Elementary Functions Trigonometry on Right Triangles Similar 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 Elementary Functions since trig functions are not just functions of angles between 0◦ and 180◦ Part 4, Trigonometry but instead have as domain the set of all real numbers. Lecture 4.4a, Trigonometry on Right Triangles The unit circle explains identities such as (cos θ)2 + (sin θ)2 = 1 Dr. Ken W. Smith and

Sam Houston State University π cos(θ) = sin(θ + 2 ). 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. Smith (SHSU) Elementary Functions 2013 1 / 22 Smith (SHSU) Elementary Functions 2013 2 / 22 Similar Triangles Similar Triangles

We can expand that triangle by the ratio r to get a triangle in which the Previously, to examine our trig functions, we displayed a typical triangle on hypotenuse has length r and the point P (x, y) is on a circle of radius r. the unit circle with central angle θ, hypotenuse 1 and point P (x, y) on the When this happens, cos(θ) will not be x but x/r. Similarly, sin(θ) will be unit circle. y/r.

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Or – imagine a triangle with hypotenuse of length H, opposite side of All of these triangles are similar and so the trig functions of the angle θ, length O, adjacent side of length A. as ratios of side lengths, are unchanged.

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

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2 2 Find tan(θ) if cos(θ) = 5 . If we are willing to draw a right triangle and use the Pythagorean theorem, 2 Solution. Draw a right triangle which has an angle with cos(θ) = 5 . then we can solve any right triangle problem in which we are given a side (The most obvious triangle will have a hypotenuse of length 5 and an and another angle. side adjacent√ to θ with length 2.) The “opposite” side will then have length 21, by the Pythagorean Theorem. Compute the tangent of Some worked problems using similar triangles. √ the angle θ. (Tangent is the ratio opp .) The answer is 21 . 3 adj 2 1 Find sec(θ) if sin(θ) = 5 .

3 Solution. Draw a right triangle which has an angle with sin(θ) = 5 . (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 5 The answer is 4 .

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7 3 Suppose sec θ = 2 and tan θ is negative. Find all six trig functions of the angle θ.

2 Solution. Since the tangent is negative and cosine (= 7 ) 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 In the next presentation, we will apply our understanding of trigonometry origin into the fourth quadrant, to a point P (2, y). By the to solving various right triangles. √Pythagorean√ theorem,√ the absolute√ value of y √is 2 2 7 − 2 = 45 = 9 · 5 = 3 5. √So y = −3 5 and our line segment ends at the point P (2, −3 5). (End) Now read oﬀ the values of the various trig functions: √ √ 2 −3 5 −3 5 cos θ = 7 , sin θ = 7 , tan θ = 2 . The reciprocals of these are √ √ sec θ = 7 , csc θ = − √7 = − 7 5 , cot θ = − √2 = − 2 5 . 2 3 5 15 3 5 15

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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. Elementary Functions We will then ﬁnd one of the acute angles of the triangle (such as θ, drawn Part 4, Trigonometry in red) and we will also be able to ﬁnd the length of one of the sides. 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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1 A radio tower is stands on a flat field. I walk 1000 feet away from the 2 I am flying a kite on the beach. The kite is attached to 3000 feet of base of the radio tower and look up at the top of the tower. I string. At the time that the string plays out the kite makes an angle measure a 71◦ angle between the horizon and the top of the tower. with the horizon of 38◦. Assuming that the 3000 feet of string is a How tall is the tower? straight line, how high is the kite? Solution. Draw a right triangle. The radio tower is a vertical line Solution. Draw a picture. The kite, the person holding the kite and perpendicular to the ground. (In the picture, this vertical line segment the ground directly below the kite form three vertices of a right has length O.) Draw the ground as a horizontal line and mark the triangle with the right angle at the point on the ground directly below length of that horizontal line as A = 1000 feet. The hypotenuse the kite. The hypotenuse of this right triangle is H = 3000 feet. The makes an angle θ = 71◦ with the ground. The tangent of 71◦ is kite is O feet above the ground. The sine of θ = 38◦ is ◦ O O ◦ O O tan 71 = A = 1000 . Solve for O: sin 38 = H = 3000 and so O is equal to O = 1000 · tan 71◦ ≈ 1000 · 2.904 = 2904 feet O = 3000 · sin(38◦) ≈ 1847 feet .

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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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Let the sun form a right angle vertex of a triangle. Set the earth and star By a standard result from geometry, this angle α (the apparent motion of as the other two vertices. If the star does not move, it would appear to the star) is also the acute angle of the triangle at the vertex given by the form a right angle with the earth in this ﬁgure. star. But if the star is “nearby” then the line of sight to the star forms an angle If A is the distance from the star to the sun and 7 A α with the anticipated line of sight. The star appears to have moved. O = 93, 000, 000 = 9.3 × 10 miles then the cotangent of α is O and so A = 9.3 × 107 × cot α.

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The ancient Greeks thought of this idea and attempted to measure The closest star to us, Proxima Centauri has a parallax of 0.77”, that is, parallax. But when they did this, the stars didn’t seem to move!! So either 0.77 arcseconds. How far away is Proxima Centauri? this picture was wrong (maybe the earth was the center of the universe?) Solution. A minute of arc is one-sixtieth of a degree; a second of arc is or the stars must be billions of miles away! Convinced that the universe one-sixtieth of a minute. So an arcsecond is 1 = 1 degrees. The could not be billions of miles in size, most Greeks agreed with Aristotle’s 602 3600 angle 0.77 arcseconds is equal to 0.77 = 0.77 degrees. The tangent of this belief that the earth was the center of the universe and that the sun 602 3600 0.77 ◦ 0.77 ◦ revolved around the earth. angle is tan( 3600 ) ≈ 0.00000373307. The cotangent of 3600 is the reciprocal of this, approximately 267876. So the distance to Proxima Now we know better – and indeed, with modern equipment, we have been Centauri is 267876 · 93000000 = (2.67876 × 105) · (9.3 × 107) miles, about able to measure the parallax of some stars. 2.49 × 1013 miles!

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The distance to Proxima Centauri is

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

13 or about 2.49 × 10 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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