Week 16 – Pre-calculus 11

This week in Math 11 we started grade 11 trigonometry and we learned how to make a right triangle and from coordinates of a point that is given. With the right triangle we can find the the opposite and adjacent side, along with the hypotenuse and the reference angle. First we find our point on the graph and draw a diagonal line from the origin to the point, then from the point we draw a vertical line connecting it to the horizontal line or x-axis. then based on our coordinates given we will know the values of our vertical and horizontal lines but we will not know the value of our hypotenuse. We can figure out our hypotenuse by using Pythagoras theorem which is a^2 + b^2 = c^2. Once we know all the lengths we can figure out all of our ratios and then find our reference angle.

Example:

Point (6, 4)

we have our point and not we draw a line between the point and origin

Then draw a vertical line between the point and x-axis

now we start to see our right triangle

The points that were given to us in the beginning tell us the value of 2 of the sides on our triangle

From here we need to find the value of our hypotenuse

  • to do this we use a^2 + b^2 = c^2
  • our hypotenuse is our c^2
  • and 6 is “a” and 4 is “b”
  • a^2 + b^2 = c^2
  • 6^2 + 4^2 = c^2
  • 36 + 16 = c^2
  • 52 = c^2
  • \sqrt{52} = \sqrt{c^2}
  • c = \sqrt{52}

Now that we know the hypotenuse we can now find out all our ratios for Sin, Cos, and Tan, along with our reference angle

based off where are reference angle is

  • For Sin it is \frac{O}{H}, where “4” is our opposite and “\sqrt{52}” is our hypotenuse.
  • For Cos it is \frac{A}{H}, where “6” is our adjacent and “\sqrt{52}” is our hypotenuse.
  • For Tan it is \frac{O}{A}, where “4” is our opposite and “6” is our adjacent

So our ratios would be

  • Sin \theta = \frac{4}{\sqrt {52}}
  • we don’t like having a square root on the bottom so what we can do is rationalise the denominator to make our answer look better
  • Sin \theta = \frac{4}{\sqrt {52}} x \frac{\sqrt {52}}{\sqrt {52}}
  • Sin \theta = \frac{4\sqrt{52}}{52}
  • We would do this for all ratios to make them look better
  • Cos \theta = \frac{6\sqrt{52}}{52}
  • Tan \theta = \frac{4}{6}

Now that we have all of our ratios we can solve one of them with a calculator to find our reference angle and we would that number to the nearest degree.

  • Tan \theta = \frac{4}{6}
  • \theta = (tan^-1)(0.667)
  • \theta = 34°
  • and this is our reference angle

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