get: x = (7-y)*Tan(30) Since y*Tan(70) and (7-y)*Tan(30) both equal x, then they must both be equal to each other, so we get: y*Tan(70) = (7-y)*Tan(30) We remove parentheses to get: y*Tan(70) = 7*Tan(30) - y*Tan(30) We add y*Tan(30) to both sides of this equation to get: y*Tan(70) + y*Tan(30) = 7*Tan(30) We factor out the y to get: y*(Tan(70) + Tan(30) = 7*Tan(30) We divide both sides of this equation by (Tan(70) + Tan(30)) to get: y = (7*Tan(30) / (Tan(70) + Tan(30)) We solve using the values indicated to get: y = 4.041451884 / 3.3248276689 which becomes: y = 1.215537244 Since x = y*Tan(70), then: x = 1.215537244 * Tan(70) = 3.339661129 To confirm this answer is good, we look at x = (7-y)*Tan(30). 7-y = 5.784462756 x = (7-y)*Tan(30) becomes x = 5.784462756 * Tan(30) = 3.339661129 This was the same answer we got with y * Tan(70) so it looks like our answer is good. We solved this problem using the trigonometric functions of a right triangle. We dropped an altitude to create the right triangles. We solved for y first, and then we were able to solve for x. A PROBLEM INVOLVING AN OBTUSE TRIANGLE You are given Triangle ABC which is an obtuse triangle. Angle BAC is 120 degrees; Angle ABC is 20 degrees; Angle C is 40 degrees. You are given that side AC is equal to 4 units in length. Side AC is given as the base of this triangle. You extend line segment AC all the way to the left. You drop an altitude from Point B to intersect with the extension of line AC at Point D. The extension of the line segment of AC stops at point D to form line segment DC passing through A. The altitude becomes line segment BD and is perpendicular to line segment DC at point D. Select the following link to see a picture of your obtuse triangle with its new altitude. Obtuse Triangle with New Altitude You want to find the height of this triangle. To find the height of this triangle, you have to find y which is the length of line segment DA. You have 2 formulas to work with. They are: Tan(40) = x/(y+4) Tan(60) = x/y You can solve for x in both of these formulas to get: x = (y+4) * Tan(40) x = y * Tan(60) Since they both equal to x, then they must be equal to each other, so you get: (y+4) * Tan(40) = y * Tan(60) Simplify by removing parentheses to get: y*Tan(40) + 4*Tan(40) = y*Tan(60) Subtract y*Tan(40) from both sides of this equation to get: 4*Tan(40) = y*Tan(60) - y*Tan(40) Factor out the y to get: 4*Tan(40) = y*(Tan(60)-Tan(40)) Divide both sides of this equation by (Tan(60)-Tan(40)) to get: 4*Tan(40)/(Tan(60)-Tan(40)) = y You can use this formula to solve for y which becomes: y = 3.758770483 Now that you have y, you can solve for x. Tan(60) = x/y = x/3.758770483 Multiply both sides of this equation by 3.758770483 to get: 3.758770483 * Tan(60) = x This makes x = 6.510381451 You can confirm this is good because Tan(40) = x/(y+4). Multiply both sides of this equation by (y+4) and you get: (y+4) * Tan(40) = x Because y = 3.758770483, this becomes: 7.758770483 * Tan(40) = x which becomes: x = 6.510381451 You solved this problem using the trigonometric functions of a right triangle. You dropped an altitude to create the right triangles. You solved for y first, and then we were able to solve for x. This time, though, you extended the base rather than intersecting with the base. This lesson has been accessed 21478 times. |
