Math 10 Week # 7 Solving Trig Equations

Following spring break, my math class began on a new unit on trigonometry. Trigonometry is the study of triangles and the relationship between angles and sides. For grade 10, we are only working with right angles where one angle is 90° so that in grade 11 and 12, it will be a lot easier to learn trigonometry with isosceles or scalene triangles. So for the past week, we have been finding the value of angles and lengths of triangles by using ratios called Sine, Cosine, and Tangent ratio. These are ratios between lengths of sides of a triangle and by using these, we only needing the value of one side to find them. Last year and in previous years, we learned about the Pythagorean Theorem and that it is a formula which can help you find the value of one side of a triangle as long as you have the other two sides. In trigonometry, you only need one side, and an angle to determine the value of another side. The most important thing we can remember to use in trigonometry is a phrase called SOH CAH TOA. These are small abbreviations to show how to find a ratio between sides. In these next few examples, we will find out what they mean and how to find missing side lengths and angle values.

For example 1, we are going to find out how to find a length of a side using the cosine ratio. Going back to SOH CAH TOA, we learn that the first letter stands for what ratio we are using and the next two is which side is going to divide in another. So in this example, the opposite side is unknown, the adjacent side is 13 cm and the hypotenuse is X. Usually, variables tell us that we need to find out their value. So because we have a value in our adjacent side and a variable which has a value for the hypotenuse, we see where this combination of letters matches in the phrase SOH CAH TOA. We then determine that the ratio we are going to use is CAH or Cosine = \frac{A}{H}, A standing for the adjacent side and H standing for the hypotenuse. So to finally begin our question, we now know what our equation is going to be. We start out with Cos (short for Cosine) 37° (we need a reference angle to determine the ratio) = \frac {13(Adjacent)}{X(Hypotenuse)}; Cos37° = \frac {13}{X}. Next what we do is we reciprocate (switch from the numbers between the numerator and denominator) 13 and X (because it is easier when the variable is in the numerator) and we do the same on the other side of the equation we end up with \frac{1}{Cos37} = \frac{X}{13}. Next we want to isolate the variable so we cancel out 13 by multiplying it and whatever we do on one side we do to the other, multiplying 13 by the 1 on the numerator. After all of this we end up with X = \frac{13}{Cos37}. We finally finish by dividing 13 by Cos37 (easiest way to do this is on the calculator) and we get 16.3 cm.

For example 2, we are going to find the length of a side using the sine ratio. Again going back to SOH CAH TOA, we see that the values we are given is the opposite side to the reference angle which is X and the hypotenuse which is 6.4 m. We again look to see which of these two letters match and find out we are going to use SOH or the Sine ratio. So since we know what ratio we are using, our equation we are using is Sine59° = \frac{X}{6.4}. So to begin, we isolate x by multiplying 6.4 to that side and to the other side to end up with X = Sin59° • 6.4. Next we multiply Sin59º with 6.4 on our calculator and we end up with X = 5.5 m.

And finally for example 3, we are going to find the missing value for our reference angle using the tangent ratio. To do this, we need have at least the value of two sides of a triangle to complete this. Once again using SOH CAH TOA, we find out that we are going to need to use the tangent ratio because we are given the values of the opposite side (or side opposite of our unknown reference angle) and the adjacent side. A and O both match with TOA so we are going to use the Equation TanX = \frac{65}{44}. First, we isolate the variable so we divide Tangent from the left and we also do it on the right so it becomes inverse tangent or Tan^-1 • \frac{65}{44}. Next we multiply Tan^-1 by \frac{65}{44} to get X = 56°.

 

 

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