Preview: Correctly fill in the missing side lengths (a, b, c) or the missing angles (α, β, γ) on the following diagrams. γ b a β c α Goal: In chapter 1 we were given information about a right triangle and asked to solve the triangle. In this section we will investigate how to solve oblique triangles (any triangle that without a right angle). The first thing to remember is that when solving a triangle we must know at least one side length. If we only know all three angles we can draw infinitely many similar triangles that satisfy the three angles. We wouldn t be able to settle on one specific triangle. Depending on the information given about the triangle we can solve the following cases using the Law of Sines or the Law of Cosines (chapter 5.2). 1. One side and any two angles (ASA or AAS) Law of Sines 2. Two sides and a nonincluded angle (SSA) Law of Sines 3. Two sides and an included angle (SAS) Law of Cosines (5.2) 4. Three sides (SSS) Law of Cosines (5.2) The four cases are illustrated in the following diagram: *The derivation of the Law of Sines is easy to follow and should be looked at. Please be sure to review the derivation in the text book on pages 254 and 255. 1
Law of Sines: In any triangle, sin α a sin β = b = sin γ c Let s take a look solving a triangle if we know two angles (ASA and AAS). Your first goal will always be to correctly sketch the triangle with the information provided. Ex) Given the information provided, solve the following triangles using the Law of Sines: a) α = 28, β = 66, and c = 8.2 b) β = 94.7, α = 30.6, and b = 3.9 2
The Ambiguous Case (SSA): It turns out that when we know two side lengths and a nonincluded angle there are several possibilities as to how the triangle may exist. Suppose we are given angle α (0 < α < 90 ) and side lengths a and b. Notice where α, a and b are positioned on the following diagram. Always draw your diagrams in a similar fashion for the SAA case. To establish the type of triangle we are dealing with we must draw our diagram correctly and find the height of the triangle h. Ex) Find a trigonometric expression for h. Depending on how the height h relates to the side lengths a and b, we can establish how many triangles we need to solve. Cases (follow the diagram): (a) If a < h, then no triangle can be formed, because a cannot reach from point C to the initial side of α. (b) If a = h, then exactly one right triangle is formed. (c) If h < a < b, then exactly two triangles are formed, because a reaches to the initial side in two places. (d) If a b, then only one triangle is formed. Ex) Given the following information, determine how many triangles can be formed. a) β = 28.6, a = 40.7, b = 52.5 b) α = 30, c = 40, a = 20 3
Solve the following triangles. First draw a proper diagram and establish the number of triangles that need to be solved for. Ex) β = 38, b = 2.9, c = 5.9 Ex) γ = 125, b = 5.7, c = 8.6 *Only two cases exist if given an obtuse angle, can you think of them? Ex) β = 38, b = 4.7, c = 5.9 4
This chapter is where we start to see some real powerful applications of trigonometry. The Law of Sines and the Law of Cosines can be used to solve many practical surveying and navigation problems. Regarding Surveying, we will need to keep in mind terms such as Angle of Elevation and Angle of Depression. Do you recall what these terms mean? Regarding Navigation, we need to be able to speak the language of navigation. Namely we need to understand how to interpret a Bearing. Traditionally there are two ways to interpret a bearing: In air navigation, bearing is typically given as a nonnegative angle less than 360 measured in a clockwise direction starting from a ray pointing due north (A bearing due north is measured at 0 ) Ex) Draw a diagram showing the following bearings: a) 100 b) 225 c) 65 In marine navigation and also in surveying, the bearing of a ray is typically given as the acute angle that ray makes with a ray pointing due north or due south. Ex) Draw a diagram showing the following bearings: a) N60 E b) S30 E c) S45 W Ex) During an important NATO exercise, an F-14 Tomcat left the carrier Nimitz on a course with a bearing of 34 and flew 400 miles. Then the F-14 flew for some distance on a course bearing 162. Finally, the plane flew back to its starting point on a course bearing 308. What distance did the plane fly on the final leg of the journey? Round answers to the nearest hundredth of a mile. 5
Ex) Joe and Jill set sail from the same point, with Joe sailing in a direction S4 E and Jill sailing in a direction S9 W. After 4 hours, Jill was 2 miles due west of Joe. How far had Jill sailed to the nearest tenth of a mile? Ex) A surveyor determines that the angle of elevation to the top of a building from a point on the ground is 30.4. He then moves back 55.4 ft and determines that the angle of elevation to the top of the tower is now 23.2. What is the height of the building to the nearest tenth of a foot? 6