Grade 10 Trig.

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This show was created for my wiki contribution in math class

This show was created for my wiki contribution in math class

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  • 1. By: semba04 (Haley) April 2007
  • 2.
    • The Pythagorean Theorem is used when you are solving for an unknown side on a right angle triangle.
    • The formula is a² + b² = c², where a and b are either the opposite and adjacent sides and c is the hypotenuse.
    • 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.
    • Note: All the angles in a right angled triangle equal 180°.
    • Ex 1)What is the hypotenuse of a right angle triangle if a=3 and b=4?
    • Ex 2) What is the missing angle on a right angled triangle if you have a 65° angle?
    • Answers on next slide.
  • 3.
    • Ex 1)
    Ex 2) NOTE: be sure you calculator is set to degrees.
  • 4.
    • Angle of Elevation – Any angle above a horizon line of sight.
    • Angle of Depression – Any angle below the horizon or line of sight.
    • Sine ( s in) = o pposite
    • h ypotenuse
    • Cosine ( c os) = a djacent
    • h ypotenuse
    • When you use sin, cos and tan your
    • Solving for an unknown angle.
    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.
    • 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.
    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.
    • The sine law states:
    • sinA = sinB = sinC
    • A b c
    • Or:
    • a = b = c
    • sinA sinB sinC
    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.
    • Ex) Solve Δ ABC (find all missing lengths) using the sine law.
    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.
    • 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.
    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.
    • Ex ) Find angle A
    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