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Chapter 7 Right Triangles and Trigonometry. 7.1 Geometric Mean.

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Presentation on theme: "Chapter 7 Right Triangles and Trigonometry. 7.1 Geometric Mean."— Presentation transcript:

1 Chapter 7 Right Triangles and Trigonometry

2 7.1 Geometric Mean

3 Means  In the past, we have thought of Means as the average. That is, the mean between 2 and 10 is 6 b/c (2+10)/2 = 6  You could notice that if you take 2 (starting value) and add 4 you will get 6 (the mean).  If you take 6 (the mean) and add 4 again, you get 10 (ending value)  Because you have added the same number twice, we call this the Arithmetic Mean.  Today we will learn about Geometric Means.

4 Geometric Means  If for the Arithmetic Mean you add the same number each time, what do you think you do to get the geometric mean?  Right, you multiply by the same number each time.  Let s be the starting number, e be the ending number and g be the geometric mean.  So, sx = g and gx = e.  Solve both for x you get x=g/s and x=e/g

5 Continue  Since both equations equal the same thing, you can combine them.  Notice that the geometric means are written twice on one diagonal!  Notice that the starting number and the ending number are on the other diagonal!  The pattern will always be the same, the GM’s are on one diagonal.

6 Find the GM between…  Find the Geometric Mean between 4 and 16.  Set up the equation 4/GM = GM/16  Cross Multiply and you get GM 2 = 64  Taking the square root of each side you get GM = + 8 but only +8 is between 4 and 16, then the GM between 4 and 16 is 8!  So, Why is Geometric Mean so important?

7 Geometric Means A C B D These three triangles are similar! ΔABC ~ ΔADB ~ ΔBDC 1 2 3 4 Rearranging the proportions: Notice AB is written twice on the diagonal? So, AB is the GM between AD and AC!

8 Geometric Means Con’t A C B ΔABC ~ ΔADB ~ ΔBDC Rearranging the proportions: Notice BC is written twice on the diagonal? So, BC is the GM between DC and AC! D

9 Geometric Mean Con’t A C B ΔABC ~ ΔADB ~ ΔBDC Rearranging the proportions: Notice BD is written twice on the diagonal? So, BD is the GM between DC and AD! D

10 7.2 Pythagorean Theorem and Converse

11 Pythagorean Theorem  Pythagoras recognized a very important relationship between the sides (legs) and the hypotenuse of a right triangle.  That is “The sum of the squares of the legs equals the square of the hypotenuse.” A B C a b c Here we have a 2 + b 2 = c 2 I like leg 2 + leg 2 = Hypot 2

12 Converse of Pythagorean Theorem  Did you notice that the only triangle that uses the Pythagorean theorem is a right triangle?  So, if the “sum of the squares of the two smaller sides equals the square of the largest side” then the triangle is a Right Triangle.  If leg 2 + leg 2 = Hypot 2, then the triangle is a right triangle.

13 Corollaries of Pythagoras  There are two corollaries of the Pythagorean theorem that allows you to classify a triangle by angles.  If a 2 + b 2 > c 2 (where a and b are the two smallest sides), then the triangle is acute.  If a 2 + b 2 < c 2 (where a and b are the two smallest sides), then the triangle is obtuse.

14 Pythagorean Triple  There are all sorts of side combinations that we can use to make a right triangle but some are more special.  When all three sides of a right triangle are integers, then these three numbers are called a “Pythagorean Triple”  Some common triples are 3, 4, 5 or 5, 12, 13, or 7, 24, 25…..  Plus all families of these… 3x, 4x, 5x etc..

15 7.3 Special Right Triangles

16 Special Right Triangles  Let’s take this equilateral triangle, all three sides and angles are congruent and all three angles measure 60° 60°  Now let us drop an altitude from the top of the triangle to the horizontal side.  Remember that when an altitude is drawn from the vertex angle of an “isosceles” triangle it is also an angle bisector and a median?

17 Special Right Triangle (Con’t) 60°  So, what do we know about the bottom side and what do we know about what is happening at the top angle?  The bottom segment is divided in ½ and so is the vertex angle. If the side is s then each piece is ½ s. Each angle is 30°

18 30°-60°-90° Right Triangles 60° 30° Short Leg Long Leg Hypotenuse Short Leg (SL) – This is the leg opposite the 30° angle. Long Leg (LL) – This is the leg opposite the 60° angle. Hypotenuse (H) – This is the side opposite the right angle. Rules: SL → H, SL x 2 = H SL → LL, SL x √3 = LL H → SL, H/ 2 = SL LL → SL, LL / √3 = SL

19 45°- 45°- 90° Right Triangles  Let us look at this square, it is equilateral and equiangular.  Let us divide the square by drawing a diagonal.  Now we have an “Isosceles, Right Triangle” or a 45-45-90 Right Triangle.  If we make the legs 1, then what is the length of the Hypotenuse? √2

20 45°- 45°- 90° Right Triangles  Rules Recap:  45 – 45 - 90  Leg to Hypotenuse: L x √2 = H  Hypotenuse to Leg: H ÷ √2 = L  30 – 60 – 90  SL to Hypotenuse: SL x 2 = H  Hypotenuse to SL: H ÷ 2 = SL  SL to LL: SL x √3 = LL  LL to SL: LL ÷ √3 = SL

21 Practice AC B a b c abc SLLLHypot 2 6 3 3√3 5 2√2 30°

22 Practice AC B a b c abc SLLLHypot 22√34 33√36 3 6 3 6 5/25√3/25 2√22√64√2 √2√62√2

23 More Practice AB C a c b Leg Hypot 3 4 6√2 7√2 8 3√2 4√3 45° a cb

24 More Practice AB C a c b Leg Hypot 333√2 444√2 666√2 777√2 4√2 8 3√2 6 4√3 4√6 45° a cb

25 7.4 Trigonometry

26 Why Trig?  Up to now we have been able to find sides and or angles of certain types of Right Triangles.  If we knew two sides, we could use the Pythagorean Theorem to find the third side.  If we saw it followed a pattern, say a Pythagorean Triple, we could find missing sides.  If it was one of the two special right triangles, we could apply the rules.  What happens if we can’t fit any of these situations?

27 Why Trig?  We can use Right Triangle trig if we have a right triangle and a known acute angle and side to find all the missing sides.  We can use Right Triangle trig if we have a right triangle and we know two sides to find any missing acute angle.

28 Trig Ratios  There are three important trig ratios we will use in geometry (there actually are six – but that will have to wait 2 years).  They are Sine (Sin), Cosine (Cos) and Tangent (Tan).  You can see the Sin, Cos and Tan buttons on your calculator.  Before we do anything, make sure you are in Degree Mode vice Radian Mode.

29 Trig Ratios (Con’t)  The sine of an acute angle is the ratio of the opposite side over the hypotenuse.  The cosine of an acute angle is the ratio of the adjacent side over the hypotenuse.  The tangent of an acute angle is the ratio of the opposite side over the adjacent side.  SOH/CAH/TOA  Some Old Hippie, Caught Another Hippie, Tripp’n Over Animals.

30 Sine and Cosine A B C a c b Sin <A = Opposite Side Hypotenuse = a b Sin <C = Opposite Side Hypotenuse = c b Cos <A = Adjacent Side Hypotenuse = c b cos <C = Adjacent Side Hypotenuse = a b

31 Tangent A B C a c b Tan <A = Opposite Side Adjacent Side = a c Tan <C = Opposite Side Adjacent Side = c a To find the missing sides of a right triangle, you must use either the Sine, Cosine or Tangent functions Depending on what they give you in the problem.

32 Calculator  Turn your calculator on… got to Mode, make sure you’re on Degree Mode.  To find the sine of 39° all you need to do is type “sin 39” and you’ll get …  sin 39° =.629320391  What does that mean?  That is the ratio of the length of the opposite side of the 39° angle over the length of the hypotenuse.

33 Calculator (Con’t)  Find the cosine of 47°  cos 47° =.6819983601  That is the ratio of the adjacent side over the hypotenuse is.6819983601  Find the tangent of 21°  tan 21° =.383864035  That is the ratio of the opposite side over the adjacent side.

34 Example A B C a c b = 20 25°  Given this triangle find a.  What do we know?  We know we have a right triangle, an acute angle that measures 25° and the length of the hypotenuse.  We want to find a That is the opposite side,  So which function?

35 Example (Con’t) A B C a c b = 20 25°  We have or want the opposite side and the hypotenuse… what function has Opp Side and Hypot?  Sine function…. So sin 25° = a/20.  Solving for a …. a = 20 sin 25° or approx 8.45  Now lets find c …. cos 25° = c/20  Solving for c …. c = 20 cos 25° or approx 18.13

36 Another Example B C a C = 20 b 35°  Given this triangle find a.  What do we know?  We know we have a right triangle, an acute angle that measures 35° and the length of the adjacent side from <A.  We want to find a That is the opposite side,  So which function?

37 Example (Con’t) A B C a c = 20 b 35°  We have or want the opposite side and the adjacent side… what function has Opp Side and Adj Side?  Tangent function…. So tan 35° = a/20.  Solving for a …. a = 20 tan 35° or approx 14.00  Now lets find b …. cos 35° = 20/b  Solving for c …. b = 20/cos 35° or approx 24.42

38 Angles?  So, all the examples we have done so far have had you find sides. You used sine, cosine and tangent functions.  To find angles, it is pretty similar except you will use Inverse Sine, Inverse Cosine and Inverse Tangent.  Inverse Sine is sin -1.  Inverse Cosine is cos -1.  Inverse Tangent is tan -1.

39 Angles (Con’t)  So to find the angle that has a sine of.8543 all you need to do is type sin -1 (.8543) and you’ll get the angle measurement.  sin -1 (.8543) = 58.7°

40 Example A B C a = 15 c = 20 b  Find the measure of <A  What do we know?  We know from Angle A, that a = 15 and c = 20.  a is the opposite side, c is the adjacent side from <A so we’re going to use Inv Tan  So, tan -1 (15/20) = 36.9°

41 7.5 Angles of Elevation and Depression

42 Angle of Elevation  How do you go about finding the height of a building that a person is standing on?  You can use right triangle trig.  So you need to draw a right triangle.

43 Angle of Elevation and Depression Horizontal Angle of Elevation Angle of Depression Horizontal < of Elevation is measured from the horizontal up to the “line of sight.” < of Depression is measured from the horizontal down to the “line of sight.”

44 Angle of Elevation and Depression  The measurements of the angle of elevation and depression are the same because they are….  Alternate Interior Angles made by a transversal “line of sight” cutting parallel lines “horizontal lines.”  Do not think that the complementary angle of the angle of elevation is the angle of depression.

45 Example  Find the height of a cliff if you are in a sail boat 2000’ from the cliff and the angle of elevation to the top is 13° 13° 2000’ y

46 7.6 Law of Sines (H)

47 What if?  What if you don’t have a right triangle?  Then depending on the information you can use either the Law of Sines or the Law of Cosines.  We will only cover the Law of Sines in this class, we will leave the Law of Cosines for Pre Calculus.

48 Law of Sines A h b c a C B D ΔACD sin A = h/b h = b sin A ΔBCD sin B = h/a h = a sin B b sin A = a sin B

49 Law of Sines A b = 10 c a C B 45° 35°

50 Example A 10 c a C B 45° 35° 100°

51 Unit Circle and Radians

52 Unit Circle Unit Circle – a circle w/ a radius = 1. A radian – an angle measurement that gives an arc length = to the radius. r = 1 Arc Length = 1 < meas = 1.

53 Unit Circle 30° or π/6 45° or π/4 60° or π/3 90° or π/2 315° or 7π/4 120° or 2π/3 135° or 3π/4 150° or 5π/6 180° or π 210° or 7π/6 225° or 5π/4 240° or 4π/3 270° or 3π/2 330° or 11π/6 300° or 5π/3 0°/360° or 2π

54 Radians  Radians is a unit that we use to measure angles. It is different unit of measure than degrees. Just like cm’s measure length and so do inches.  Conversion factor π r = 180°  The “r” is to show you that this is a radian measure (not the ratio of Circumference divided by the diameter)  Convert degrees to radians mult by (π/180°)  Convert radians to degrees mult by (180°/π)

55 Examples:  Given 3π/4 convert to degrees.  Take (3π/4)(180°/π)  You get 135°  Given 225° convert to π form radians.  Take (225°)(π/180°)  You get 5π/4


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