Angles
Recall the following definitions from elementary geometry:
(a) An angle isacuteif it is between 0 ◦ and 90 ◦
(b) An angle is aright angleif it equals 90 ◦
(c) An angle isobtuseif it is between 90 ◦ and 180 ◦
(d) An angle is astraight angleif it equals 180 ◦
(a) acute angle (b) right angle (c) obtuse angle (d) straight angle
In elementary geometry, angles are always considered to be positive and not larger than
360 ◦ For now we will only consider such angles 2 The following definitions will be used throughout the text:
1 Ahmes claimed that he copied the papyrus from a work that may date as far back as 3000 B C
2 Later in the text we will discuss negative angles and angles larger than 360 ◦
(a) Two acute angles arecomplementaryif their sum equals 90 ◦ In other words, if 0 ◦ ≤
∠ A ,∠ B ≤90 ◦ then∠ A and∠ B are complementary if∠ A +∠ B ◦
(b) Two angles between 0 ◦ and 180 ◦ aresupplementaryif their sum equals 180 ◦ In other words, if 0 ◦ ≤∠ A ,∠ B ≤180 ◦ then∠ A and∠ B are supplementary if∠ A +∠ B 0 ◦
(c) Two angles between 0 ◦ and 360 ◦ areconjugate(orexplementary) if their sum equals
360 ◦ In other words, if 0 ◦ ≤∠ A ,∠ B ≤360 ◦ then∠ A and∠ B are conjugate if∠ A +∠ B 360 ◦
Figure 1.1.2 Types of pairs of angles
In geometry, angles can be represented not only by the angle notation ∠ A but also by capital letters (e.g., A, B, C) or lowercase variables (e.g., x, y, t) Additionally, it is customary to use letters from the Greek alphabet, as illustrated in the accompanying table, to denote angles.
Letters Name Letters Name Letters Name
B β beta K κ kappa Σ σ sigma Γ γ gamma Λ λ lambda T τ tau
H η eta O o omicron Ψ ψ psi Θ θ theta Π π pi Ω ω omega
In elementary geometry, you learned that the sum of the angles in a triangle is 180° An isosceles triangle has two sides of equal length, while a right triangle contains a right angle measuring 90° In a right triangle, the remaining two angles are acute and together sum to 90°, making them complementary angles.
For each triangle below, determine the unknown angle(s):
In this article, we will occasionally refer to the angles of a triangle using their vertex points for simplicity For instance, the angle ∠BAC will be referred to simply as angle A.
Solution: For triangle △ ABC, A = 35 ◦ and C = 20 ◦ , and we know that A + B + C = 180 ◦ , so
For the right triangle △ DEF , E = 53 ◦ and F = 90 ◦ , and we know that the two acute angles D and E are complementary, so
For triangle △ X Y Z, the angles are in terms of an unknown number α, but we do know that X + Y +
Z = 180 ◦ , which we can use to solve for α and then use that to solve for X, Y , and Z: α + 3α + α = 180 ◦ ⇒ 5α = 180 ◦ ⇒ α = 36 ◦ ⇒ X = 36 ◦ , Y = 3 × 36 ◦ = 108 ◦ , Z = 36 ◦
Thales’ Theorem states that if A, B, and C are (distinct) points on a circle such that the line segment
AB is a diameter of the circle, then the angle∠ ACB is a right angle (see Figure 1.1.3(a)) In other words, the triangle △ ABC is a right triangle.
To prove this, let O be the center of the circle and draw the line segment OC, as in Figure 1.1.3(b).
Let α represent the angle ∠BAC and β denote the angle ∠ABC Given that AB is the diameter of the circle, the lengths OA and OC are equal, both representing the radius of the circle Consequently, triangle OAC is isosceles, leading to the conclusion that ∠OCA is equal to α.
∠ O AC = α Likewise, △ OBC is an isosceles triangle and∠ OCB =∠ OBC = β So we see that
∠ ACB = α + β And since the angles of △ ABC must add up to 180 ◦ , we see that 180 ◦ = α + (α + β) + β =
In a right triangle, the side opposite the right angle is called thehy- potenuse, and the other two sides are called itslegs For example, in
Figure 1.1.4 the right angle is C, the hypotenuse is the line segment
AB, which has length c, and BC and AC are the legs, with lengthsa andb, respectively The hypotenuse is always the longest side of a right triangle (see Exercise 11).
By knowing the lengths of two sides of a right triangle, the length of the third side can be determined by using thePythagorean Theorem:
Theorem 1.1 Pythagorean Theorem: The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of its legs.
Thus, if a right triangle has a hypotenuse of length cand legs of lengths aandb, as in
Figure 1.1.4, then the Pythagorean Theorem says: a 2 + b 2 = c 2 (1.1)
In the right triangle △ABC, illustrated in Figure 1.1.5(a), drawing a perpendicular line segment from vertex C to point D on the hypotenuse AB creates two smaller triangles, △CBD and △ACD Both of these triangles are similar to the original triangle △ABC, demonstrating the properties of triangle similarity in right triangles.
Figure 1.1.5 Similar triangles △ ABC, △ CBD, △ ACD
Triangles are considered similar when their corresponding angles are equal, which leads to the conclusion that their corresponding sides are proportional Therefore, since triangle ABC is similar to triangle CBD, we can deduce that the sides of these triangles maintain a proportional relationship.
ABis toCB(hypotenuses) as BCis toBD (vertical legs) ⇒ c a = a d ⇒ cd = a 2 Since△ABCis similar to△ACD, comparing horizontal legs and hypotenuses gives b c−d = c b ⇒ b 2 = c 2 − cd = c 2 − a 2 ⇒ a 2 + b 2 = c 2 QED
Note: The symbols⊥and∼denote perpendicularity and similarity, respectively For exam- ple, in the above proof we had CD⊥AB and△ABC∼ △CBD∼ △ACD.
For each right triangle below, determine the length of the unknown side:
Solution: For triangle △ ABC, the Pythagorean Theorem says that a 2 + 4 2 = 5 2 ⇒ a 2 = 25 − 16 = 9 ⇒ a = 3 For triangle △ DEF, the Pythagorean Theorem says that e 2 + 1 2 = 2 2 ⇒ e 2 = 4 − 1 = 3 ⇒ e = p
3 For triangle △ X Y Z, the Pythagorean Theorem says that
A 17 ft ladder leaning against a wall has its foot 8 ft from the base of the wall At what height is the top of the ladder touching the wall?
To determine the height at which the ladder touches the wall, we consider a right triangle formed by the ladder, the wall, and the ground The ladder, measuring 17 feet, serves as the hypotenuse, while the legs of the triangle are 8 feet and the height h Applying the Pythagorean Theorem, we find that h² + 8² = 17² This simplifies to h² = 289 - 64, resulting in h = 15 feet.
For Exercises 1-4, find the numeric value of the indicated angle(s) for the triangle △ ABC.
1 Find B if A = 15 ◦ and C = 50 ◦ 2 Find C if A = 110 ◦ and B = 31 ◦
3 Find A and B if C = 24 ◦ , A = α, and B = 2α 4 Find A, B, and C if A = β and B = C = 4β.
For Exercises 5-8, find the numeric value of the indicated angle(s) for the right triangle △ ABC, with
5 Find B if A = 45 ◦ 6 Find A and B if A = α and B = 2α.
7 Find A and B if A = φ and B = φ 2 8 Find A and B if A = θ and B = 1/θ.
9 A car goes 24 miles due north then 7 miles due east What is the straight distance between the car’s starting point and end point?
A rope measuring 150 feet is secured to the top of a 100-foot tall pole, creating a scenario where the maximum distance from the base of the pole to the point where the rope can be attached on the ground can be calculated Assuming the pole stands vertically, the maximum horizontal distance can be determined using the Pythagorean theorem, resulting in a distance of approximately 130 feet from the base of the pole.
11 Prove that the hypotenuse is the longest side in every right triangle (Hint: Is a 2 + b 2 > a 2 ?)
12 Can a right triangle have sides with lengths 2, 5, and 6? Explain your answer.
A Pythagorean triple consists of positive integers a, b, and c that satisfy the equation a² + b² = c², representing the lengths of the sides of a right triangle Common examples of Pythagorean triples include (3, 4, 5) and (5, 12, 13).
(a) Show that (6,8,10) is a Pythagorean triple.
(b) Show that if (a,b,c) is a Pythagorean triple then so is (ka,kb,kc) for any integer k > 0 How would you interpret this geometrically?
(c) Show that (2mn,m 2 − n 2 ,m 2 + n 2 ) is a Pythagorean triple for all integers m > n > 0.
(d) The triple in part(c) is known as Euclid’s formula for generating Pythagorean triples Write down the first ten Pythagorean triples generated by this formula, i.e use: m = 2 and n = 1; m = 3 and n = 1, 2; m = 4 and n = 1, 2, 3; m = 5 and n = 1, 2, 3, 4.
This exercise illustrates how to construct a tangent line through a point located outside a circle, ensuring that the line intersects the circle at precisely one point This concept of a tangent line is fundamental and will be referenced throughout the text.
To begin, draw a circle with a radius of 1 inch and label its center as O Select a point P located 2.5 inches from O, then create a circle using OP as the diameter Identify point A where this new circle intersects the original circle By drawing a line through points P and A, we can utilize the property that the tangent line at any point on a circle is perpendicular to the radius at that point to establish that this line is indeed the tangent line at A Additionally, we can calculate the length of segment PA based on this relationship.
Does it match the physical measurement of P A?
In triangle △ ABC, where side AB serves as the diameter of a circle centered at O and vertex C lies on the circle, a 180° rotation of the circle around its center positions the triangle in a new orientation This transformation illustrates Thales' theorem, demonstrating that the angle ∠ACB is a right angle, as the rotation maintains the relationship between the triangle and the circle.
Trigonometric Functions of an Acute Angle • Section 1.2 7
Trigonometric Functions of an Acute Angle
B b adjacent a o p p o si te hy po te c nus
In a right triangle △ABC, where the right angle is located at C, the sides are designated as lengths a, b, and c For the acute angle A, the leg BC serves as the opposite side, while leg AC is identified as the adjacent side The hypotenuse of the triangle is the longest side, opposite the right angle.
AB The ratios of sides of a right triangle occur often enough in prac- tical applications to warrant their own names, so we define the six trigonometric functionsofAas follows:
Table 1.2 The six trigonometric functions of A
The trigonometric functions are fundamental in mathematics, defined as follows: sine (sin A) is the ratio of the opposite side to the hypotenuse (a/c), cosine (cos A) is the ratio of the adjacent side to the hypotenuse (b/c), and tangent (tan A) is the ratio of the opposite side to the adjacent side (a/b) Additionally, cosecant (csc A) is the reciprocal of sine (c/a), secant (sec A) is the reciprocal of cosine (c/b), and cotangent (cot A) is the reciprocal of tangent (b/a) Understanding these relationships is essential for solving various mathematical problems involving right triangles.
In trigonometry, the abbreviated names of functions are commonly used, as shown in Table 1.2 The pairs sin A and csc A, cos A and sec A, and tan A and cot A are reciprocals of each other, meaning csc A equals 1/sin A, sec A equals 1/cos A, and cot A equals 1/tan A.
For the right triangle △ ABC shown on the right, find the values of all six trigono- metric functions of the acute angles A and B.
Solution: The hypotenuse of △ ABC has length 5 For angle A, the opposite side
BC has length 3 and the adjacent side AC has length 4 Thus: sin A = opposite hypotenuse = 3
3 For angle B, the opposite side AC has length 4 and the adjacent side BC has length 3 Thus: sin B = opposite hypotenuse = 4
Notice in Example 1.5 that we did not specify the units for the lengths This raises the possibility that our answers depended on a triangle of a specific physical size.
For example, suppose that two different students are reading this textbook: one in the United States and one in Germany The American student thinks that the lengths 3, 4, and
In Example 1.5, measurements are given in inches, but a German student assumes they are in centimeters Since 1 inch is approximately 2.54 centimeters, this results in the students utilizing triangles of varying physical sizes.
If the American triangle is △ABC and the German triangle is △A ′ B ′ C ′ , then we see from Figure 1.2.1 that △ABC is similar to △A ′ B ′ C ′ , and hence the corresponding angles
Trigonometric functions of acute angles demonstrate that the ratios of corresponding sides in similar triangles are equal For example, the sides of triangle ABC are approximately 2.54 times longer than those of triangle A'B'C' Consequently, both an American student calculating sin A and a German student calculating sin A' arrive at the same result.
The trigonometric functions of angles A and A' are identical, and our reasoning applies universally to any comparable right triangles This observation leads us to a significant conclusion.
When calculating the trigonometric functions of an acute angle A, you may useanyright triangle which has Aas one of the angles.
Trigonometric functions are defined as ratios of the sides of a triangle, resulting in unitless values since the units cancel out For instance, when an American student calculates sinA as 3/5, it is equivalent to the 3/5 calculated by a German student, regardless of the different units used for the side lengths.
Find the values of all six trigonometric functions of 45 ◦
To solve the problem, we can utilize a right triangle with a 45° angle by starting with a square that has sides measuring 1 unit each and dividing it diagonally This division creates an isosceles triangle, △ ABC, where the legs are equal in length, leading to angles A and B being equal as well Given that A + B equals 90°, we conclude that both angles measure 45° Applying the Pythagorean Theorem, the hypotenuse length c is calculated as c² = 1² + 1² = 2, resulting in c = √2.
2 Thus, using the angle A we get: sin 45 ◦ = opposite hypotenuse = 1 p 2 cos 45 ◦ = adjacent hypotenuse = 1 p 2 tan 45 ◦ = opposite adjacent = 1
Note that we would have obtained the same answers if we had used any right triangle similar to
△ ABC For example, if we multiply each side of △ ABC by p
2, then we would have a similar triangle with legs of length p
2 and hypotenuse of length 2 This would give us sin 45 ◦ = p 2
2 = p 1 2 as before The same goes for the other functions.
3 We will use the notation AB to denote the length of a line segment AB.
Find the values of all six trigonometric functions of 60 ◦
To solve the problem, we can use a right triangle with a 60° angle Consider an equilateral triangle where all sides measure 2 units By drawing the bisector from one vertex to the opposite side, we create two triangles In this case, triangle ABC has angle A at 60° and angle B at 30°, resulting in angle C being 90° Therefore, triangle ABC is a right triangle with a hypotenuse of length c = AB = 2 and a leg of length b = AC = 1.
By the Pythagorean Theorem, the length a of the leg BC is given by a 2 + b 2 = c 2 ⇒ a 2 = 2 2 − 1 2 = 3 ⇒ a = p
3 Thus, using the angle A we get: sin 60 ◦ = opposite hypotenuse = p 3
1 = p 3 csc 60 ◦ = hypotenuse opposite = 2 p 3 sec 60 ◦ = hypotenuse adjacent = 2 cot 60 ◦ = adjacent opposite = 1 p 3
Notice that, as a bonus, we get the values of all six trigonometric functions of 30 ◦ , by using angle
B = 30 ◦ in the same triangle △ ABC above: sin 30 ◦ = opposite hypotenuse = 1
2 tan 30 ◦ = opposite adjacent = 1 p 3 csc 30 ◦ = hypotenuse opposite = 2 sec 30 ◦ = hypotenuse adjacent = 2 p 3 cot 30 ◦ = adjacent opposite = p 3
A is an acute angle such that sin A = 2 3 Find the values of the other trigonometric functions of A.
To solve problems involving trigonometric functions, it is helpful to draw a right triangle In this case, we are given the sine function, where sin A = 2/3 This means that in triangle ABC, the length of the side opposite angle A is 2, and the length of the hypotenuse is 3 The length of the adjacent side, denoted as b, can be determined using the Pythagorean Theorem.
Trigonometric Functions of an Acute Angle • Section 1.2 11
We now know the lengths of all sides of the triangle △ ABC, so we have: cos A = adjacent hypotenuse = p 5
3 tan A = opposite adjacent = 2 p 5 csc A = hypotenuse opposite = 3
2 sec A = hypotenuse adjacent = 3 p 5 cot A = adjacent opposite = p 5 2
The relationships between sine and cosine, secant and cosecant, as well as tangent and cotangent of complementary angles can be observed in previous examples This observation leads us to a generalized theorem that encapsulates these connections.
Theorem 1.2 Cofunction Theorem: IfAandBare the complementary acute angles in a right triangle△ABC, then the following relations hold: sinA = cosB sec A = cscB tan A = cotB sinB = cos A secB = csc A tanB = cot A
We say that the pairs of functions { sin, cos }, { sec, csc }, and { tan, cot } arecofunctions.
So sine and cosine are cofunctions, secant and cosecant are cofunctions, and tangent and cotangent are cofunctions That is how the functions cosine, cosecant, and cotangent got the
“co” in their names The Cofunction Theorem says that any trigonometric function of an acute angle is equal to its cofunction of the complementary angle.
Write each of the following numbers as trigonometric functions of an angle less than 45 ◦ : (a) sin 65 ◦ ;
Solution: (a) The complement of 65 ◦ is 90 ◦ − 65 ◦ = 25 ◦ and the cofunction of sin is cos, so by the Cofunction Theorem we know that sin 65 ◦ = cos 25 ◦
(b) The complement of 78 ◦ is 90 ◦ − 78 ◦ = 12 ◦ and the cofunction of cos is sin, so cos 78 ◦ = sin 12 ◦
(c) The complement of 59 ◦ is 90 ◦ − 59 ◦ = 31 ◦ and the cofunction of tan is cot, so tan 59 ◦ = cot 31 ◦ a a p a
Figure 1.2.2 Two general right triangles (any a > 0)
The angles 30 ◦ , 45 ◦ , and 60 ◦ arise often in applications We can use the Pythagorean Theorem to generalize the right triangles in Examples 1.6 and 1.7 and see whatany45−
45−90 and 30−60−90 right triangles look like, as in Figure 1.2.2 above.
Find the sine, cosine, and tangent of 75 ◦
Solution: Since 75 ◦ = 45 ◦ + 30 ◦ , place a 30 − 60 − 90 right triangle
△ ADB with legs of length p
3 and 1 on top of the hypotenuse of a 45 − 45 − 90 right triangle △ ABC whose hypotenuse has length p
3, as in the figure on the right From Figure 1.2.2(a) we know that the length of each leg of △ ABC is the length of the hypotenuse divided by p
2 Draw DE perpendicular to AC, so that △ ADE is a right triangle Since
∠ B AC = 45 ◦ and∠ D AB = 30 ◦ , we see that∠ D AE = 75 ◦ since it is the sum of those two angles Thus, we need to find the sine, cosine, and tangent of∠ D AE
Notice that∠ ADE = 15 ◦ , since it is the complement of∠ D AE
And∠ ADB = 60 ◦ , since it is the complement of∠ D AB Draw
BF perpendicular to DE, so that △ DFB is a right triangle.
Then∠ BDF = 45 ◦ , since it is the difference of∠ ADB = 60 ◦ and
∠ ADE = 15 ◦ Also,∠ DBF = 45 ◦ since it is the complement of
∠ BDF The hypotenuse BD of △ DFB has length 1 and △ DFB is a 45 − 45 − 90 right triangle, so we know that DF = FB = p 1 2
In the given geometric configuration, lines DE and BC are both perpendicular to line AC, establishing that FE is parallel to BC Additionally, since both FB and EC are perpendicular to DE, FB is parallel to EC Consequently, the quadrilateral FBCE forms a rectangle due to the right angle at ∠BCE This leads to the conclusions that EC equals FB, represented as p1 2, and FE equals BC, represented as q3.
2 Note: Taking reciprocals, we get csc 75 ◦ = p 6 4 + p 2 , sec 75 ◦ = p 6 4 − p 2 , and cot 75 ◦ = p 6 − p p 2
For Exercises 1-10, find the values of all six trigonometric functions of angles A and B in the right triangle △ ABC in Figure 1.2.3.
For Exercises 11-18, find the values of the other five trigonometric functions of the acute angle A given the indicated value of one of the functions.
Trigonometric Functions of an Acute Angle • Section 1.2 13
For Exercises 19-23, write the given number as a trigonometric function of an acute angle less than
19 sin 87 ◦ 20 sin 53 ◦ 21 cos 46 ◦ 22 tan 66 ◦ 23 sec 77 ◦
For Exercises 24-28, write the given number as a trigonometric function of an acute angle greater than 45 ◦
24 sin 1 ◦ 25 cos 13 ◦ 26 tan 26 ◦ 27 cot 10 ◦ 28 csc 43 ◦
29 In Example 1.7 we found the values of all six trigonometric functions of 60 ◦ and 30 ◦
(a) Does sin 30 ◦ + sin 30 ◦ = sin 60 ◦ ? (b) Does cos 30 ◦ + cos 30 ◦ = cos 60 ◦ ?
(c) Does tan 30 ◦ + tan 30 ◦ = tan 60 ◦ ? (d) Does 2 sin 30 ◦ cos 30 ◦ = sin 60 ◦ ?
30 For an acute angle A, can sin A be larger than 1? Explain your answer.
31 For an acute angle A, can cos A be larger than 1? Explain your answer.
32 For an acute angle A, can sin A be larger than tan A? Explain your answer.
33 If A and B are acute angles and A < B, explain why sin A < sin B.
34 If A and B are acute angles and A < B, explain why cos A > cos B.
35 Prove the Cofunction Theorem (Theorem 1.2) (Hint: Draw a right triangle and label the angles and sides.)
36 Use Example 1.10 to find all six trigonometric functions of 15 ◦ p 3
37 In Figure 1.2.4, CB is a diameter of a circle with a radius of
2 cm and center O, △ ABC is a right triangle, and CD has length p
(a) Find sin A (Hint: Use Thales’ Theorem.)
(b) Find the length of AC.
(c) Find the length of AD.
To accurately assess angle A in Figure 1.2.4, utilize a protractor for measurement, then calculate the sine of that angle using a calculator Compare the calculator's result with your previous answer from part (a) to determine their proximity.
Note: Make sure that your calculator is in degree mode.
38 In Exercise 37, verify that the area of △ ABC equals 1 2 AB ã CD Why does this make sense?
39 In Exercise 37, verify that the area of △ ABC equals 1 2 AB ã AC sin A.
40 In Exercise 37, verify that the area of △ ABC equals 1 2 (BC) 2 cot A.
Applications and Solving Right Triangles
Trigonometry, initially developed for indirect measurement of large distances through angles and known small distances, is now essential in fields such as physics, astronomy, engineering, navigation, surveying, and mathematics This section will explore various applications of trigonometry, and it is recommended to use a calculator in degree mode for the examples provided.
A person positioned 150 feet from a flagpole measures an angle of elevation of 32 degrees to the top of the pole With their eyes 6 feet above the ground, the height of the flagpole can be calculated using trigonometric principles This scenario illustrates the application of basic trigonometry in determining the height of objects based on distance and angle measurements.
Solution: The picture on the right describes the situation We see that the height of the flagpole is h + 6 ft, where h
We determined that tan 32° equals 0.6249 using a calculator Since the initial numbers provided were whole numbers, we rounded the height (h) to the nearest integer Consequently, the total height of the flagpole is calculated as h + 6, resulting in a final height of 100 feet.
A person positioned 400 feet away from the base of a mountain observes an angle of elevation of 25 degrees to the peak After walking 500 feet further back, the angle of elevation decreases to 20 degrees This scenario can be used to calculate the height of the mountain.
To analyze the height of a mountain given a flat ground, we define the height as \( h \) and the horizontal distance from the mountain's base to the point directly beneath its peak as \( x \) Utilizing trigonometric relationships, we establish two equations: \( h = (x + 400) \tan 25^\circ \) and \( h = (x + 900) \tan 20^\circ \) These equations allow us to calculate the mountain's height based on the specified angles of elevation.
To solve for x in the equation (x + 400) tan 25° = (x + 900) tan 20°, we can rearrange it to x tan 25° - x tan 20° = 900 tan 20° - 400 tan 25° This simplifies to x = (900 tan 20° - 400 tan 25°) / (tan 25° - tan 20°), resulting in x = 1378 ft Substituting x back into the original equation for h gives us the height of the mountain: h = (1378 + 400) tan 25° = 1778 (0.4663), which equals 829 ft.
Applications and Solving Right Triangles • Section 1.3 15
A blimp flying at an altitude of 4,280 feet observes an angle of depression of 24 degrees to the base of a house on the ground Assuming a flat terrain, the distance from the blimp to the house along the ground can be calculated using trigonometric principles.
To solve the problem, let x represent the distance on the ground from the blimp to the house Given that the horizontal line of sight from the blimp is parallel to the ground, elementary geometry tells us that the angle of elevation θ from the base of the house to the blimp is equal to the angle of depression from the blimp to the base of the house, which is θ = 24°.
An observer positioned at the summit of a mountain 3 miles above sea level measures an angle of depression of 2.23 degrees to the ocean horizon This angle can be utilized to estimate the radius of the Earth, providing valuable insights into the curvature of our planet By applying trigonometric principles, one can determine the Earth's radius based on the observer's altitude and the measured angle.
To solve the problem, we will consider the Earth as a sphere with radius r In this scenario, point A denotes the summit of the mountain, while point H represents the horizon of the ocean as seen from that vantage point.
In Figure 1.3.1, let O represent the center of the Earth, while point B lies on the horizontal line of sight from point A, which is perpendicular to line OA The angle formed at point O, denoted as θ, corresponds to the angle ∠ AOH.
Since A is 3 miles above sea level, we have O A = r + 3 Also,
In this geometric scenario, we establish that OH equals the radius r Given that line segment AB is perpendicular to OA, we determine that the angle OAB measures 90 degrees Consequently, the angle OAH is calculated to be 87.77 degrees, derived from subtracting 2.23 degrees from 90 degrees The line connecting points A and H serves as a tangent to the Earth's surface, which can be represented as a circle with radius r According to Exercise 14 in Section 1.1, we confirm that AH is perpendicular to OH, affirming that the angle OHA is also 90 degrees.
Since the angles in the triangle △ O AH add up to 180 ◦ , we have θ = 180 ◦ − 90 ◦ − 87.77 ◦ = 2.23 ◦ Thus, cos θ = OH
O A = r r + 3 ⇒ r r + 3 = cos 2.23 ◦ , so solving for r we get r = (r + 3) cos 2.23 ◦ ⇒ r − r cos 2.23 ◦ = 3 cos 2.23 ◦
⇒ r = 3958.3 miles Note: This answer is very close to the earth’s actual (mean) radius of 3956.6 miles.
The Earth is not a perfect sphere; it is an ellipsoid, resembling an egg shape, with an ellipticity of 1/297, while a true sphere has an ellipticity of 0 For more detailed information, refer to W.H Munk and G.J.F MacDonald's work, "The Rotation of the Earth: A Geophysical Discussion," published by Cambridge University Press in 1960.
In this article, we explore the application of trigonometry in astronomy by calculating the distance from the Earth to the Sun We define point O as the center of the Earth, point A as a location on the equator, and point B as a celestial object, such as a star By analyzing the Earth's position and the angle formed, we can determine this astronomical distance.
The angle α = ∠ OBA, where ∠ OAB = 90°, represents the equatorial parallax of an object The equatorial parallax of the sun is approximately α = 0.00244° This value can be utilized to estimate the distance from the center of the Earth to the sun.
Solution: Let B be the position of the sun We want to find the length of OB.
We will use the actual radius of the earth, mentioned at the end of Example
1.14, to get O A = 3956.6 miles Since∠ O AB = 90 ◦ , we have
OB = sin α ⇒ OB = O A sin α = 3956.6 sin 0.00244 ◦ = 92908394 , so the distance from the center of the earth to the sun is approximately 93 million miles
Note: The earth’s orbit around the sun is an ellipse, so the actual distance to the sun varies.
Trigonometric Functions of Any Angle
To define the trigonometric functions ofanyangle - including angles less than 0 ◦ or greater than 360 ◦ - we need a more general definition of an angle We say that anangleis formed by rotating a ray−−→
O A about the endpointO(called thevertex), so that the ray is in a new position, denoted by the ray−−→
O Ais called theinitial sideof the angle, and−−→
OB is theterminal sideof the angle (see Figure 1.4.1(a)).
(a) angle∠ AOB counter-clockwise direction ( + ) clockwise direction ( − ) A
Figure 1.4.1 Definition of a general angle
The angle created by the rotation is represented as ∠AOB, ∠O, or simply O A counter-clockwise rotation indicates a positive angle, while a clockwise rotation signifies a negative angle.
One full counter-clockwise rotation of−−→
O A back onto itself (called a revolution), so that the terminal side coincides with the initial side, is an angle of 360 ◦ ; in the clockwise direction this would be−360 ◦ 6 Not rotating−−→
O Aconstitutes an angle of 0 ◦ More than one full rotation creates an angle greater than 360 ◦ For example, notice that 30 ◦ and 390 ◦ have the same terminal side in Figure 1.4.2, since 30+36090.
6 The system of measuring angles in degrees, such that 360 ◦ is one revolution, originated in ancient Babylonia.
The number 360 may have originated not from the Babylonian belief in a 360-day year, but rather from their measurement of distance In ancient times, a person could travel 12 Babylonian miles in one day, corresponding to one full rotation of the Earth on its axis The Babylonian mile, approximately equivalent to seven of our miles, was conveniently divided into 30 equal parts, resulting in a total of 360 parts for a complete rotation.
Trigonometric Functions of Any Angle • Section 1.4 25
Trigonometric functions can be defined for any angle using Cartesian coordinates, which consist of points represented by pairs of real numbers (x, y) In this system, the first number, x, indicates the point's x-coordinate, while the second number, y, represents the y-coordinate These coordinates are measured along the x-axis and y-axis, respectively, determining the point's location within the plane Consequently, the Cartesian coordinate plane is divided into four quadrants (QI, QII, QIII, QIV), categorized by the signs of x and y.
An angle θ is considered to be in standard position when its initial side aligns with the positive x-axis and its vertex is located at the origin (0, 0) For any point (x, y) on the terminal side of θ, with a distance r > 0 from the origin, the distance can be expressed as r = √(x² + y²) The trigonometric functions of θ are defined as follows: sine (sin θ) is equal to y/r, cosine (cos θ) is equal to x/r, and tangent (tan θ) is equal to y/x Additionally, the reciprocal functions are defined as cosecant (csc θ) equal to r/y, secant (sec θ) equal to r/x, and cotangent (cot θ) equal to x/y.
The definitions of sine and cosine using similar triangles are consistent and independent of the chosen point (x, y) on the terminal side of angle θ It is important to note that the absolute values of sine and cosine are always less than or equal to one, as the values of y and x are constrained by the radius r in these definitions.
In the context of acute angles, the definitions of sine and cosine align with previous definitions related to right triangles By constructing a right triangle with angle θ, where x represents the adjacent side, y denotes the opposite side, and r is the hypotenuse, we can express the sine and cosine functions as sinθ = y/r (opposite over hypotenuse) and cosθ = x/r (adjacent over hypotenuse), reinforcing the established relationships.
0 θ r hy po te nus e (x, y) x adjacent side y opposite side
In Figure 1.4.4(b) we see in which quadrants or on which axes the terminal side of an angle
In the range of 0° to 360°, trigonometric functions can yield negative values, as illustrated in Figure 1.4.3(a) and supported by formulas (1.2) and (1.3) Specifically, the sine function, sinθ, is negative when the y-coordinate is less than zero Figure 1.4.5 provides a comprehensive overview of the signs of trigonometric functions based on the quadrant in which the angle resides.
QI sin + cos + tan + csc + sec + cot +
QII sin + cos − tan − csc + sec − cot −
QIII sin − cos − tan + csc − sec − cot +
QIV sin − cos + tan − csc − sec + cot −
Figure 1.4.5 Signs of the trigonometric functions by quadrant
Trigonometric Functions of Any Angle • Section 1.4 27
Find the exact values of all six trigonometric functions of 120 ◦
Solution: We know 120 ◦ = 180 ◦ − 60 ◦ By Example 1.7 in Section 1.2, we see that we can use the point ( − 1, p
In the second quadrant, the terminal side of a 120° angle forms a basic right triangle with a 60° angle, where the adjacent side measures 1 and the opposite side is denoted as p.
3, and hypotenuse of length 2, as in the figure on the right.
Drawing that triangle in QII so that the hypotenuse is on the terminal side of 120 ◦ makes r = 2, x = − 1, and y = p
Find the exact values of all six trigonometric functions of 225 ◦
Solution: We know that 225 ◦ = 180 ◦ + 45 ◦ By Example 1.6 in Section
At an angle of 225°, the point (-1, -1) lies on the terminal side in the third quadrant (QIII) In this context, a basic right triangle with a 45° angle has an adjacent side of length 1, an opposite side of length 1, and a hypotenuse of length √2.
2, as in the figure on the right.
Drawing that triangle in QIII so that the hypotenuse is on the terminal side of 225 ◦ makes r = p
2, x = − 1, and y = − 1 Hence: sin 225 ◦ = y r = − 1 p 2 cos 225 ◦ = x r = − 1 p 2 tan 225 ◦ = y x = − 1
Find the exact values of all six trigonometric functions of 330 ◦
Solution: We know that 330 ◦ = 360 ◦ − 30 ◦ By Example 1.7 in Section
1.2, we see that we can use the point ( p
3, − 1) on the terminal side of the angle 225 ◦ in QIV, since we saw in that example that a basic right triangle with a 30 ◦ angle has adjacent side of length p
In a right triangle situated in the fourth quadrant, with one side measuring 1 and the hypotenuse measuring 2, the triangle's configuration aligns with the terminal side of 330 degrees This results in a radius (r) of 2 and an x-coordinate (x) equivalent to p.
Find the exact values of all six trigonometric functions of 0 ◦ , 90 ◦ ,
The angles discussed here differ from previous examples as their terminal sides align with the x-axis or y-axis, eliminating the possibility of forming right triangles Despite this, calculating the values of the trigonometric functions remains straightforward by selecting the simplest points on the terminal sides and applying the definitions in the relevant formulas.
For an angle of 0°, the corresponding point on its terminal side is (1, 0), which lies on the positive x-axis This can be visualized as a degenerate right triangle with a height of 0, where both the hypotenuse and base measure 1 In this case, we set r = 1, x = 1, and y = 0, leading to the conclusion that sin 0° = y/r = 0.
Note that csc 0 ◦ and cot 0 ◦ are undefined, since division by 0 is not allowed.
At 90 degrees, the terminal side aligns with the positive y-axis, represented by the point (0, 1) This configuration can be visualized as a degenerate right triangle, where the base measures 0 and the height corresponds to the hypotenuse's length In this case, the radius (r) is 1, the x-coordinate is 0, and the y-coordinate is 1, leading to the conclusion that sin 90° equals y divided by r, which simplifies to sin 90° = 1.
1 = 0 Likewise, for 180 ◦ use the point ( − 1, 0) so that r = 1, x = − 1, and y = 0 Hence: sin 180 ◦ = y r = 0
0 = undefined Lastly, for 270 ◦ use the point (0, − 1) so that r = 1, x = 0, and y = − 1 Hence: sin 270 ◦ = y r = − 1
Trigonometric Functions of Any Angle • Section 1.4 29
The following table summarizes the values of the trigonometric functions of angles be- tween 0 ◦ and 360 ◦ which are integer multiples of 30 ◦ or 45 ◦ :
Table 1.3 Table of trigonometric function values
Angle sin cos tan csc sec cot
Since 360 ◦ represents one full revolution, the trigonometric function values repeat every
Angles that differ by an integer multiple of 360° are known as coterminal angles, meaning they share the same initial and terminal sides For instance, sin 360° equals sin 0°, cos 390° equals cos 30°, and tan 540° equals tan 180° Additionally, sin(-45°) is equivalent to sin 315° In general, all trigonometric functions yield the same values for these coterminal angles.
In Examples 1.20-1.22, we demonstrated how to determine the values of trigonometric functions for angles greater than 90 degrees by utilizing a corresponding acute angle within a right triangle This specific acute angle is referred to as the reference angle When dealing with a nonacute angle θ, the reference angle is defined as the acute angle formed between the terminal side of θ and either the positive x-axis or the negative x-axis.
Rotations and Reflections of Angles
In this section, we will explore how geometric operations, specifically rotation and reflection, can simplify the use of trigonometric functions for any angle Understanding these operations will also reveal fundamental relationships between various trigonometric functions.
To rotate an angle means to turn its terminal side around the origin while in standard position For instance, rotating an angle θ by 90° counterclockwise results in the new angle θ + 90° in the second quadrant (QII) In this scenario, the complement of θ in the first quadrant (QI) is equal to the supplement of θ + 90° in QII, as the sum of θ, its complement, and 90° equals 180° Consequently, this relationship indicates that the remaining angle in the right triangle formed in QII is also θ.
Figure 1.5.1 Rotation of an angle θ by 90 ◦
The right triangle in Quadrant I (QI) is similar to the right triangle in Quadrant II (QII) due to their identical angles Rotating the angle θ by 90 degrees maintains the length of the terminal side, ensuring that the hypotenuses of the similar triangles are equal, which implies that the corresponding sides are also equal By analyzing the corresponding sides in Figure 1.5.1, we find that the point (-y, x) lies on the terminal side of θ + 90° when (x, y) is on the terminal side of θ Consequently, we can define the trigonometric functions as follows: sin(θ + 90°) = x/r = cosθ, cos(θ + 90°) = -y/r = -sinθ, and tan(θ + 90°) = x.
In all quadrants, the trigonometric relationships for any angle θ can be expressed as follows: the sine of an angle plus 90 degrees equals the cosine of that angle, represented mathematically as sin(θ + 90°) = cos θ Additionally, the cosine of an angle plus 90 degrees is equal to the negative sine of that angle, which can be written as cos(θ + 90°) = -sin θ Lastly, the tangent of an angle plus 90 degrees is equal to the negative cotangent of that angle, denoted as tan(θ + 90°) = -cot θ.
Rotations and Reflections of Angles • Section 1.5 33
In the x-y coordinate plane, any nonvertical line can be expressed in the form y = mx + b, where m represents the slope, calculated as the rise over the run, and b denotes the y-intercept, the point where the line intersects the y-axis It is important to note that the slopes of perpendicular lines are negative reciprocals of each other Therefore, for two nonvertical and nonhorizontal perpendicular lines represented by the equations y = m1x + b1 and y = m2x + b2, the relationship between their slopes can be defined as m2 = -1/m1.
0 y = mx + b b run rise m = rise run
When a line represented by the equation y = mx + b has a nonzero slope, it intersects the x-axis at a certain point Let θ denote the angle formed between the positive x-axis and the section of the line that lies above the x-axis.
Figure 1.5.3 For m > 0 we see that θ is acute and tan θ = rise run = m. x y
When the slope \( m \) is less than zero, the angle \( \theta \) is obtuse, indicating a negative rise Since the run is always positive, the tangent of \( \theta \) can be expressed as \( \tan \theta = \frac{-\text{rise}}{\text{run}} = m \) This relationship holds true even when the line is shifted horizontally to pass through the origin, keeping \( \theta \) unchanged while placing the point \((- \text{run}, -\text{rise})\) on the terminal side of \( \theta \).
For a line y = mx + b with m 6= 0, the slope is given by m = tan θ , where θ is the angle formed by the positive x-axis and the part of the line above the x-axis.
In Figure 1.5.2(b), we observe that two lines, represented by the equations y = m₁x + b₁ and y = m₂x + b₂, are perpendicular By rotating one line counterclockwise by 90 degrees around their intersection point, we obtain the second line If θ denotes the angle that the line y = m₁x + b₁ forms with the positive x-axis, then the line y = m₂x + b₂ makes an angle of θ + 90 degrees with the positive x-axis Consequently, we find that m₁ = tan(θ) and m₂ = tan(θ + 90°) According to the tangent addition formula, tan(θ + 90°) equals -cot(θ), leading to the conclusion that m₂ = -cot(θ) = -tan(1/θ) = -m₁.
Rotating an angle θ by 90° clockwise transforms it into θ − 90° Instead of deriving trigonometric relations for θ − 90° through geometric arguments, we can simply substitute θ with θ − 90° in the existing formulas This substitution leads to the following trigonometric identities: sin(θ − 90°) = −cosθ, cos(θ − 90°) = sinθ, and tan(θ − 90°) = −cotθ.
We now consider rotating an angleθ by 180 ◦ Notice from Figure 1.5.4 that the angles θ±180 ◦ have the same terminal side, and are in the quadrant oppositeθ. x y θ + 180 ◦ θ − 180 ◦
Since (−x,−y) is on the terminal side ofθ±180 ◦ when (x,y) is on the terminal side ofθ, we get the following relations, which hold for allθ: sin (θ±180 ◦ ) = −sinθ (1.10) cos (θ±180 ◦ ) = −cosθ (1.11) tan (θ±180 ◦ ) = tanθ (1.12) x y c c c c
Areflectionis simply the mirror image of an object For example, in
In the coordinate plane, the original object located in Quadrant I (QI) has its reflection in Quadrant II (QII) when reflected around the y-axis, and in Quadrant IV (QIV) when reflected around the x-axis If we first reflect the object in QI across the y-axis and then across the x-axis, the resulting image appears in Quadrant III (QIII) This final image represents the reflection of the original object around the origin, which is equivalent to a 180° rotation around the origin Additionally, reflecting an object around the y-axis can be viewed as performing a reflection around the x-axis followed by a 180° rotation around the origin.
Rotations and Reflections of Angles • Section 1.5 35
Applying this to angles, we see that the reflection of an angle θ around thex-axis is the angle− θ , as in Figure 1.5.6. x y
Figure 1.5.6 Reflection of θ around the x-axis
So we see that reflecting a point (x,y) around thex-axis just replacesyby−y Hence: sin (− θ ) = −sinθ (1.13) cos (− θ ) = cosθ (1.14) tan (− θ ) = −tanθ (1.15)
Notice that the cosine function does not change in formula (1.14) because it depends onx, and not ony, for a point (x,y) on the terminal side ofθ.
A function f(x) is classified as an even function if it satisfies the condition f(−x) = f(x) for all x, while it is considered an odd function if f(−x) = −f(x) for all x In this context, the cosine function is identified as an even function, whereas the sine and tangent functions are categorized as odd functions.
Replacingθby− θ in formulas (1.4)−(1.6), then using formulas (1.13)−(1.15), gives: sin (90 ◦ − θ ) = cosθ (1.16) cos (90 ◦ − θ ) = sinθ (1.17) tan (90 ◦ − θ ) = cotθ (1.18)
Note that formulas (1.16)−(1.18) extend the Cofunction Theorem from Section 1.2 to all θ , not just acute angles Similarly, formulas (1.10)−(1.12) and (1.13)−(1.15) give: sin (180 ◦ − θ ) = sinθ (1.19) cos (180 ◦ − θ ) = −cosθ (1.20) tan (180 ◦ − θ ) = −tan θ (1.21)
Notice that reflection around the y-axis is equivalent to reflection around thex-axis ( θ 7→
− θ ) followed by a rotation of 180 ◦ (− θ 7→ − θ +180 ◦ 0 ◦ − θ ), as in Figure 1.5.7. x y θ
Figure 1.5.7 Reflection of θ around the y-axis = 180 ◦ − θ
While calculators can evaluate trigonometric functions, understanding geometric operations and formulas is essential for two key reasons Firstly, these formulas apply to any angle, making them valuable for proving general mathematical concepts across various fields Secondly, they assist in identifying specific angles that correspond to particular trigonometric function values.
Find all angles 0 ◦ ≤ θ < 360 ◦ such that sin θ = − 0.682.
When inputting −0.682 into a calculator, we find that θ = −43°, which falls outside the range of 0° to 360° Since θ = −43° is located in Quadrant IV, its reflection around the y-axis, calculated as 180° − θ, places it in Quadrant III with the same sine value This reflection gives us 180° − (−43°) = 223° Additionally, both −43° and −43° + 360° = 317° yield identical trigonometric function values Therefore, the only angles between 0° and 360° that correspond to a sine value of −0.682 are θ = 223° and θ = 317°.
Figure 1.5.8 Reflection around the y-axis: − 43 ◦ and 223 ◦
7 In Chapter 5 we will discuss why the
Rotations and Reflections of Angles • Section 1.5 37
1 Let θ = 32 ◦ Find the angle between 0 ◦ and 360 ◦ which is the
(a) reflection of θ around the x-axis
(b) reflection of θ around the y-axis
(c) reflection of θ around the origin
2 Repeat Exercise 1 with θ = 248 ◦ 3 Repeat Exercise 1 with θ = − 248 ◦
4 We proved formulas (1.4)-(1.6) for any angle θ in QI Mimic that proof to show that the formulas hold for θ in QII.
5 Verify formulas (1.4)-(1.6) for θ on the coordinate axes, i.e for θ = 0 ◦ , 90 ◦ , 180 ◦ , 270 ◦
In Example 1.26, we demonstrated that the slopes of perpendicular lines are negative reciprocals using the formulas related to θ + 90° This article will show that the same conclusion can be reached by applying the formulas involving θ - 90° To achieve this, only the final paragraph of the previous example requires modification.
For Exercises 7 - 14, find all angles 0 ◦ ≤ θ < 360 ◦ which satisfy the given equation:
In our exploration of the Pythagorean Theorem, we demonstrated that in a right triangle △ ABC, a line segment CD can be drawn from the right angle vertex C to a point D on the hypotenuse AB, ensuring that CD is perpendicular to AB To validate this claim, we can analyze the triangle's position on the x-y coordinate plane, focusing on the slope of the hypotenuse.
What would be the slope of a line perpendicular to it?) Also, find the (x, y) coordinates of the point D in terms of a and b.