Download Lecture Notes 4.4 Right Triangle Trig

4.4
Right Triangle Trigonometry
Trigonometry is introduced to students in two different forms, as functions on the unit circle and as
functions on a right triangle. The unit circle approach is the most natural setting for the trig functions
since they are not just functions of angles between 0◦ and 180◦ but instead have as domain the set of all
real numbers. The unit circle explains identities such as (cos θ)2 + (sin θ)2 = 1 and cos(θ) = sin(θ + π2 ).
However, we would also like to apply trigonometry to right triangles which have a hypotenuse of
length different than one. We may do this by using similar triangles.
4.4.1
Similar Triangles
Previously, to examine our trig functions, we displayed a typical triangle on the unit circle with central
angle θ, hypotenuse 1 and point P (x, y). We can expand that triangle by the ratio r to get a triangle in
which the hypotenuse has length r and the point P (x, y) is on a circle of radius r. When this happens,
cos(θ) will not be x but x/r. Similarly, sin(θ) will be y/r.
Or – to repeat an idea from section 4.2 of these notes – we could imagine a triangle with hypotenuse
of length H, opposite side of length O, adjacent side of length A. The point P (x, y) sits on the circle of
radius H and sin(θ) = O
A.
All of these triangles are similar and so the trig functions of the angle θ, as ratios of side lengths, are
unchanged.
P (x, y)
1
θ
y
x
P (x, y)
P (x, y)
r
H
y
θ
O
x
θ
A
Figure 17. Three similar triangles we can use when we examine trig functions
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If we are willing to draw a right triangle and use the Pythagorean theorem, then we can solve any
right triangle problem in which we are given a side and another angle. Here are some examples.
Some worked problems using similar triangles.
1. Find sec(θ) if sin(θ) = 35 .
Solution. Draw a right triangle which has an angle with sin(θ) = 53 . (A 3-4-5 triangle will do.) Then
hyp
H
compute the secant of the angle θ. The secant is the reciprocal of cosine and so sec(θ) =
= .
adj
A
The answer is 45 .
2. Find tan(α) if cos(α) = 25 .
Solution. Draw a right triangle which has an angle with cos(θ) = 25 . (The most obvious triangle
will have a hypotenuse√of length 5 and an side adjacent to θ with length 2.) The “opposite” side
will then have length 21, by the Pythagorean
Theorem. Then compute the tangent of the angle
√
21
.)
The
answer
is
.
θ. (Tangent is the ratio opp
adj
2
3. Suppose sec θ =
the angle θ.
7
2
and tan θ is negative. Identify the angle θ and then find all six trig functions of
Solution. Since the tangent is negative and cosine (= 27 ) is positive, then the angle θ must be in the
fourth quadrant. Draw a line segment of length 7 from the origin into
√ the fourth√quadrant,
√ to a point
√
P (2, y). By√the Pythagorean theorem, the absolute value of y√is 72 − 22 = 45 = 9 · 5 = 3 5.
So y = −3 5 and our line segment ends at the point P (2, −3 5).
Now read off the values of the various trig functions:
√
−3 5
7 , tan θ
cos θ = 72 , sin θ =
=
√
−3 5
2 .
The reciprocals of these are
√
√
7
2
sec θ = 72 , csc θ = − 3√
= − 7155 , cot θ = − 3√
= − 2155 .
5
5
4.4.2
Applications with right triangles
Anytime we have a right triangle, then, if we can measure one of the acute angles and also know the
length of a side, then we know everything about the triangle. Consider the figure below.
H
O
θ
A
Figure 18. A typical right triangle
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In most of the applications in this section, we will be able to draw a triangle like that in the figure above.
(The right triangle may have a different orientation; the right angle may be a the top of our figure or
on the leftside... but we should be able to draw a right triangle when we start these problems!) We will
then find one of the acute angles of the triangle (such as θ, drawn in red) and we will also be able to find
the length of one of the sides. Once we have this information, the lengths of the other two sides can be
computed using our trig functions.
Here is are some examples involving practical applications.
1. A radio tower is stands on a flat field. I walk 1000 feet away from the base of the radio tower and
look up at the top of the tower. I measure a 71◦ angle between the horizon and the top of the
tower. How tall is the tower?
Solution. We begin by drawing a picture much like Figure 18, above. The radio tower is a vertical
straight line perpendicular to the ground. We want to know the length of this vertical line segment.
(Call its length O.) Draw the ground as a horizontal line and mark the length of that horizontal line
as A = 1000 feet. The hypotenuse of this right triangle makes an angle θ = 71◦ with the ground.
A
The tangent of 71◦ is tan 71◦ = O
A = 1000 . Solving for O we have
O = 1000 · tan 71◦ ≈ 2, 904 feet.
So the radio tower is 2904 feet tall.
2. I am flying a kite on the beach. The kite is attached to 3000 feet of string. At the time that the
string plays out and I have used all the string, the kite makes an angle with the horizon of 38◦ .
Assuming that the 3000 feet of string is a straight line, how high is the kite?
Solution. Again, as before, begin by drawing a picture! (See Figure 18.) The kite and the person
holding the kite and the ground directly below the kite form the three vertices of a right triangle
with the right angle being at the point on the ground directly below the kite. The hypotenuse
of this right triangle is H = 3000 feet long and the kite is O feet above the ground. The sine of
O
O
θ = 38◦ is sin 38◦ = H
= 3000
and so the height O is equal to
O = 3000 · sin(38◦ ) ≈ 1847 feet.
So the kite is at heigth 1847 feet (which is quite high!!)
4.4.3
Parallax
Astronomers use simple right triangles to find the distance to nearby stars. Here is how they do this.
As the earth revolves around the sun, it marks out a circle2 of radius 93 million miles. (This is a fact
worth remembering: the distance from the earth to the sun is about 93,000,000 miles or about 8 light
minutes.) Over the course of the year, a nearby star should appear to move back and forth in the night
sky as the earth revolves around the sun and so one should be able to measure that angle of apparent
motion and use a right triangle (with one side equal to 93,000,000 miles) to compute that distance.
In the figure below, we let the sun from a right angle vertex of a triangle and we set the earth and
star as the other two vertices. If the star does not move, it would appear to form a right angle with the
earth in this figure. But if the star is “nearby” then the line of sight to the star forms an angle α with
the anticipated line of sight. The star appears to have moved.
2 To be precise, the earth’s orbit is really an ellipse but the shape of this ellipse – fortunately for us! – is very close to
the shape of a circle.
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α
A
α
Earth
O
Sun
Earth
Figure 19. Parallax.
By a standard result from geometry, this angle α (the apparent motion of the star) is also the acute
angle of the triangle at the vertex given by the star. If A is the distance from the star to the sun and
A
O = 93, 000, 000 = 9.3 × 107 miles then the cotangent of α is O
and so A = 9.3 × 107 × cot α.
The ancient Greeks thought of this idea and attempted to measure parallax. But when they did this,
the stars didn’t seem to move!! So either this picture was wrong (maybe the earth was the center of the
universe?) or the stars must be billions of miles away! Convinced that the universe could not be billions
of miles in size, most Greeks agreed with Aristotle’s belief that the earth was the center of the universe
and that the sun revolved around the earth.
Now we know better – and indeed, with modern equipement, we have been able to measure the
parallax of some stars.
A worked problem.
The closest star to us, Proxima Centauri3 has a parallax of 0.77”, that is, 0.77 arcseconds. How far
away is Proxima Centauri?
Solution. A minute of arc is one-sixtieth of a degree; a second of arc is one-sixtieth of a minute. So an
1
0.77
degrees. The angle 0.77 arcseconds is equal to 0.77
arcsecond is 6012 = 3600
602 = 3600 degrees. The tangent of
0.77 ◦
0.77 ◦
this angle is tan( 3600 ) ≈ 0.00000373307. The cotangent of 3600 is the reciprocal of this, approximately
267876. So the distance to Proxima Centauri is 267876 · 93000000 = (2.67876 × 105 ) · (9.3 × 107 ) miles or
3 “Proxima”
means “close”
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about 2.49 × 1013 miles!
The fastest rocket ever made reaches speeds of 25,000 miles per hour. A rocket traveling at that speed
would take over one hundred thousands years to reach Proxima Centauri. Light travels about 5.879×1012
miles in a year, so this distance is about 4.24 light years. And this is our closest star.... It is no wonder
that the Greeks could not measure parallax; the universe is indeed unbelievably large!
4.4.4
Other resources on the trigonometry of right triangles
In the free textbook, Precalculus, by Stitz and Zeager (version 3, July 2011, available at stitz-zeager.com)
this material is covered in section 10.3
In the free textbook, Precalculus, An Investigation of Functions, by Lippman and Rassmussen (Edition
1.3, available at www.opentextbookstore.com) this material is covered in section 5.5.
In the textbook by Ratti & McWaters, Precalculus, A Unit Circle Approach, 2nd ed., c. 2014 much of
this material appears in section 6.1 In the textbook by Stewart, Precalculus, Mathematics for Calculus,
6th ed., c. 2012 (here at Amazon.com) this material appears in chapter 5.
Many online sources introduce trigonometry by beginning with a right triangle. Khan Academy
has a number of videos on this topic.
Homework.
As class homework, please complete Worksheet 4.4, Trig Functions and Right Triangles available through the class webpage.
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