anonymous
  • anonymous
Using ∡C of △CBA shown below, create three different trigonometric ratios using sine, cosine, and tangent. You will need to create your own angle and side measures for this task. For full credit, please provide both the ratio and work used to solve for a missing piece of the triangle. Round all solutions to the nearest hundredth.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
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Mertsj
  • Mertsj
\[\sin C=\frac{AB}{AC}\]
Mertsj
  • Mertsj
\[\cos C=\frac{BC}{AC}\]

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campbell_st
  • campbell_st
Sin C = AB/AC cosC = BC/AC Tan C = AC/BC
Mertsj
  • Mertsj
\[\tan C=\frac{AB}{BC}\]
anonymous
  • anonymous
Here goes: Sin C: AB/AC Cos C:BC/AC Tan C:AB/BC Cosec C: AC/AB Sec C: AC/BC Cot C: BC/AB
campbell_st
  • campbell_st
so draw a triangle, meausre the length of one side and the angle C
anonymous
  • anonymous
@CountryGirl: Do you have any kind of parameters such as lengths or angles or something?
anonymous
  • anonymous
no..
anonymous
  • anonymous
Okay I got the answer I think. Let the angles of the triangle be: \[90^{o}, 45^{o}, 45^{o}\] So, Sin C= Sin 45 = \[1/\sqrt{2}\] Cos C=Cos 45 =\[1/\sqrt{2}\] Tan C=Cos 45 =1 These are given the ratios IN TERMS OF ANGLES. Now, Let: AB= 2 cm BC= 2 cm AC= 4cm Sin C: AB/AC = 2/4 = 1/2 Cos C:BC/AC = 2/4 = 1/2 Tan C: AB/BC = 2/2 = 1 These are the ratios IN TERMS OF SIDES. Hope that helps. We took some angles and sides and based on that we found the ratios :)

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