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Trigonometry on Right Triangles
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 trig functions 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
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Part 4, Trigonometry
Lecture 4.4a, Trigonometry on Right Triangles
(cos θ)2 + (sin θ)2 = 1
Dr. Ken W. Smith
and
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cos(θ) = sin(θ + π2 ).
2013
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.
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Similar Triangles
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) on the
unit circle.
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.
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Similar Triangles
Similar Triangles
Or – imagine a triangle with hypotenuse of length H, opposite side of
length O, adjacent side of length A.
All of these triangles are similar and so the trig functions of the angle θ,
as ratios of side lengths, are unchanged.
The point P (x, y) sits on the circle of radius H and sin(θ) =
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O
A.
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Similar Triangles
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Similar Triangles
2
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.
Some worked problems using similar triangles.
1
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Find sec(θ) if sin(θ) = 53 .
Find tan(θ) if cos(θ) = 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. Compute the
√ tangent of
opp
the angle θ. (Tangent is the ratio adj .) The answer is 221 .
Solution. Draw a right triangle which has an angle with sin(θ) = 53 .
(A 3-4-5 triangle will do.) Then compute the secant of the angle θ.
hyp
H
The secant is the reciprocal of cosine and so sec(θ) =
= .
adj
A
The answer is 54 .
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Similar Triangles
3
Suppose sec θ =
the angle θ.
Trig on right triangles
7
2
and tan θ is negative. Find all six trig functions of
Solution. Since the tangent is negative and cosine (= 27 ) is positive,
then we know x is positive and y is negative and so the angle θ points
into 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,
is
√
√ the absolute
√ value of y √
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:
cos θ = 27 , sin θ =
√
−3 5
7 ,
tan θ =
In the next presentation, we will apply our understanding of trigonometry
to solving various right triangles.
(End)
√
−3 5
2 .
The reciprocals of these are
√
√
7
2
sec θ = 72 , csc θ = − 3√
= − 7155 , cot θ = − 3√
= − 2155 .
5
5
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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.
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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.
Part 4, Trigonometry
Lecture 4.4b, Applications of Right Triangles
Once we have this information, the lengths of the other two sides can be
computed using our trig functions.
Dr. Ken W. Smith
Sam Houston State University
2013
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Some worked problems
1
Similar Triangles
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. Draw a right triangle. The radio tower is a vertical line
perpendicular to the ground. (In the picture, this vertical line segment
has length O.) Draw the ground as a horizontal line and mark the
length of that horizontal line as A = 1000 feet. The hypotenuse
makes an angle θ = 71◦ with the ground. The tangent of 71◦ is
O
tan 71◦ = O
A = 1000 . Solve for O:
O = 1000 · tan 71◦ ≈ 1000 · 2.904 = 2904 feet
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Parallax
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 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. Draw a picture. The kite, the person holding the kite and
the ground directly below the kite form three vertices of a right
triangle with the right angle at the point on the ground directly below
the kite. The hypotenuse of this right triangle is H = 3000 feet. The
kite is O feet above the ground. The sine of θ = 38◦ is
O
O
sin 38◦ = H
= 3000
and so O is equal to
O = 3000 · sin(38◦ ) ≈ 1847 feet .
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Parallax
Astronomers use simple right triangles to find the distance to nearby stars.
As the earth revolves around the sun, it marks out a ellipse (almost a
circle) of radius 93 million miles. 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 we 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.
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Parallax
Parallax
Let the sun form a right angle vertex of a triangle. 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.
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
7
A = 9.3 × 10 × cot α.
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Parallax
A worked problem.
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.
The closest star to us, Proxima Centauri has a parallax of 0.77”, that is,
0.77 arcseconds. How far away is Proxima Centauri?
Now we know better – and indeed, with modern equipment, we have been
able to measure the parallax of some stars.
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Solution. A minute of arc is one-sixtieth of a degree; a second of arc is
1
one-sixtieth of a minute. So an arcsecond is 6012 = 3600
degrees. The
0.77
0.77
angle 0.77 arcseconds is equal to 602 = 3600 degrees. The tangent of this
0.77 ◦
0.77 ◦
) ≈ 0.00000373307. The cotangent of 3600
is the
angle is tan( 3600
reciprocal of this, approximately 267876. So the distance to Proxima
Centauri is 267876 · 93000000 = (2.67876 × 105 ) · (9.3 × 107 ) miles, about
2.49 × 1013 miles!
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A worked problem.
Trig on Right Triangles
The distance to Proxima Centauri is
267876 · 93000000 = (2.67876 × 105 ) · (9.3 × 107 ) miles
or about 2.49 × 1013 miles!
In the next presentation, we will look at graphs of the six trig functions.
The fastest rocket ever made reaches speeds of 25,000 miles per hour.
(End)
A rocket traveling at that speed would take over one hundred thousand
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!
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