Right Triangle Model and the Pythagorean Theorem Help (page 2)

By — McGraw-Hill Professional
Updated on Oct 25, 2011

Range of Angles

In the right-triangle model, the values of the circular functions are defined only for angles between, but not including, 0° and 90° (0 rad and π/2 rad). All angles outside this range are better dealt with using the unit-circle model.

Using the right-triangle scheme, a trigonometric function is undefined whenever the denominator in its “side ratio” (according to the formulas above) is equal to zero. The length of the hypotenuse (side f ) is never zero, but if a right triangle is “squashed” or “squeezed” flat either horizontally or vertically, then the length of one of the adjacent sides ( d or e ) can become zero. Such objects aren’t triangles in the strict sense, because they have only two vertices rather than three.

The Right Triangle Model Practice Problems

Practice 1

Suppose there is a triangle whose sides are 3, 4, and 5 units, respectively. What is the sine of the angle θ opposite the side that measures 3 units? Express your answer to three decimal places.

Solution 1

If we are to use the right-triangle model to solve this problem, we must first be certain that a triangle with sides of 3, 4, and 5 units is a right triangle. Otherwise, the scheme won’t work. We can test for this by seeing if the Pythagorean theorem applies. If this triangle is a right triangle, then the side measuring 5 units is the hypotenuse, and we should find that 3 2 + 4 2 = 5 2 . Checking, we see that 3 2 = 9 and 4 2 = 16. Therefore, 3 2 + 4 2 = 9 + 16 = 25, which is equal to 5 2 . It’s a right triangle, indeed!

It helps to draw a picture here, after the fashion of Fig. 12-9. Put the angle θ , which we are analyzing, at lower left (corresponding to the vertex point Q ). Label the hypotenuse f = 5. Now we must figure out which of the other sides should be called d , and which should be called e . We want to find the sine of the angle opposite the side whose length is 3 units, and this angle, in Fig. 12-9, is opposite side PR , whose length is equal to e . So we set e = 3. That leaves us with no other choice for d than to set d = 4.

According to the formulas above, the sine of the angle in question is equal to e / f . In this case, that means sin θ = 3/5 = 0.600.

Practice 2

What are the values of the other five circular functions for the angle θ as defined in Problem 12-5? Express your answers to three decimal places.

Solution 2

Plug numbers into the formulas given above, representing the ratios of the lengths of sides in the right triangle:

cos θ = d / f = 4/5 = 0.800 tan θ = e / d = 3/4 = 0.750 csc θ = f / e = 5/3 = 1.667 sec θ = f / d = 5/4 = 1.250 cot θ = d / e = 4/3 = 1.333

Find practice problems and solutions for these concepts at: Trigonometry Practice Test.

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