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a b c 2 c a b 2 2 This is a right triangle: We call it a right triangle because it contains a right angle. The measure of a right o angle is 90 90o The little square in the angle tells you it is a right angle. 90o About 2,500 years ago, a Greek mathematician named Pythagorus discovered a special relationship between the sides of right triangles. Pythagorus realized that if you have a right triangle, 5 3 4 and you square the lengths of the two sides that make up the right angle, 5 3 2 4 3 4 2 and add them together, 5 3 3 4 2 4 2 you get the same number you would get by squaring the other side. 5 3 3 4 5 2 4 2 2 Is that correct? 2 ? 3 4 5 2 ? 2 9 16 25 It is. And it is true for any right triangle. 6 8 10 2 2 2 10 8 36 64 100 6 The two sides which come together in a right angle are called The two sides which come together in a right angle are called The two sides which come together in a right angle are called The lengths of the legs are usually called a and b. a b The side across from the right angle is called the a b And the length of the hypotenuse is usually labeled c. a c b The relationship Pythagorus discovered is now called The Pythagorean Theorem: a c b The Pythagorean Theorem says, given the right triangle with legs a and b and hypotenuse c, a c b then a b c . 2 a 2 c b 2 You Suppose can use youThe drive Pythagorean directly Theorem west for 48 to miles, solve many kinds of problems. 48 Then turn south and drive for 36 miles. 48 36 How far are you from where you started? 48 36 ? Using The Pythagorean Theorem, 2 2 48 + 36 = c 2 36 48 c Why? Can you see that we have a right triangle? 2 2 48 + 36 = c 2 36 48 c Which sides side isare thethe hypotenuse? legs? 2 2 48 + 36 = c 2 36 48 c Then all we need to do is calculate: 48 36 2304 1296 2 2 3600 c 2 2 Andsince So, you end c isup 3600, 60 miles c is 60. from where you started. 48 36 60 Find the length of a diagonal of the rectangle: 15" ? 8" Find the length of a diagonal of the rectangle: 15" b=8 c ? a = 15 8" 15 225 acc 17 8b 64 289 c 2 b=8 c a = 15 2 2 Find the length of a diagonal of the rectangle: 15" 17 8" Practice using The Pythagorean Theorem to solve these right triangles: c = 13 5 12 b 10 26 b = 24 a b c 2 2 2 10 b 26 2 100 b 676 2 b 676 100 2 b 576 2 2 2 10 (a) 26 (c) b 24 12 b= 9 15 7.2 The Converse of the Pythagorean Theorem • If c2 = a2 + b2, then ∆ABC is a right triangle. B c a C b A Verify a Right Triangle • Is ∆ABC a right triangle? C 16 12 A B c a b 2 2 20 2 20 12 16 2 2 2 400 144 256 400 400 • Yes, it is a right triangle. Classifying Triangles In ABC with longest side c: C If c2 < a2 + b2, a B If c2 = a2 + b2, then B If c2 > a2 + b2, then B A c ABC is right. c a C b then A b ABC is obtuse. c a C b A ABC is acute. Acute Triangles • Show that the triangle is an acute triangle. 35 5 c a b 2 35 2 2 2 4 5 2 2 35 16 25 35 41 • Because c2 < a2 + b2, the triangle is acute. 4 Obtuse Triangles • Show that the triangle is an obtuse triangle. 15 12 8 c a b 2 2 2 15 8 12 2 2 2 225 64 144 225 208 • Because c2 > a2 + b2, the triangle is obtuse. Classify Triangles • Classify the triangle as acute, right, or obtuse. 8 6 5 c a b 2 2 2 8 5 6 2 2 2 64 25 36 64 61 • Because c2 > a2 + b2, the triangle is obtuse. Classify Triangles • Classify the triangle with the given side lengths as acute, right, or obtuse. • A. 4, 6, 7 c a b 2 2 2 7 4 6 2 2 2 49 16 36 49 52 • Because c2 < a2 + b2, the triangle is acute. Classify Triangles • Classify the triangle with the given side lengths as acute, right, or obtuse. • B. 12, 35, 37 c a b 2 2 2 37 12 35 1369 144 1225 2 2 2 1369 1369 • Because c2 = a2 + b2, the triangle is right. 7.3 Similar Right Triangles Geometry Mr. Lopiccolo Objectives/Assignment • Solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle. • Use a geometric mean to solve problems such as estimating a climbing distance. • Assignment: pp. 453-454 (4-26) even Proportions in right triangles • In Lesson 6.4, you learned that two triangles are similar if two of their corresponding angles are congruent. For example P ∆PQR ~ ∆STU. Recall that the corresponding side lengths of similar triangles are in proportion. S U R T Q Activity: Investigating similar right triangles. Do in pairs or threes 1. 2. 3. 4. Cut an index card along one of its diagonals. On one of the right triangles, draw an altitude from the right angle to the hypotenuse. Cut along the altitude to form two right triangles. You should now have three right triangles. Compare the triangles. What special property do they share? Explain. Tape your group’s triangles to a piece of paper and place in labwork. Theorem 7.5 • If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to A each other. C D ∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC, ∆CBD ~ ∆ACD B A plan for proving thm. 7.5 is shown below: • Given: ∆ABC is a right triangle; altitude CD is drawn to hypotenuse AB. • Prove: ∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC, ∆CBD ~ ∆ACD • Plan for proof: First prove that ∆CBD ~ ∆ABC. Each triangle has a right triangle and each includes B. The triangles are similar by the AA Similarity Postulate. You can use similar reasoning to show that ∆ACD ~ ∆ABC. To show that ∆CBD ~ ∆ACD, begin by showing that ACD B because they are both complementary to DCB. Then you can use the AA Similarity Postulate. C A D B Ex. 1: Finding the Height of a Roof • Roof Height. A roof has a cross section that is a right angle. The diagram shows the approximate dimensions of this cross section. • A. Identify the similar triangles. • B. Find the height h of the roof. Solution: Y • You may find it helpful to sketch the three similar 3.1 m h triangles so that the corresponding angles and X W sides have the same orientation. Mark the congruent angles. Notice that some sides appear in more than one triangle. For 5.5 m instance XY is the hypotenuse in ∆XYW and the shorter leg in ∆XZY. Y h ∆XYW ~ ∆YZW ~ ∆XZY. Z 6.3 m Z X W 3.1 m 5.5 m Y Solution for b. • Use the fact that ∆XYW ~ ∆XZY to write a proportion. YW ZY = XY XZ Corresponding side lengths are in proportion. h 5.5 = 3.1 6.3 Substitute values. 6.3h = 5.5(3.1) h ≈ 2.7 Cross Product property Solve for unknown h. The height of the roof is about 2.7 meters. Using a geometric mean to solve problems • In right ∆ABC, altitude CD is drawn A to the hypotenuse, forming two smaller right triangles that C are similar to ∆ABC From Theorem 9.1, you know that ∆CBD ~ ∆ACD ~ ∆ABC. C D B B C D D A A B C Write this down! C A D B B C D C A D B Notice that CD is the longer leg of ∆CBD and the shorter leg of ∆ACD. When you write a proportion comparing the legs lengths of ∆CBD and ∆ACD, you can see that CD is the geometric mean of BD and AD. Longer leg of ∆CBD. Shorter leg of ∆CBD. BD CD A C Shorter leg of ∆ACD = CD AD Longer leg of ∆ACD. Copy this down! C A D B B C D C A Sides CB and AC also appear in more than one triangle. Their side lengths are also geometric means, as shown by the proportions below: D B AB CB Hypotenuse of ∆CBD A Shorter leg of ∆ABC. Hypotenuse of ∆ABC. C = CB DB Shorter leg of ∆CBD. Copy this down! C A D B B C D C A Sides CB and AC also appear in more than one triangle. Their side lengths are also geometric means, as shown by the proportions below: D B AB AC Hypotenuse of ∆ACD A Longer leg of ∆ABC. Hypotenuse of ∆ABC. C = AC AD Longer leg of ∆ACD. Geometric Mean Theorems • Theorem 7.6: In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the geometric mean of the lengths of the two segments • Theorem 7.7: In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg. A C D BD = CD CD AD AB CB = CB DB AB AC = AC AD B What does that mean? 2 x 6 6 = x 18 = x2 √18 = x 5 y 3 x 5+2 3 y 7 y = y = 2 y 2 √9 ∙ √2 = x 14 = y2 3 √2 = x √14 = y Ex. 3: Using Indirect Measurement. • MONORAIL TRACK. To estimate the height of a monorail track, your friend holds a cardboard square at eye level. Your friend lines up the top edge of the square with the track and the bottom edge with the ground. You measure the distance from the ground to your friend’s eye and the distance from your friend to the track. In the diagram, XY = h – 5.75 is the difference between the track height h and your friend’s eye level. Use Theorem 9.2 to write a proportion involving XY. Then you can solve for h. *You will be able to find the lengths of sides of special right triangles 45-45-90 And 30-60-90 45 45 90 Leg:Leg:Hypotenuse 30 60 90 1:1: 2 x: x: x 2 1: 3 : 2 x : x 3 : 2x Short Leg:Long Leg:Hypotenuse In a 45-45-90 triangle… We will use a reference triangle to set up a proportion then solve. 45-45-90 Right Triangle 45 2 1 45 1 This is our reference triangle for the 45-45-90. 45-45-90 Right Triangle 45 x 2 x 45 x EX: 1 Solve for x 3 Let’s set up a proportion by using our reference triangle. x 2 1 3 3 1 x 2 x3 2 1 EX: 2 Solve for x 5 x 2 1 5 5 1 x 2 x 5 2 1 EX: 3 Solve for x 45 3 3 2 x 1 2 1 1 x 3 x 2 3 2 x 2 2 3 2 x 2 30-60-90 Right Triangle 60 2 1 30 3 This is our reference triangle for the 30-60-90 triangle. We will use a reference triangle to set up a proportion then solve. 30-60-90 Right Triangle 60 x 2x 30 x 3 Ex: 1 60 8 60 x 2 1 30 y x 1 8 2 2x 8 x4 30 y 3 8 3 2 8 3 2y 4 3y Ex: 2 Solve for x 30 60 x 1 2 24 60 30 24 2 x 1 2x = 24 x = 12 3 Ex: 3 30 60 14 2 1 y 30 3 60 x 14 2 x 1 2x = 14 x=7 14 2 y 3 2y = 14 3 y=7 3 Ex: 4 x 5 3 60 2 1 60 30 y 30 3 5 3 3 x 1 x=5 3 5 3 y 2 y = 10 • To find trigonometric ratios using right triangles, and • To solve problems using trigonometric ratios. Great Chief Soh Cah Toa A young brave, frustrated by his inability to understand the geometric constructions of his tribe's battle dress, kicked out in anger against a stone and crushed his big toe. Fortunately, he learned from this experience, and began to use study and concentration to solve his problems rather than violence. This was especially effective in his study of math, and he went on to become the wisest man of his tribe. He studied many aspects of trigonometry; and even today we remember many of the functions by his name. When he became an adult, the tribal priest gave him a name that reflected his special nature -- one that reminded them of his great discoveries and of the event which changed his life. Because he was troubled throughout his life by the problematic foot, he was constantly at the edge of the river, soaking his aches in the cooling waters. For that behavior, he was named Chief Soh Cah Toa. •Sine: Opposite side over hypotenuse. Cosine: Adjacent side over hypotenuse. •Tangent: Opposite side over adjacent. Find the sin S, cos S, tan S, sin E, cos E and tan E. Express each ratio as a fraction and as a decimal. M Sin S = ME/SE = 3/5 or 0.6 Cos S = SM/SE = 4/5 or 0.8 Tan S = ME/SM = ¾ or 0.75 Sin E = SM/SE = 4/5 or 0.8 S Cos E = ME/SE = 3/5 or 0.6 Tan E = SM/ME = 4/3 or 1.3 4 3 E 5 Find each value using a calculator. Round to the nearest ten thousandths. 1. Cos 41 2. Sin 78 A plane is one mile about sea level when it begins to climb at a constant angle of 2 for the next 70 ground miles. How far above sea level is the plane after its climb? 2 h 1 mi Sea level 70 mi Tan 2 = h/70 70 tan 2 = h h = 2.44 37 Show that the sin and cos give you the same hypotenuse. 4 53 3 -1 SINE Pronounced “sign inverse” -1 COSINE Pronounced “co-sign inverse” -1 TANGENT Pronounced “tan-gent inverse” Greek Letter q Prounounced “theta” Represents an unknown angle Opp Leg Sin Hyp Adj Leg Cos Hyp Opp Leg Tan Adj Leg hypotenuse q adjacent opposite opposite We need a way to remember all of these ratios… Some Old Hippie Came A Hoppin’ Through Our Old Hippie Apartment SOHCAHTOA Old Hippie Sin Opp Hyp Cos Adj Hyp Tan Opp Adj Finding an angle. (Figuring out which ratio to use and getting to use the 2nd button and one of the trig buttons.) Ex. 1: Find q. Round to four decimal places. nd 2 17.2 q 9 17.2 tan q 9 tan 17.2 9 ) q 62.3789 Shrink yourself down and stand where the angle is. Now, figure out which trig ratio you have and set up the problem. Make sure you are in degree mode (not radians). Ex. 2: Find q. Round to three decimal places. 7 q 23 nd 2 7 cos q 23 cos 7 23 q 72.281 Make sure you are in degree mode (not radians). ) Ex. 3: Find q. Round to three decimal places. q 200 sin q 400 200 nd 2 sin 200 400 ) q 30 Make sure you are in degree mode (not radians). When we are trying to find a side we use sin, cos, or tan. When we are trying to find an angle we use sin-1, cos-1, or tan-1.