Download - Elementary Functions Trigonometry on Right …kws006/Precalculus/4.4_Right_Triangle_Trig...Trigonometry on Right Triangles Trigonometry is introduced to students in two di erent forms,

Elementary FunctionsPart 4, Trigonometry

Lecture 4.4a, Trigonometry on Right Triangles

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 1 / 22

Trigonometry on Right Triangles

Trigonometry is introduced to students in two different forms, as functionson the unit circle and as functions on a right triangle.

The unit circle approach is the most natural setting for the trig functionssince 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

(cos θ)2 + (sin θ)2 = 1

and

cos(θ) = sin(θ + π2 ).

However, we would also like to apply trigonometry to right triangles whichhave a hypotenuse of length different than one.

We may do this by using similar triangles.Smith (SHSU) Elementary Functions 2013 2 / 22

Similar Triangles

Previously, to examine our trig functions, we displayed a typical triangle onthe unit circle with central angle θ, hypotenuse 1 and point P (x, y) on theunit circle.


Similar Triangles

We can expand that triangle by the ratio r to get a triangle in which thehypotenuse 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 bey/r.


Similar Triangles

Or – imagine a triangle with hypotenuse of length H, opposite side oflength O, adjacent side of length A.

The point P (x, y) sits on the circle of radius H and sin(θ) = OA .


Similar Triangles

All of these triangles are similar and so the trig functions of the angle θ,as ratios of side lengths, are unchanged.


Similar Triangles

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 sideand another angle.

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) = 35 .

(A 3-4-5 triangle will do.) Then compute the secant of the angle θ.

The secant is the reciprocal of cosine and so sec(θ) =hyp

adj=H

A.

The answer is 54 .


Similar Triangles

2 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 anside adjacent to θ with length 2.) The “opposite” side will then havelength

√21, by the Pythagorean Theorem. Compute the tangent of

the angle θ. (Tangent is the ratio oppadj .) The answer is

√212 .


Similar Triangles

3 Suppose sec θ = 72 and tan θ is negative. Find all six trig functions of

the angle θ.

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 pointsinto the fourth quadrant. Draw a line segment of length 7 from theorigin into the fourth quadrant, to a point P (2, y). By thePythagorean 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:

cos θ = 27 , sin θ =

−3√5

7 , tan θ = −3√5

2 .

The reciprocals of these are

sec θ = 72 , csc θ = −

73√5= −7

√5

15 , cot θ = −2

3√5= −2

√5

15 .


Trig on right triangles

In the next presentation, we will apply our understanding of trigonometryto solving various right triangles.

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Elementary FunctionsPart 4, Trigonometry

Lecture 4.4b, Applications of Right Triangles

Dr. Ken W. Smith

Sam Houston State University

2013


Applications with right triangles

Anytime we have a right triangle, then, if we can measure one of the acuteangles and also know the length of a side, then we know everything aboutthe triangle.

We will then find one of the acute angles of the triangle (such as θ, drawnin 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 becomputed using our trig functions.


Some worked problems

1 A radio tower is stands on a flat field. I walk 1000 feet away from thebase of the radio tower and look up at the top of the tower. Imeasure 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 lineperpendicular to the ground. (In the picture, this vertical line segmenthas length O.) Draw the ground as a horizontal line and mark thelength of that horizontal line as A = 1000 feet. The hypotenusemakes an angle θ = 71◦ with the ground. The tangent of 71◦ istan 71◦ = O

A = O1000 . Solve for O:

O = 1000 · tan 71◦ ≈ 1000 · 2.904 = 2904 feet


Similar Triangles

2 I am flying a kite on the beach. The kite is attached to 3000 feet ofstring. At the time that the string plays out the kite makes an anglewith the horizon of 38◦. Assuming that the 3000 feet of string is astraight line, how high is the kite?Solution. Draw a picture. The kite, the person holding the kite andthe ground directly below the kite form three vertices of a righttriangle with the right angle at the point on the ground directly belowthe kite. The hypotenuse of this right triangle is H = 3000 feet. Thekite is O feet above the ground. The sine of θ = 38◦ issin 38◦ = O

H = O3000 and so O is equal to

O = 3000 · sin(38◦) ≈ 1847 feet .


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 acircle) of radius 93 million miles. Over the course of the year, a nearby starshould appear to move back and forth in the night sky as the earthrevolves around the sun and so we should be able to measure that angle ofapparent motion and use a right triangle (with one side equal to93,000,000 miles) to compute that distance.


Parallax


Parallax

Let the sun form a right angle vertex of a triangle. Set the earth and staras the other two vertices. If the star does not move, it would appear toform 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 angleα with the anticipated line of sight. The star appears to have moved.


Parallax

By a standard result from geometry, this angle α (the apparent motion ofthe star) is also the acute angle of the triangle at the vertex given by thestar.If A is the distance from the star to the sun andO = 93, 000, 000 = 9.3× 107 miles then the cotangent of α is A

O and soA = 9.3× 107 × cotα.


Parallax

The ancient Greeks thought of this idea and attempted to measureparallax. But when they did this, the stars didn’t seem to move!! So eitherthis picture was wrong (maybe the earth was the center of the universe?)or the stars must be billions of miles away! Convinced that the universecould not be billions of miles in size, most Greeks agreed with Aristotle’sbelief that the earth was the center of the universe and that the sunrevolved around the earth.

Now we know better – and indeed, with modern equipment, we have beenable to measure the parallax of some stars.


A worked problem.

The closest star to us, Proxima Centauri 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 isone-sixtieth of a minute. So an arcsecond is 1

602= 1

3600 degrees. Theangle 0.77 arcseconds is equal to 0.77

602= 0.77

3600 degrees. The tangent of this

angle is tan( 0.773600

◦) ≈ 0.00000373307. The cotangent of 0.77

3600

◦is the

reciprocal of this, approximately 267876. So the distance to ProximaCentauri is 267876 · 93000000 = (2.67876× 105) · (9.3× 107) miles, about2.49× 1013 miles!


A worked problem.

The distance to Proxima Centauri is

267876 · 93000000 = (2.67876× 105) · (9.3× 107) miles

or 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 thousandyears to reach Proxima Centauri. Light travels about 5.879× 1012 miles ina year, so this distance is about 4.24 light years. And this is our closeststar.... It is no wonder that the Greeks could not measure parallax; theuniverse is indeed unbelievably large!


Trig on Right Triangles

In the next presentation, we will look at graphs of the six trig functions.

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