MATH Week 8. Ferenc Balogh Winter. Concordia University. Based on the textbook

Transcription

1 MATH Week 8 Ferenc Balogh Concordia University 2008 Winter Based on the textbook J. Stuart, L. Redlin, S. Watson, Precalculus - Mathematics for Calculus, 5th Edition, Thomson

2 Solving Triangles Law of Sines (Section 6.4) The Law of Sines Solving a Triangle using Law of Sines The Ambiguous Case The Law of Cosines (Section 6.5) The Law of Cosines The Area of a Triangle: Heron s Formula Navigation Some applications

3 The data to be known about a triangle: angles α β γ sides a b c Solving a triangle means that we have to find all angles and all sides from the data provided. Cases depending on the given data ASA or SAA - one side and two angles are known SSA - two sides and the angle opposite to one of those sides are known SAS - two sides and the included angle are known SSS - all three sides are known

4 data solution law to use ASA or SAA unique law of sines SSA ambiguous law of sines SAS unique law of cosines SSS unique law of cosines

5 The Law of Sines The lengths of the sides of a triangle are proportional to the sines of the corresponding opposite angles: sin α a = sin β b = sin γ c. Proof. The area of a triangle is expressible as A = 1 2 ab sin γ = 1 2 bc sin α = 1 ca sin β. 2 Dividing through by 1 2abc we get sin γ c = sin α a = sin β b.

6 How to solve a triangle in the ASA or SAA case? 1. Find the third angle using α + β + γ = Find the two other sides using the Law of Sines. There is no ambiguity, the solution is always uniquely determined.

7 Example. Solve the triangle if c = 12m, α = 23, β = 42. Solution. One side and two angles are given (ASA). Since α + β + γ = 180, the third angle is given by By the Law of Sines, we have γ = = 115. Therefore Similarly, b = c sin β sin γ a = c sin α sin γ sin β b = sin γ c. 12m sin 42 = sin m. = 12m sin 23 sin m.

8 Example. Solve the triangle if b = 3.2m, γ = 31, β = 74. Solution. One side and two angles are given (SAA). Since α + β + γ = 180, the third angle is given by By the Law of Sines, we have α = = 75. Therefore Similarly, c = b sin γ sin β a = b sin α sin β sin β b = sin γ c. 3.2m sin 31 = sin m. = 3.2m sin 75 sin m.

9 If two sides and the angle opposite to one of those sides is given (SSA) (let s say, a, b and α) then the following cases are possible depending on a: there is no solution (no intersection point) there is exactly one solution (right triangle case) there are two solutions (two intersection points) there is only one solution (two intersection points)

10 The method of solution for SSA: 1. Try to determine the angle β from sin β = b a sin α 2. Find the third angle using α + β + γ = 180, and the third side c using the Law of Sines.

11 Example. Solve the triangle if α = 53, b = 6m, a = 1.1m. Solution. The sine of β is given by sin β = b a sin α = 6m 1.1m sin But we should have sin β 1 for an angle β! This means that there is no angle β satisfying the equation above. Therefore there is no solution in this case.

12 Example. Solve the triangle if α = 30, b = 3.2m, a = 1.6m. Solution. The sine of β is given by sin β = b a sin α = 3.2m 1.6m sin 30 = 1 There is only one angle 0 β 180 such that sin β = 1: β = 90. There is a unique solution in this case. The Law of Sines gives γ = = 60. c = b sin γ sin β = 3.2m sin 60 sin m.

13 Example. Solve the triangle if α = 53, b = 6m, a = 5m. Solution. The sine of β is given by sin β = b a sin α = 6m 5m sin There are two angles 0 β 180 such that sin β = 0.96: Since β and β β γ 1 = 180 α β and γ 2 = 180 α β are both allowed angles, we have two different solutions in this case.

14 The Law of Sines gives c 1 = a sin γ 1 sin α = 5m sin 53.4 sin m c 2 = a sin γ 2 5m sin 20.4 = sin α sin m. There are two different triangles corresponding to this data of SSA type.

15 Example. Solve the triangle if α = 53, b = 6m, a = 7m. Solution. The sine of β is given by sin β = b a sin α = 6m 7m sin There are two angles 0 β 180 such that sin β = 0.68: Then β and β β γ 1 = 180 α β and γ 2 = 180 α β γ 1 is allowed but γ 2 is not: we have a unique solution corresponding to γ 1 in this case.

16 The Law of Sines gives c = a sin γ 1 sin α = 7m sin 83.8 sin m There is one triangle corresponding to this data of SSA type.

17 For a right triangle with sides a, b and c, where γ = 90 we have the Pythagorean Theorem: c 2 = a 2 + b 2. The following generalization holds for an oblique triangle: The Law of Cosines c 2 = a 2 + b 2 2ab cos γ. And we have two other equations corresponding to α and β: a 2 = b 2 + c 2 2bc cos α. b 2 = c 2 + a 2 2ca cos β.

18 Example. Solve the triangle if a = 2m, b = 6m, c = 7m. Solution. This is a problem of the SSS type, we have to find the angles. The Law of Cosines gives cos α = b2 + c 2 a 2 2bc cos β = c2 + a 2 b 2 2ca cos γ = a2 + b 2 c 2 2ab = = = 3 8 Therefore 1 27 α = cos β = cos ( γ = cos 1 3 )

19 Example. Solve the triangle if a = 2m, b = 6m and γ = 49. Solution. This is a problem of the SAS type. The Law of Cosines gives the third side: Therefore c 2 = a 2 + b 2 2ab cos γ. c = a 2 + b 2 2ab cos γ = cos 49 m 4.925m cos α = b2 + c 2 a 2 2bc cos β = c2 + a 2 b 2 2ca = Therefore α = cos β = cos 1 ( 0.4)

20 If we know the sides a, b and c of a triangle, then it is uniquely determined and therefore its area is expressible in terms of the lengths of the sides: Heron s Formula The area of a triangle of sides a, b, c is given by A = s(s a)(s b)(s c), where s = a + b + c 2 is the semiperimeter of the triangle.

21 Example. Suppose that a triangle has sides a = 5cm, b = 4cm, c = 7cm. Calculate the area of the triangle. Solution. The semiperimeter is given by s = a + b + c 2 = 5cm + 4cm + 7cm 2 = 8cm. Using Heron s formula, we get the area: A = s(s a)(s b)(s c) = 8(8 5)(8 4)(8 7)cm 2 = cm 2 = 96cm cm 2

22 Navigation In navigation, a direction is usually given as a bearing, for example: N 30 W The first letter is either N or S indicating north or south. The last letter is either E or W, either east or west. The acute angle between the letters indicates the direction measured from N/S to E/W. The goal is to avoid the use of negative angles and angles exceeding 90.

23 Example. An airplane takes off from airport A heading to N 52 E. After flying 300 kms, it makes a course correction above the point B and heads to the new direction N 11 W. Flying 120 kms more, it lands at point C. Find the distance between the points A and C. Find the bearing from A to C. Solution. First we realize that we have to solve a triangle from data of the type SAS because c = 300km, a = 120km and the angle β is given by β = = 117.

24 The Law of Cosines gives the third side: Therefore b = c 2 + a 2 2ca cos β b 2 = c 2 + a 2 2ca cos β. = cos 117 km km cos γ = a2 + b 2 c ab cos α = b2 + c 2 a 2 = bc α = cos β = cos The bearing from A to C is approximately N 5.78 E

25 Example. An Unidentified Flying Object is observed from observatories A and B simultaneously. The angles of elevations are α = 43 and β = 29 respectively. How far is the object from A and B if we know that the distance between A and B is c = 134km? Solution. We have to solve a triangle from data of the form ASA. By the Law of Sines, we have Therefore Similarly, b = c sin β sin γ a = c sin α sin γ γ = = 108. sin β b = sin γ c. 134km sin 29 = sin km. = 134km sin 43 sin km.

26 Example. What is the area of the triangle-shaped area where Norman Bethune s statue stands near Metro Guy-Concordia? (The sides are approximately 77.3m, 18.7m and 80.6m.) Solution.We can use Heron s Formula to calculate the approximate area. s = a + b + c 2 = 77.3m m cm 2 = 88.3m. Using Heron s formula, we get the area: A = s(s a)(s b)(s c) = 88.3( )( )( )m 2 = m 2 = 721.5m 2