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trigonometry

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trigonometry
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  • Slajd: 1
  • Today I'm going to be teaching a few different things but they all connect in a certain way
  • Slajd: 2
  • over here on the board we have 2 triangles and we have missing sides !! can you help me with them ?
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  • 10
  • ?
  • 6
  • ?
  • 4
  • In order to find a missing side it depends on what side youre looking for. If youre looking for the hypotenuse in this example it would be 6^2+4^2=c^2 or 36+16=c^2. 52=c^2 so, we would use the square root to find c. c=7.2. The hypotenuse can be a decimal if a and b are not perfect squares. To find the leg in the second triangle, it would be 100+b^2=169. b^2=69. If you take the square root of 69 you get b=8.3
  • Slajd: 3
  • The next part of this is Solving for a side of a triangle using trigonometry. Remember SOHCAHTOA!! S=sin, C= cos, T=tan, O=opposite, A=adjacent, H=hypotenuse
  • x
  • 19
  • 69°
  • In this example its different, now we have an angle! but now our sides have names : the tallest side (x) is called a hypotenuse , and the side right next to the angle is called adjacent ,and finally the one OPPOSITE the angle is called opposite . now we try to find what is common between the sides , we have they hypotenuse and the opposite which will make us use the SIN ! put on the calculator the following : 19/sin(69) and we'll get how much is X !
  • Slajd: 4
  • and now our final part ! a bit harder but i'm sure you can do it !
  • 31
  • ?
  • 13
  • For this you would use cos. Usually for x we would use something called theta(θ). So for this problem it would be Cosθ=13/31. Since we're finding an angle we have to use inverse cos. θ=cos^-1(13/31). Put the second equation into a calculator and you'll get 65.2 degrees for your missing angle.
  • Slajd: 5
  • now your turn to solve ! take it slowly step by step and i am sure you'll find the way , if you got stuck just take a quick look at the examples we did before !
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  • Slajd: 6
  • THE END
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