

Basic Trigonometric Ratios: Examples (page 2 of 2)
For either angle, the hypotenuse has length 9.7. For the angle α, "opposite" is 6.5 and "adjacent" is 7.2, so the sine of α will be 6.5/9.7 = 0.6701030928... and the cosine of α will be 7.2/9.7 = 0.7422680412.... For the angle β, "opposite" is 7.2 and "adjacent" is 6.5, so the sine of β will be 7.2/9.7 = 0.7422680412... and the tangent of β will be 7.2/6.5 = 1.107692308.... Rounding to three decimal places, I get: sin(α) = 0.670, cos(α) = 0.742, sin(β) = 0.742, tan(β) = 1.108 Once you've memorized the trig ratios, you can start using them to find other values. You'll likely need to use a calculator. If your calculator does not have keys or menu options with "SIN", "COS", and "TAN", then now is the time to upgrade! Make sure you know how to use the calculator, too; the owners manual should have clear instructions.
They've given me an angle measure and the length of the side "opposite" this angle, and have asked me for the length of the hypotenuse. The sine ratio is "opposite over hypotenuse", so I can turn what they've given me into an equation: sin(20°) = 65/x I have to plug this into my calculator to get the value of x: x = 190.047286... x = 190.047 Note: If your calculator displayed a value of 71.19813587..., then check the "mode": your calculator is set to "radians" rather than to "degrees". You'll learn about radians later.
r/60 = tan(30°) I'm supposed to the nearest whole number, so r = 35. Now that I have the value of r, I can use r and the other base angle, 55°, to find the length of the other base, s, by using r/s = tan(55°): 35/s = tan(55°) r = 35, s = 25 Note: Since the sine and cosine ratios involve dividing a leg (one of the shorter two sides) by the hypotenuse, the values will never be more than 1, because (some number) / (a bigger number) from a right triangle is always going to be smaller than 1. But you can have really wide and short or really tall and skinny right triangles, so "opposite" and "adjacent" can have very different values. This tells you that the tangent ratio, being (opposite) / (adjacent), can have very large and very small values, depending on the triangle. Original URL: http://www.purplemath.com/modules/basirati2.htm Copyright 2009 Elizabeth Stapel; All Rights Reserved. 