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Lesson 9.1 Using Similar Right Triangles Students need scissors, rulers, and note cards. Today, we are going to… …use geometric mean to solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle With a straight edge, draw one diagonal of the note card. Draw an altitude from one vertex of the note card to the diagonal. Cut the note card into three triangles by cutting along the segments. C A short leg B B short leg B long long leg long leg C A D short Color code all 3 sides of all 3 triangles on the front and back. D Arrange the small and medium triangles on top of the large triangle like this.? Theorems 9.1 – 9.3 If the altitude is drawn to the hypotenuse of a right triangle, then… Theorem 9.1 the three triangles formed are similar to each other. … B C BD AD = CD BD D A BD is a side of the medium and a side of the small BD AD = BD CD B x n = m x C x m D n A Theorem 9.2 …the altitude is the geometric mean of the two segments of the hypotenuse. B x n = m x C x m D n A When you do these problems, always tell yourself… ____ is the geometric mean of ____ and ____ 1. Find x. x 3 = x 8 x ≈ 4.9 C 8 B x D 3 A 2. Find x. 4 x = 4 8 x=2 C 8 B 4 D x A B C CB CA = CD CB D A CB is a side of the large and a side of the medium CB CA = CB CD B x h = m x x C m D h A B C AB AC = AD AB D A AB is a side of the large and a side of the small AB AC = AB AD B x h = n x x C D h n A Theorem 9.3 …the leg of the large triangle is the geometric mean of the “adjacent leg” and the hypotenuse. B x h = m x y h = C n y x y m n D h A When you do these problems, always tell yourself… ____ is the geometric mean of ____ and ____ 3. Find x. x 14 = x 9 x x ≈ 11.2 C B D 9 14 A 4. Find x. x 10 = x 4 B x x ≈ 6.3 C D 10 4 A 5. Find x, y, z x x 13 = x 9 x ≈ 10.8 y 4 = y 9 y=6 y z 9 4 z 13 = z 4 z ≈ 7.2 3 5 = 3 x x = 1.8 5 6. Find h. h 3.2 = 1.8 h h = 2.4 3.2 4 x1.8 h 3 Lesson 9.2 & 9.3 The Pythagorean Theorem & Converse Today, we are going to… …prove the Pythagorean Theorem …use the Pythagorean Theorem and its Converse to solve problems Theorem 9.4 Pythagorean Theorem 2 hyp 2 leg 2 leg = + 2 2 2 (c = a + b ) 102 = y2 + 82 2. Find x. 2 + 64 100 = y Why can’t we use a geo mean proportion? 36 = y2 y=6 x 10 8 x2 = 142 + 82 x2 = 196 + 64 y 20 14 x2 = 260 x ≈ 16.1 A Pythagorean Triple is a set of three positive integers that satisfy the 2 2 2 equation c = a + b . The integers 3, 4, and 5 form a Pythagorean Triple because 52 = 32 + 42. Theorem 9.5 Converse of the Pythagorean Theorem 2 c 2 a 2 b, If = + then the triangle is a right triangle. The “hypotenuse” is toois The hypotenuse long for opposite thethe perfect length for anglethe to be 90˚ opposite angle to be The c2 >90˚ a2 +“hypotenuse” b2 2 = a 2 + bfor 2 the isctoo short opposite angle to be 90˚ c2 < a2 + b2 Theorem 9.6 If 2 c < 2 a 2 b, + then the triangle is an acute triangle. Theorem 9.7 If 2 c > 2 a 2 b, + then the triangle is an obtuse triangle. How do we know if 3 lengths can represent the side lengths of a triangle? L<M+S Can a triangle be formed? What kind of triangle? L=M+S No can be formed L2 = M2 + S2 Right L<M+S Yes, can be formed L2 < M2 + S2 Acute L>M+S No can be formed L2 > M2 + S2 Obtuse Do the lengths represent the lengths of a triangle? Is it a right triangle, acute triangle, or obtuse triangle? 262 = 102 + 242 right triangle 26 < 10 + 24? 3. 10, 24, 26 4. 3, 5, 7 7 < 5 + 3? 5. 5, 8, 9 9 < 8 + 5? 72 > 32 + 52 obtuse triangle 92 < 52 + 82 acute triangle Do the lengths represent the lengths of a triangle? Is it a right triangle, acute triangle, or obtuse triangle? 6. 5, 56 , 9 9 < 5 + 56 ? 7. 23, 44, 70 70 < 23 + 44? 8. 12, 80, 87 87 < 12 + 80? 92 = 52 + 56 2 right triangle not a triangle 872 > 122 + 802 obtuse triangle Find the area of the triangle. 9. 92 = x2 + 62 6 81 = x2 + 36 9 A ≈ ½ (6)(6.7) ≈ 20.12 units2 45 = x2 6.7 = x Find the area of the triangle. 10. 13 202 = x2 + 132 400 = x2 + 169 20 A ≈ ½ (13)(15.2) ≈ 98.8 units2 231 = x2 15.2 = x Find the area of the triangle. 11. 7 x 72 = x2 + 22 49 = x2 + 4 2 A ≈ ½ (4)(6.7) ≈ 13.4 units2 45 = x2 6.7 = x Which method requires less ribbon? How much ribbon is needed using Method 1? (3+12+3+12) + (3+6+3+6) = 48 in. The diagram shows the 30 ? ribbon for Method 2. How much ribbon is 18 ? needed to wrap the box? 302 + 182 = 35 in Project ideas… A 20-foot ladder leans against a wall so that the base of the ladder is 8 feet from the base of the building. How far up on the building will the ladder reach? A 50-meter vertical tower is braced with a cable secured at the top of the tower and tied 30 meters from the base. How long is the cable? The library is 5 miles north of the bank. Your house is 7 miles west of the bank. Find the distance from your house to the library. Project ideas… Playing baseball, the catcher must throw a ball to 2nd base so that the 2nd base player can tag the runner out. If there are 90 feet between home plate and 1st base and between 1st and 2nd bases, how far must the catcher throw the ball? While flying a kite, you use 100 feet of string. You are standing 60 feet from the point on the ground directly below the kite. Find the height of the kite. Lesson 9.4 Special Right Triangles Today, we are going to… …find the side lengths of special right triangles 1. Find x. Use the Pythagorean Theorem to find y. Leave y in simplest radical form. y5 2 45˚ x=5 y x 45˚ 5 2. Find x. Use the Pythagorean Theorem to find y. Leave y in simplest radical form. y9 2 45˚ x=9 y x 45˚ 9 Do you notice a pattern? 45˚ 45˚ 5 2 5 9 2 9 45˚ 5 45˚ 9 Theorem 9.8 45˚- 45˚- 90˚ Triangle Theorem 45˚ x 2 x 45˚ x In a 45˚- 45˚- 90˚ Triangle, hypotenuse = leg 2 3. Find x and y. y=7 2 45˚ x=7 y x 45˚ 7 4. Find x and y. 45˚ x=3 y=3 3 2 x 45˚ y 5. Find x and y. 2 10 x= 2 45˚ 10 x x =5 2 45˚ y y =5 2 6. Label the measures of all angles. Find x. Use the Pythagorean Theorem to find 30˚ 30˚ y in simplest radical form. 6 y=3 3 x=3 6 y x 60˚ 60˚ 6 7. Label the measures of all angles. Find x. Use the Pythagorean Theorem to find 30˚ 30˚ y in simplest radical form. 8 y=4 3 x=4 8 y x 60˚ 60˚ 8 Do you notice a pattern? 6 3 3 8 4 3 60˚ 60˚ 3 4 Theorem 9.9 30˚-60˚-90˚ Triangle Theorem 2s s 3 In a 30˚-60˚-90˚ Triangle, hypotenuse = 2 short leg long leg = short leg 3 60˚ s 8. Find x and y. 30˚ x=5 y=5 3 y 10 60˚ x 9. Find x and y. 30˚ x = 20 y = 10 3 y x 60˚ 10 10. Find x and y. 30˚ x = 12 y 3 12 y = 24 60˚ x 11. Find x and y. 30˚ 12 y 3 x=8 3 x 12 y= 4 3 60˚ y 12. In regular hexagon ABCDEF, find x and y. 360 y= 12 y = 30 x = 60 13. Find x and y. x = 24 y = 12 3 14. Find x and y. x= 8 y=8 2 Lesson 9.5 & 9.6 Trigonometric Ratios Today, we are going to… …find the sine, cosine, and tangent of an acute angle …use trigonometric ratios to solve problems Trigonometric Ratios sine cosine tangent B opposite A C hypotenuse adjacent to A A AC is adjacent BC AB ____ the hypotenuse opposite to A A B adjacent to B C hypotenuse opposite B A ____ BC is adjacent AC AB opposite to B B the hypotenuse sin A = leg opposite A hypotenuse cos A = leg adjacent to A hypotenuse tan A = leg opposite A leg adjacent to A SOH CAH TOA Sine is Opp / Hyp Cosine is Adj / Hyp Tangent is Opp / Adj B 1. 13 5 5 sin A = 13 12 cos A = 13 5 tan A = 12 12 = 0.3846 = 0.9231 = 0.4167 A B 2. 13 5 12 sin B = 13 5 cos B = 13 12 tan B = 5 12 = 0.9231 = 0.3846 = 2.4000 A Find the Sine, Cosine, and Tangent. 3. sin 32° = 0.5299 4. cos 58°= 0.5299 5. tan 32° = 0.6249 6. Find x and y to the nearest tenth. cos 42˚ = x 12 1 x = 12 cos 42˚ ADJ OPP x ≈ 8.9 sin 42˚ = y 12 1 y = 12 sin 42˚ HYP y ≈ 8.0 7. How tall is the tree? tan 65˚ = x 30 1 x x = 30 tan 65˚ 64.34 ft 65° 30 ft 8. Find x to the nearest tenth. tan 42˚ = 12 x 1 12 = x tan 42˚ 12 x= tan 42˚ x = 13.3 x 42° 12 48˚ ? tan 48˚ = x 12 1 x = 12 tan 48˚ x = 13.3 Use Inverse Sine, Inverse Cosine, and Inverse Tangent. 16 m A = 40° 9. sin A = 25 45 m A = 32° 10. cos A = 53 11. tan A = 0.4402 m A = 24° Use Inverse Sine, Inverse Cosine, and Inverse Tangent. 12. sin-1(0.7660)= A 13. cos-1 5 =A 13 m A = 50° m A = 67° 14. tan-1(11.4300) = A m A = 85° Calculator language: 5 =A cos-1 13 Human language: 5 cos A = 13 15. Solve the right triangle. mA = 28° 30 adj C A opp 16 16 tan A = 30 mB = 62° AB = 34 (AB)2 = 302 + 162 B 16. Solve the right triangle. C 6 sin A = 10 opp A mA = 37° mB = 53° AC = 8 6 10 hyp 102 = (AC)2 + 62 B angle of elevation A C B angle of depression C A B x tan 20°= 8 x = 8 tan 20° x = 2.9 ft ? 30 sin 45°= 26+x 30 = (26+x) sin45° sin 45° sin 45° 42.43 = 26+x 16.43 = x For the most comfortable height, the handle should be 16.43 inches. 60 sin 28° = x x 127.8 300 cos 22° = x x 323.56 x x 44˚ 12 cos 44° = x 12 miles x 16.68 44˚ x tan 44° = 12 x 11.59 12 miles x tan 25° = 250 50 tan 58° = x x 116.6 x 31.2 x cos 32° = 100 x 84.8 30 sin 40° = x x 46.7 Lesson 9.7 Vectors Today, we are going to… …find the magnitude and the direction of a vector …add two vectors The magnitude of a vector PQ is the distance from the initial point to the terminal point and is written | PQ |. WeInuse absolute other words,value it is symbols because “the length of the length vector” cannot be negative. ? To find the magnitude of a vector… Step 1: Identify the vector component form X, Y Step 2: Find the magnitude by simplifying (X)2 + (Y)2 For example, to make AB, we go left 5 units and down 3 units. The component form of AB is -5, -3 __________ To find the magnitude of AB, we simplify (-5)2 + (-3)2 and get 34 5.8 1. Find the magnitude of the vector. B First, the Now, write find the componentusing form the of magnitude the vector. formula. ? A +5 4,5 +4 2 (4) + 2 (5) | AB | ≈ 6.4 2. Find the magnitude of the vector. B component form? magnitude formula? -7,5 +5 (-7)2 + (5)2 -7 A | AB | ≈ 8.6 The direction of a vector is determined by the angle it makes with a horizontal line. 45˚ 45˚ southeast Identify the direction of the vector. 3. 30˚ 30˚ southwest 4. 50˚ 50˚ northwest 5. Find the direction of the vector 5 4 m A = 51˚ sin, cos, tan A = or tan? +5 ? markthe the mark hypotenuse adjacent leg opposite +4 51° northeast opposite 6. Find the direction of the vector sin, cos, ortan tan?A = 5 7 m A = 36˚ +5 ? -7 opposite leg? hypotenuse? 36° adjacent leg? northwest Use +7 because it is length 7. Find the magnitude and direction of AB if A (1,2) and B (4, 6). Step 1: Component Form We doxthis, without drawing the vector! xcan – y – y 2 1 2 1 = X , Y Step 4 –2: 1 ,Magnitude 6 – 2 = 3 , 4 | AB | = 2 (X) + 2 (Y) Step 3: Direction | AB | = (3)2 + (4)2|Y=| 25 5 4 tan A = tan A = 3 |X| 53 northeast How do we know if the vector is north or south, east or west without sketching it? LOOK at the component form! +,+ is right and up -,+ is left and up -,- is left and down +,- is right and down northeast northwest southwest southeast northwest northeast -,+ right,down left+,+ ,down right,up left ,up +,- -,- southwest southeast It might help to think about a map of the US! 8. Find the magnitude and direction of AB if A (-3,3) and B (4, -5). x2 – x1 , y2 – y1 = X , Y – 4 – 3 , – 5 – 3 = 7 , - 8 | AB | = 2 (X) + 2 (Y) = (7)2 + (-8)2 | AB | = 113 10.6 8 |Y| tan A = 7 tan A = |X| 49 southeast Component Form x2 – x1 , y2 – y1 = X , Y Magnitude | AB | = 2 (X) + 2 (Y) Direction |Y| tan A = |X| A north/south - east/west Adding Vectors in Component Form 9. 2,-3 + 5,4 = 7,1 See a pattern? Pick any starting point from movethat right point, 2 and godown right35 and up 4 7,1 2,-3 5,4 instead of taking this route, you could take a short cut 10. -3,5 + 9,-2 = 6,3 See a pattern? from Pickthat aleft starting point, point 9 and down 2 move 3 andgo upright 5 9,-2 short cut? -3,5 6,3 u 3, 4 , v 6,5 , w 2,1 11. uv 12. uw 1,-3 9,1 60 sin 28° = x x 127.8 300 cos 22° = x x 323.56 x x 44˚ 12 cos 44° = x 12 miles x 16.68 44˚ x tan 44° = 12 x 11.59 12 miles x tan 25° = 250 x 116.6 x cos 32° = 100 x 84.8 50 tan 58° = x x 31.2 30 sin 40° = x x 46.7