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Chapter 7 Right trianlges and Trigonometry SECTION 7.1: Geometric Mean A Geometric Mean is the multiplicative average between two numbers. The Geometric Mean, x, is proportionally between the numbers a and b. = Find the Geometric Mean between each pair of numbers. 1. 4 and 9 2. 6 and 15 Find the Geometric Mean between each pair of numbers. 1. 4 and 9 2. 6 and 15 4 1. = 9 4(9) = x(x) 36 = x2 36 = 2 6=x 6 2. = 15 6(15) = x(x) 90 = x2 90 = 2 9.49 = x When the Altitude of a Right Triangle is drawn from the Right Angle to the Hypotenuse, it divides the Triangle into 3 Similar Triangles… ∆ACD ~ ∆CBD ~ ∆ABC Baby ~ Momma ~ Poppa B B C A D C D A C Therefore there are 3 different Geometric Means formed • #1 The altitude splits the hypotenuse, h is the GM between x and y = • #2 & #3 The legs each connect twice to the hypotenuse, • b is the GM between x and c • a is the GM between y and c = = In ∆PQR, RS = 3 m and QS = 14 m. Find PS. In ∆PQR, RS = 3 m and QS = 14 m. Find PS. PS is the GM between RS and QS. Let PS = x. = 42 = x2 = x ≈ 6.48, PS ≈ 6.48 m Find x and y in ∆PQR. Find x and y in ∆PQR. SECTION 7.2: The Pythagorean Theorem and its Converse Solve for x. Solve for x. inches Groups of INTEGERS that form Right Triangles are called PYTHAGOREAN TRIPLES • 3, 4, 5 • 5, 12, 13 • 7, 24, 25 • 8, 15, 17 • 9, 40, 41 • 11, 60, 61 You can use them with multipliers too…. 3, 4, 5 doubled is 6, 8, 10 • They’re worth remembering… Determine if ∆PQR is a right Triangle. Determine if ∆PQR is a right Triangle. Since the sum of the squares of the shorter two sides is equal to the square of the longest side, ∆PQR is a right triangle. SECTION 7.3: Special Right Triangles In your groups, solve each triangle for “x”. a. b. x 1 1 c. x 2 2 x 3 3 Answer the following questions about the ∆s: • What type of triangle are these? Be as specific as possible. • What are the three angles measures of the triangles? • What pattern do you see? • Are there any conclusions your group can draw about this type of triangle? A ∆ABC is an equilateral triangle with side length of 2x units. 1 is an Altitude and Median. Find: • The value of a • The value of b • The measure of ∠1 • The measure of ∠2 • The measure of ∠3 b C 2 3 D a B A is an Altitude and Median. Find: • The value of a = x • • • • 1 The value of b = x 3 The measure of ∠1 =30° The measure of ∠2 =60° The measure of ∠3 =90° b C 2 3 D a B 45° - 45°- 90° Triangles Angle Measure 45° 45° 90° Side Across n n 2 Find x. Find x. 30° - 60°- 90° Triangles Angle Measure Side Across 30° 60° 90° n n 3 2n Find the length of segment AC. Find the length of segment AC. Angle Measure 30° 60° 90° Side Across n n 3 2n AB AC BC Find the length of segment AC. Angle Measure 30° Side Across n 60° n 3 90° 2n in SECTION 7.4: Trigonometric Ratios Trigonometric Ratios define relationships between 1 Angle and 2 Sides of a Right Triangle: • The Sine of an angle is the ratio of Opposite Side to Hypotenuse • The Cosine of an angle is the ratio of Adjacent Side to Hypotenuse • The Tangent of an angle is the ratio of Opposite Side to Adjacent Signs Of Happiness - Come After Having Taken On Algebra •Soh •Cah •Toa •sin Ꝋ = ℎ •cos Ꝋ = ℎ •tan Ꝋ = The “sides” depend on the angle: • Angle A • Angle B B B 5 5 A A C C 4 4 • sin A = 3 3 3 5 • tan A = cos A = 3 5 • sin B = 4 5 • tan A = cos B = 4 5 Steps for Solving Soh Cah Toa Problems: • 1. Identify the Angle • 2. Label/Identify the Sides (hypotenuse, opposite, adjacent) • 3. Match with Soh, Cah, Toa • 4. Plug in values • 5. Solve (3 methods – multiply, switch or inverse) Method 1: “x” on top… MULTIPLY B 10 x A 41° C Method 1: “x” on top… MULTIPLY B 10 x A 41° C • Angle is A, 41° • Hypotenuse = 10, Opposite = x • Soh, Cah, Toa…. H and O = SINE sin 41° = 10(sin 41° ) = ( )10 10(sin 41° ) = x ≈ 6.56 u Method 2: “x” on bottom… SWITCH B x 62° 12 A C Method 2: “x” on bottom… SWITCH B x 62° 12 • Angle is B, 62° • Hypotenuse = x, Adjacent = 12 • Soh, Cah, Toa…. H and A = COS cos 62° = SWITCH! A C x= cos 62° x ≈ 25.56 u Method 3: “x” is the angle… INVERSE B 11 A x° C 18 Method 3: “x” is the angle… INVERSE B 11 • Angle is A, x° • Opposite = 11, Adjacent = 18 • Soh, Cah, Toa…. O and A = TAN tan x° = A x° C 18 INVERSE! 2nd Button − (tan x°) = − ( ) x ≈ 31.43° SECTION 7.5: Angles of Elevation and Depression Angles of Elevation and Depression •Angles formed by “line of sight” and a HORIZONTAL line •Elevation is “looking up” •Depression is “looking down” •∠Elevation ≅ ∠Depression by “Z-Rule” •Just SOH CAH TOA Word Problems Mrs. Evans is standing on cliff, 40 meters above the ground. She sees a group of students in need of tutoring, at an angle of depression of 34°. How far from the base of the cliff are they standing? Since the angle of depression is equal to the angle of elevation, we can use either. The angle of elevation is easier to use, since the “triangle” is clear. Once you’ve drawn the diagram and labelled the figure, then just use our steps for Soh Cah Toa problems. Angle is 34° Opposite side = 40, Adjacent Side = x Soh, Cah, Toa…. O and A = TANGENT tan 34° = x= tan 34° x ≈ 59.30 m SECTION 7.6: Law of Sines Find b. Round to the nearest hundredth. Find b. Round to the nearest hundredth. b ≈ 4.62 u • Draw a picture to help: • Draw a picture to help: We need to find C, a, and c to Solve the Triangle. C is easy, using the 180° Rule. 33 + 47 = 80. 180 – 80 = 100. Therefore, ∠C = 100° Finding sides a and c, use the Law of Sines. SECTION 7.7: Law of Cosines Solve for a side if you know 2 sides and the angle between them: = + − Or solve for an angle if you know all three sides: + − = a ≈ 9.17 u + − = + − = () = = − ( ) R ≈ 29.09°