Basic Trigonometric Ratios: Examples (page 1 of 2)

• List the values of sin(α), cos(α), sin(β), and tan(β) for the triangle below, accurate to three decimal places:

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.

• In the triangle shown below, find the value of x, accurate to three decimal places.

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
x
= 65/sin(20°)

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.

 For the triangle shown, find the value of y, accurate to four decimal places. They've given me an angle, a value for "adjacent", and a variable for "opposite", so I can form an equation: tan(55.3°) = y/10 10tan(55.3°) = y Plugging this into my calculator, I get y = 14.44183406.... y = 14.4418
 Find the angles and sides indicated by the letters in the diagram. Give each answer correct to the nearest whole number. At first, this looks fairly intimidating. But then I notice that, to find the length of the height r, I can use the base angle 30° and the full base length of 60, because r/60 is "opposite" over "adjacent", which is the tangent.

r/60 = tan(30°)
r = 60tan(30°) = 34.64101615...

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°)
35/tan(55°) = s = 24.50726384...

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.

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 Cite this article as: Stapel, Elizabeth. "Basic Trigonometric Ratios: Examples." Purplemath. Available from     http://www.purplemath.com/modules/basirati.htm. Accessed [Date] [Month] 2014

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