Right Triangle Trigonometry

Transcription

1 Right Triangle Trigonometry 1

2 The six trigonometric functions of a right triangle, with an acute angle, are defined by ratios of two sides of the triangle. hyp opp The sides of the right triangle are: the side opposite the acute angle, the side adjacent to the acute angle, θ adj and the hypotenuse of the right triangle.

3 The trigonometric functions are hyp adj opp sine, cosine, tangent, cotangent, secant, and cosecant. opp sin = cos = adj opp tan = hyp hyp adj θ 3

4 Given sides of a right triangle you should be able to find the value of all 6 trigonometric functions. Example: 1 5 4

5 Geometry of the triangle Consider an isosceles right triangle with two sides of length The Pythagorean Theorem implies that the hypotenuse is of length. 5

6 Calculate the trigonometric functions for a 30 angle. 1 opp 1 sin 30 = = hyp 30 3 adj 3 cos 30 = = hyp opp 1 tan 30 = = = adj

7 Calculate the trigonometric functions for a 60 angle opp sin 60 = = hyp 3 1 adj 1 cos 60 = = hyp opp 3 tan 60 = = = adj 1 3 7

8 Using Trigonometry to Solve a Right Triangle A surveyor is standing 115 feet from the base of the Washington Monument. The surveyor measures the angle of elevation to the top of the monument as How tall is the Washington Monument? Figure 4.33

9 Applications Involving Right Triangles The angle you are given is the angle of elevation, which represents the angle from the horizontal upward to an object. For objects that lie below the horizontal, it is common to use the term angle of depression.

10 Solution where x = 115 and y is the height of the monument. So, the height of the Washington Monument is y = x tan (4.888) 555 feet.

11 Law of Sines

12 Introduction The Law of Sines can also be written in the reciprocal form:.

13 Given Two Angles and One Side AAS For the triangle below C = 10, B = 9, and the remaining angle and sides. b = 8 feet. Find

14 Example AAS - Solution The third angle of the triangle is A = 180 B C = = 49. By the Law of Sines, you have.

15 Example AAS Solution cont d Using b = 8 produces and

16 Law of Sines For non right triangles B Law of sines c a a sin A b sin B c sin C A b C

17 Let s look at this: Example 1 Given a triangle, demonstrate using the Law of Sines that it is a valid triangle (numbers are rounded to the nearest tenth so they may be up to a tenth off): a = 5 b = 7 c = 7.5 Is it valid?? A = 44 o B = 64.1 o C = 71.9 o

18 The Ambiguous Case (SSA)

19 The Ambiguous Case (SSA) (1) no such triangle exists, () one such triangle exists, or (3) two distinct triangles may satisfy the conditions.

20 The Ambiguous Case (SSA)

21 Example ssa For the triangle below, a = inches, b = 1 inches, and A = 4. Find the remaining side and angles.

22 Example Solution SSA By the Law of Sines, you have sin B sin A b a Reciprocal form sin ( sin A B b a ) Multiply each side by b. ( sin 4 sin B 1 ) o B Substitute for A, a, and b. B is acute.

23 Example Solution SSA cont d Now, you can determine that B = > 180 therefore only 1 triangle can be created. Then, the remaining side is Angle C = = c sin C a sin A c a sin C sin A c sin(4) sin(116.59) 9.40 inches

24 Area of an Oblique Triangle

25 Area of an Oblique Triangle The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle. Referring to the triangles below, that each triangle has a height of h = b sin A. A is acute. A is obtuse.

26 Area of a Triangle - SAS SAS you know two sides: b, c and the angle between: A c h B a Remember area of a triangle is ½ base height A b C Base = b Height = c sin A Area = ½ bc(sina) Looking at this from all three sides: Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)

27 Area of an Oblique Triangle

28 Example Finding the Area of a Triangular Lot Find the area of a triangular lot having two sides of lengths 90 meters and 5 meters and an included angle of 10. Solution: Consider a = 90 meters, b = 5 meters, and the included angle C = 10 Then, the area of the triangle is Area = ½ ab sin C = ½ (90)(5)(sin10) 89 square meters.

29 Law of Cosines

30 Law of Cosines: Introduction Two cases remain in the list of conditions needed to solve an oblique triangle SSS and SAS. If you are given three sides (SSS), or two sides and their included angle (SAS), none of the ratios in the Law of Sines would be complete. In such cases, you can use the Law of Cosines.

31 Law of Cosines Side, Angle, Side B c a a b c A bccosa b C b a c accosb c a b abcosc

32 An Application of the Law of Cosines The pitcher s mound on a women s softball field is 43 feet from home plate and the distance between the bases is 60 feet (The pitcher s mound is not halfway between home plate and second base.) How far is the pitcher s mound from first base?

33 Solution In triangle HPF, H = 45 (line HP bisects the right angle at H), f = 43, and p = 60. Using the Law of Cosines for this SAS case, you have h = f + p fp cos H = (43)(60) cos So, the approximate distance from the pitcher s mound to first base is 4.43 feet.