Right Triangle Problems and Exact Values

Parts of a Degree

We get degree measure from the ancient Babylonians.   They defined a degree as 1/360 of a circle.  We also get 60 minutes in an hour and 60 seconds in a minute from them, as they developed a "sexagesimal" numbering system around 2,000 B.C.  This Babylonian positional system is still in wide use today, and degrees-minutes-seconds are frequently used for angular measure. Angles that contain fractions of a degree are given in terms similar to time: 1/60 of a degree is called a "minute" and 1/60 of a minute is called a "second."  You are already familiar with the superscript o for "degree."  The symbol for a "minute" is a single quote and the symbol for a "second" is a double quote.  For example, 32o 22' 48" represents an angle of 32 degrees, 22 minutes, and 48 seconds.  In terms of degrees,

When degrees are given in decimal form, they can be converted to degrees, minutes and seconds, as the following example illustrates:

Example:  Convert 54.78o to degrees, minutes and seconds.

Solution:  54.78o = 54o + 0.78(60') = 54o + 46.8' = 54o + 46' + 0.8(60") = 54o + 46' + 48"

That is, 54.78o = 54o 46' 48" (54 degrees, 46 minutes, 48 seconds).

Finding Angles

When two sides of a right triangle are known, we can use the trig functions to find the angles of the triangle.  With a calculator, we need to "undo" what we might have done if we had known the angle.  Fortunately, calculators include "inverses" of the trig functions to do this.  The standard notation for the inverse of a function is to use a –1 as an exponent after its name.  The following are the definitions of the inverses of the sine, cosine and tangent, as found on any standard scientific calculator:

For example,

So if you need to find an angle in a triangle whose sides are given, you can use one of the inverse functions, as in the following example:

Example:  Find the measure of the smallest angle in a 3-4-5 right triangle.

Solution:

The smallest angle is the angle opposite the smallest side:

We could use any of the trig functions to find this angle.  If we use the sine, then we have (using a calculator):

If instead we use the tangent, we have:

The Airplane Problem

An airplane is flying at an altitude of 31,000 feet when it begins its descent for landing.  If the airport is 112 miles away, at what angle does the airplane descend?

Solution:  The angle of descent, θ, is the same as the angle of elevation from the airport to the airplane:

 

We can set up a ratio using the tangent function:

 

Most large airplanes descend at about 3 degrees.  Notice we had to convert 112 miles to feet in order to solve this problem.

Exact Values

As mentioned in the previous activity, the exact values of the trig functions of certain "special angles" can be found.  In addition, if a ratio of two sides of a right triangle is known, then this ratio can be used to find the exact values of sides of any similar triangles.

First let us summarize the special angles.

The trig functions of 30o and 60o can be found by considering any 30-60-90 right triangle, since the ratios reduce to the same value.  In any such triangle, the leg opposite the 30o angle is half the hypotenuse and the other leg is the square root of 3 times as long as the short leg.  Thus if we label the hypotenuse as 2a then we have:

From this triangle we can reduce the ratios for the trig functions.  Here are the first three functions of 30o and 60o:

We could just as well have taken a to be 1.  Then we would get the same results.

For 45o, we could use any isosceles right triangle.  Since the legs have same length, we could choose a for that common length.  Then the hypotenuse would have a length of a times the square root of 2:

The exact values of sine, cosine and tangent of 45o are:

Sometimes a special angle or a ratio is given and we need to find the exact length of a side.  The following problems illustrate this.

Problem 1

Find the exact value of x:

Solution:

Relative to the 60o angle, 9 is opposite and x is the hypotenuse, so we can use the sine ratio:

Replacing sin 60o with its exact value, we have:

Then we cross-multiply and find x:

Problem 2

In the following triangle, sin θ = ¼ .  Find the exact length of side AC.

Solution:

First we find the exact value of BC and then use the Pythagorean Theorem to find AC:

Problem 3

In a right triangle, one leg is twice as long as the other leg.  Find the exact value of the sine of the smallest angle in this triangle.

Solution:

We can use any length for the shortest leg, so let us make it 1 unit.  Then the other leg is 2 units long, and we can use the Pythagorean Theorem to find the hypotenuse:

Now we can give the value of sin θ:

Problem 4

Find the exact values of cos θ and tan θ if

Solution:

We can draw any right triangle in which

Then we use the Pythagorean Theorem to find the missing side:

Now we can use this triangle to find the exact values of cos θ and tan θ :

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