Grade 10 Trig.


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  • Grade 10 Trig.

    1. 1. By: semba04 (Haley) April 2007
    2. 2. <ul><li>The Pythagorean Theorem is used when you are solving for an unknown side on a right angle triangle. </li></ul><ul><li>The formula is a² + b² = c², where a and b are either the opposite and adjacent sides and c is the hypotenuse. </li></ul><ul><li>If you have a missing angle and need to find the third (your using a right angle) you can subtract your known angle and 90 from 180. </li></ul><ul><li>Note: All the angles in a right angled triangle equal 180°. </li></ul><ul><li>Ex 1)What is the hypotenuse of a right angle triangle if a=3 and b=4? </li></ul><ul><li>Ex 2) What is the missing angle on a right angled triangle if you have a 65° angle? </li></ul><ul><li>Answers on next slide. </li></ul>
    3. 3. <ul><li>Ex 1) </li></ul>Ex 2) NOTE: be sure you calculator is set to degrees.
    4. 4. <ul><li>Angle of Elevation – Any angle above a horizon line of sight. </li></ul><ul><li>Angle of Depression – Any angle below the horizon or line of sight. </li></ul><ul><li>Sine ( s in) = o pposite </li></ul><ul><li> h ypotenuse </li></ul><ul><li>Cosine ( c os) = a djacent </li></ul><ul><li>h ypotenuse </li></ul><ul><li>When you use sin, cos and tan your </li></ul><ul><li>Solving for an unknown angle. </li></ul>Tangent ( t an) = o pposite a djacent NOTE: Angles may be represented by capital letters.Another way to remember cos, sin and tan equations is to remember soh cah toa.
    5. 5. <ul><li>Ex) A 10m ladder rests against the side of a school. The bottom of the ladder is 2m from the wall. What is the angle formed between the ladder and the ground. </li></ul>We know the adjacent and the length of the ladder so we can use cosine to find the unknown angle. 10m Ground Cos Ø = a/h Cos Ø = 2/10 Cos Ø = .2 Now using a scientific calculator you push .2 * 2ndF cos which will give you 78.46. (depending on your calculator you may need to push 2ndF before multiplying it by .2) The answer is 78° If you were to find a length based of an angle you would not use 2ndF (sin^-1)
    6. 6. <ul><li>The sine law states: </li></ul><ul><li>sinA = sinB = sinC </li></ul><ul><li>A b c </li></ul><ul><li>Or: </li></ul><ul><li>a = b = c </li></ul><ul><li>sinA sinB sinC </li></ul>The sine law can be used when you have all the angles and need to find segment lengths. NOTE: Capital letters are used to label angles. Lower case letters are used for line segments.
    7. 7. <ul><li>Ex) Solve Δ ABC (find all missing lengths) using the sine law. </li></ul>85° 68° 27° NOTE: Segment c is opposite angle C. Segment b is opposite angle B and , so on. Segment b: Sin68 = sin27 b 18 18(sin 68) = b(sin27) 16.689 = b(sin27) sin27 sin27 36.67m = b NOTE: Cross multiplied here (at second step). Segment a: a = 18 Sin85 sin27 a(sin27) = 18(sin85) a(sin27) = 17.931 sin27 sin27 a = 39.496 NOTE: I used either or types of formulas however you cannot (when solving for a segment) have a segment as the numerator on one side of the equal sign and a segment as the denominator on the other side of the equals sign.
    8. 8. <ul><li>The cosine law is useful to find the measurement of angles in a non-right angle triangle when we know side lengths but not angle measurements, or two sides adjacent to a single angle. </li></ul>The formulas are: a² = b² + c² - 2bc (cosA) b² = a² + c² - 2ac (cosB) c² = a² + b² - 2ab (cosC) In these formulas you are solving for the side length.
    9. 9. <ul><li>Ex ) Find angle A </li></ul>a² = b² + c² - 2bc (cosA) 3² = 5² + 6² - 2(5)(6) (cosA) 9 = 25 + 36 – 60 (cosA) 9 = 61 – 60 (cosA) 9 = 61 – 60 (cosA) – 61 -52 = -60(cosA) -60 -60 .867 = cosA .867 = 29.89° NOTE: In the fourth step you cannot combine 61 to -60. You must follow the order of operations , your multiplying -60 to cosA. Ex ) Find b b² = a² + c² - 2ac (cosB) b² = 8² + 6² - 2(8)(6) (cos40) b² = 64 + 36 – 96 (cos40) b² = 100 – 93.5 SQRT(b²) = SQRT(26.5) b = 5.1 NOTE: SQRT() is computer language saying ‘square rooting whatever is in the brackets’. A B C 5 3 Ø 6 C b 8 40° A 6 B