Solution Guide for Chapter 6: The Geometry of Right Triangles


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1 Solution Guide for Chapter 6: The Geometry of Right Triangles 6. THE THEOREM OF PYTHAGORAS E. Another demonstration: (a) Each triangle has area ( ). ab, so the sum of the areas of the triangles is 4 ab One ( ) square has area b, and the other has area a. So the total area is b + a + 4 ab. (b) Each triangle has area ( ). ab, so the sum of the areas of the triangles is 4 ab The ( ) square has area c. So the total area is c + 4 ab. (c) Equating the two expressions gives ( ) ( ) b + a + 4 ab = c + 4 ab, ( ) and canceling the common term of 4 ab leaves a + b = c. E. Yet another demonstration: (a) The distance between the parallel sides is a + b, and the average of the lengths of those sides is (a + b). The area is the product of these two numbers, so it equals (a + b) (a + b) = (a + b). (b) Two of the triangles have area ab, and the third has area c. (c) Expanding the expression in Part (a) using the rules of algebra gives ( a + ab + b ).
2 490 Solution Guide for Chapter 6 By Part (b) the area of the trapezoid is ( ) ab + c. Equating these two expressions gives ( a + ab + b ) ( ) = ab + c. ( ) Canceling the common term of ab and then multiplying through by gives a + b = c. E3. Other demonstrations: Answers will vary. S. A right triangle: Now = 4, so the sides satisfy the equation in the Pythagorean theorem. Hence this is a right triangle. S. A right triangle: Now = 85, so the sides satisfy the equation in the Pythagorean theorem. Hence this is a right triangle. S3. Find the missing side: By the Pythagorean theorem the hypotenuse has length = 58 = 7.6. S4. Find the missing side: By the Pythagorean theorem the hypotenuse has length = 97 = S5. Find the missing side: By the Pythagorean theorem the other leg has length 7 4 = 33 = S6. Find the missing side: By the Pythagorean theorem the other leg has length 9 5 = 56 = S7. Finding area: The triangle has area (4 5) = 0. S8. Finding area: The triangle has area (7 0) = 35.
3 SECTION 6. The Theorem of Pythagoras 49 S9. Finding perimeter: By the Pythagorean theorem the hypotenuse has length + 3 = 3. Thus the perimeter is = 8.6. S0. Finding perimeter: By the Pythagorean theorem the other leg has length 3 = 5. Thus the perimeter is = 7.4. S. A funny right triangle? By the Pythagorean theorem the other leg would have length 3 3 = 0. No such triangle is possible.. Area and perimeter: By the Pythagorean theorem a = 8 5 = 39 and b = 7 5 = 4. Thus the area is and the perimeter is ( 5 ) = 7.86, = Area and perimeter: By the Pythagorean theorem a = 3 0 = 69 and b = 7 0 = 89. Thus the area is and the perimeter is ( 0 ) = 0.7, = Area and perimeter: By the Pythagorean theorem the indicated altitude has length 3 6 = 33. Applying the Pythagorean theorem again gives b = 7 33 = 56. Thus the area is and the perimeter is ( ) = 06.6, =
4 49 Solution Guide for Chapter 6 4. Area and perimeter: By the Pythagorean theorem the indicated altitude has length 9 5 = 56. Applying the Pythagorean theorem again gives b = 7 56 = 33. Thus the area is ( ) = 75.8, and the perimeter is = A formula for Pythagorean triples: (a) To verify the formula (a b ) +(ab) = (a +b ), we expand the terms on each side to see if they are equal. The lefthand side of the formula is (a b ) + (ab) = (a ) (a )(b ) + (b ) + 4a b = a 4 a b + b 4 + 4a b = a 4 + a b + b 4. The righthand side of the formula is (a + b ) = a 4 + a b + b 4. Thus the two expressions are equal. (b) Using the formula in Part (a), we see that if r = a b, s = ab, and t = a + b for some a and b, then (r, s, t) is a Pythagorean triple. For example, if b = and a is any positive integer larger than, then we see that (a, a, a + ) is a Pythagorean triple, and so we can generate many such Pythagorean triples, as shown below. a Pythagorean triple (3,4,5) 3 (8,6,0) 4 (5,8,7) 5 (4,0,6) 6 (35,,37)
5 SECTION 6. The Theorem of Pythagoras Verifying the area formula: Refer to the accompanying figure. Altitude Extension of Base Base We use b to denote the length of the base, h the length of the altitude, and t the length of the extension of the base. Then the large triangle is a right triangle with base of length t + b, so its area is h(t + b). Inside the large triangle is another right triangle, one with base of length t and area ht. Subtracting these two areas gives the area of the original triangle: This is the desired formula. h(t + b) ht = hb. 7. An isosceles triangle: The altitude divides the isosceles triangle into two right triangles, and the altitude is a leg of each of these. Also, each of these has hypotenuse a and another leg on length b, so (by the Pythagorean theorem) the altitude has length a b. Then the area of the isosceles triangle is 4 b a b 4, which is the desired formula. 8. An equilateral triangle: Every equilateral triangle is isosceles, so the formula in Exercise 7 applies. In this case the base length b equals a, so the equilateral triangle has area as desired. a a a 4 = a 3a 3a 4 =, 4
6 494 Solution Guide for Chapter 6 9. Area and perimeter: (a) By the Pythagorean theorem the hypotenuse has length a + b. perimeter is a + b + a + b. Hence the (b) Since the area is 4, we get Solving for b gives Then the perimeter is ab = 4. b = 8 a. a + 8 a + a + ( ) 8. a (c) We want to know when the perimeter is 5. So we need to solve the equation a + 8 a + a + ( ) 8 = 5. a We do this using the crossinggraphs method. From the figure on the left below we see that there are two crossing points and thus two possible values for a. (We used a horizontal span from 0 to 0 and a vertical span from 0 to 5.) From this figure we read a value of a =.6. It turns out that the second crossing point occurs at a = Both of these values of a produce a triangle with a perimeter of 5 units. (d) We want to find when the graph we have made reaches a minimum. This is shown in the figure on the right above, where we see that the minimum perimeter of 9.66 occurs at a =.83.
7 SECTION 6. The Theorem of Pythagoras Area and perimeter: (a) By the Pythagorean theorem the hypotenuse has length a + b. perimeter is a + b + a + b. Hence the (b) Since the area is 0, we get ab = 0. Solving for b gives b = 0 a. Then the perimeter is a + 0 a + a + ( ) 0. a (c) We want to know when the perimeter is 5. So we need to solve the equation a + 0 a + a + ( ) 0 = 5. a We do this using the crossinggraphs method. From the figure on the left below we see that there are two crossing points and thus two possible values for a. (We used a horizontal span from 0 to 5 and a vertical span from 0 to 30.) From this figure we read a value of a =.73. It turns out that the second crossing point occurs at a =.57. Both of these values of a produce a triangle with a perimeter of 5 units. (d) We want to find when the graph we have made reaches a minimum. This is shown in the figure on the right above, where we see that the minimum perimeter of 5.7 occurs at a = A funny triangle? By the solution to Part (d) of the example, the minimum possible perimeter for such a triangle is Hence it is not possible to make such a triangle with a perimeter of 5 units.
8 496 Solution Guide for Chapter 6. A funny triangle? According to Part (b) of the example, the perimeter of such a triangle is a + 0 a + a + ( ) 0. a Here a denotes a leg. So we want to know if there is a solution to the equation a + 0 a + a + ( ) 0 = 500. a We do this using the crossinggraphs method. In the figure below we see one crossing point. (We used a horizontal span from 0 to and a vertical span from 0 to 600.) The corresponding value is a = It turns out that there is another crossing point, one that gives the value a = Both of these values of a produce a triangle with a perimeter of 500 units. 3. Finding a side: Let a denote the length of the other leg. By the Pythagorean theorem the hypotenuse has length 64 + a. Thus in terms of a the perimeter is 8 + a a. So we want to solve the equation 8 + a a = 0. We do this using the crossinggraphs method. In the figure below we used a horizontal span from 0 to 0 and a vertical span from 0 to 30.
9 SECTION 6. The Theorem of Pythagoras 497 We see that the corresponding value is a = This is the length of the other leg. 4. Finding the lengths of legs: (a) By the Pythagorean theorem b = 8 a = 64 a. (b) The perimeter is 8 + a + b, and by Part (a) this equals 8 + a + 64 a. (c) By Part (b) we want to solve the equation 8 + a + 64 a = 7. We do this using the crossinggraphs method. From the figure below we see that there are two crossing points and thus two possible values for a. (We used a horizontal span from 0 to 8 and a vertical span from 0 to 5.) From this figure we read a value of a =.07. The corresponding value of b is then b = = (It turns out that the second crossing point occurs at a = 7.93 no surprise and of course the corresponding value of b is then.07.) 5. Finding the lengths of legs: (a) By the Pythagorean theorem b = 5 a = 5 a. (b) The perimeter is 5 + a + b, and by Part (a) this equals 5 + a + 5 a. (c) By Part (b) we want to solve the equation 5 + a + 5 a = 35. We do this using the crossinggraphs method. From the figure below we see that there are two crossing points and thus two possible values for a. (We used a horizontal span from 0 to 5 and a vertical span from 0 to 40.)
10 498 Solution Guide for Chapter 6 From this figure we read a value of a = The corresponding value of b is then b = = (It turns out that the second crossing point occurs at a = 3.54 no surprise and of course the corresponding value of b is then 6.46.) 6. Finding the lengths of legs: (a) In general, the sum of the lengths of the legs must be larger than the hypotenuse because the shortest distance between two points is along a line. Here is another way to see this: If the hypotenuse has length c and the legs have lengths a and b, then (a + b) = a + b + ab > a + b = c, where the last equality holds by the Pythagorean theorem. Thus a + b > c. Of course, in this exercise c = 5. Hence the sum of the lengths of the legs must be at least 5. (b) No, because by Part (a) the sum of the lengths of the legs must be larger than 5, so the perimeter must be larger than 30. (c) In general, the hypotenuse is the longest side by the Pythagorean theorem: If the hypotenuse has length c and the legs have lengths a and b, then c = a + b > a, so c is larger than a. Of course, in this exercise c = 5. Hence the length of each leg is at most 5. (d) No, because by Part (c) the length of each leg is less than 5, so the perimeter is less than Maximizing area: (a) By the Pythagorean theorem b = 0 a = 400 a. (b) The area is ab, and by Part (a) this equals a 400 a.
11 SECTION 6. The Theorem of Pythagoras 499 (c) By Part (b) we want to find where the function a 400 a attains its maximum value. From the figure below we see that this occurs where a = 4.4. (We used a horizontal span from 0 to 0 and a vertical span from 0 to 0.) The corresponding value of b is then b = = 4.4. It makes sense that the maximum area occurs when the two legs are of equal length. 8. Maximizing area: (a) By the Pythagorean theorem b = 65 a = 45 a. (b) The area is ab, and by Part (a) this equals a 45 a. (c) By Part (b) we want to find where the function a 45 a attains its maximum value. From the figure below we see that this occurs where a = (We used a horizontal span from 0 to 65 and a vertical span from 0 to 00.) The corresponding value of b is then b = = It makes sense that the maximum area occurs when the two legs are of equal length.
12 500 Solution Guide for Chapter 6 9. Cartography: The line segment from our observation point to the peak is the hypotenuse of a right triangle. The hypotenuse has length. miles, and one leg (from our observation point to the base) has length.6 miles. By the Pythagorean theorem the other leg is..6 =.5 miles in length. Thus the height of the wall is about.5 miles. 0. Cartography: The line segment from our observation point to the peak is the hypotenuse of a right triangle. The hypotenuse has length miles, and one leg (the wall itself) has length 0.6 mile. By the Pythagorean theorem the other leg is 0.6 =.99 miles in length. That is the distance from our observation point to the base of the wall. 6. ANGLES E. The Pythagorean theorem: (a) The leg of length c in the smaller triangle corresponds to the leg of length a in the larger triangle. The hypotenuse of the smaller triangle on the left has length a, and the hypotenuse of the larger triangle has length c + c. By similarity we have c a = a c + c. (b) The leg of length c in the smaller triangle corresponds to the leg of length b in the larger triangle. The hypotenuse of the smaller triangle on the right has length b, and the hypotenuse of the larger triangle has length c + c. By similarity we have c b = b c + c. (c) In the equation from Part (a) we crossmultiply and recall that c + c = c. The result is a = cc. Doing the same thing in the equation from Part (b) yields b = cc. Adding these equations gives a + b = cc + cc = c(c + c ) = c, as desired.
13 SECTION 6. Angles 50 E. Isosceles right triangle: (a) Consider the smaller triangle on the left. It shares an angle with the larger triangle (the angle whose vertex is at the lower left). Both triangles have a right angle. This means that the third pair of angles must also be equal, so the triangles are similar. The same argument applies to the smaller triangle on the right. (b) The hypotenuse of the larger triangle has length, and the hypotenuse of the smaller triangle has length x. For the angle in common the corresponding lengths are x for the larger triangle and for the smaller triangle. By similarity we have and so x =. Hence x =. S. Conversion: This is 3 80 π S. Conversion: This is π 7 80 π S3. Conversion: This is 7 S4. Conversion: This is π x = x, = 7.89 degrees. = 5.7 degrees. π =.6 radians. 80 π = 0.05 radian. 80 S5. Angle sum: The radian measure of the third angle is π π 0.8 = S6. Angle sum: The degree measure of the third angle is = 0. S7. Angle sum: Because π 4 is 45 degrees, the degree measure of the third angle is The radian measure is = 05. π 80 =.83. S8. Angle sum: Because 0 degrees is 0 π 80 = π radian, the radian measure of the third 9 angle is π π 0.3 =.49. The degree measure is 4.8. (Here we converted to degree 9 measure before rounding the radian measure.)
14 50 Solution Guide for Chapter 6 S9. Angle sum: Because 70 degrees is 70 third angle is π 80 = 7π 8 π 7π 0.4 =.5. 8 radian, the radian measure of the The degree measure is (Here we converted to degree measure before rounding the radian measure.) S0. Angle sum: All three angles are equal, and they add up to 80 degrees, so the degree measure of an angle is 80 3 = 60. S. Angle sum: All three angles are equal, and they add up to π radians, so the radian measure of an angle is π 3 =.05.. Area and arc length: The arc length is π 6 5 =.6 units. The area is π 6 5 = 6.54 square units.. Area and arc length: The arc length is.4 9 =.6 units. The area is.4 9 = 56.7 square units. 3. Area and arc length: The arc length is.8 square units. 5 π 0 80 = 4.36 units. The area is 5 π = 4. Area and arc length: The arc length is 7 π 3 = 0.89 unit. The area is square units. 7 π = 5. Area and arc length: Let a denote the radian measure of the angle and r the radius. We know that ar = 5 and that ar = 9. Dividing the second equation by the first gives r = 9 5. Thus the radius is r = 9 = 3.6 units. Putting this result into the equation 5 ar = 5 gives that 3.6a = 5, so the angle is a = 5 =.39 radians Area and arc length: Let a denote the radian measure of the angle and r the radius. We know that ar = and that ar = 7. Dividing the second equation by the first gives r = 7 7. Thus the radius is r = = 7 =.83 units. Putting this result into 6 the equation ar = gives that the angle is a = 7 6 = 7 radians. This is degrees. 7. Doubling: If we consider the circle formed by the bottom of the pie, then the bigger piece represents twice the area of the smaller. Now in general the area is proportional to the angle, so doubling the area requires doubling the angle. Hence the bigger piece (that of your friend) has twice the central angle of the smaller one (yours). To be explicit, let a denote the radian measure of the smaller angle and A that of the bigger angle. Let
15 SECTION 6. Angles 503 r denote the radius. The area for the smaller angle is ar, and the area for the bigger angle is Ar Ar. The second of these is twice the first, so = ar. Cancelling common terms gives A = a. 8. More on doubling: If we consider the circle formed by the bottom of each pie, then the circle for the larger pie has twice the diameter of the smaller. Let r denote the radius of the smaller circle. Then the radius of the larger circle is r. Let a denote the radian measure corresponding to a piece. (This is the same for both circles because the number of pieces is the same.) The area for a piece of the smaller pie is ar, and the area for a piece of the larger pie is a(r) = a r = 4 ar. Thus the pieces of the larger pie are 4 times those of the smaller. 9. Latitude: The distance from Fort Worth to Wichita is the arc length cut by an angle of 37 3 = 5 degrees out of a circle of radius 3950 miles. That length is 5 π 3950 = miles. 0. More latitude: Now we are given the arc length (80 miles), and we need to find the angle. The radius is still 3950 miles. If the angle is d degrees then d π 3950 = Solving for d gives d = 80 π 3950 = 7. degrees. Thus the latitude of Winnipeg 80 is = 49., or about 49, degrees north.. Similarity: By similarity we have B 8 = 3 8 and thus B = =.33. In a similar way we find c = 8 3 and thus c = 8 3 =.. Similarity: By similarity we have B 0 = 3 0 and thus B = =.5. In a similar way we find c = 0 3 and thus c = 0 3 = Similarity: By similarity we have y = x and thus y = x. 4. Similarity: By similarity we have y = x y and thus y = x. Hence y = x. 5. Grads: Because 90 degrees represents onequarter of a circle, the grad measure of this angle is 400 = 00. Because π radians represents onehalf of a circle, the grad 4 measure of this angle is = 00. An angle of 3 degrees represents 360 of a circle, so the grad measure of this angle is =
16 504 Solution Guide for Chapter 6 6. Mils: (a) Now mil is 80 radian, and radian is degrees. Hence mil is 000 π = degree. π (b) Let r denote the distance, in yards, from the soldier to the target. We know that an angle of mil cuts an arc length of yard out of a circle of radius r. Because mil is radian, the formula for arc length says 000 Hence the distance is r = 000 yards. 000 r =. (c) Now there are 000 mils in a radian, and there are π radians in a circle. Hence there are 000 π = mils in a circle. 7. Shadows: There are two right triangles here. One has the pole and its shadow as legs, and the other has the tree and its shadow. Because the position of the sun is the same for both shadows, these right triangles are similar. Let t denote the height of the tree. Then by similarity t 0 = 6. Thus t = 0 6 = 35, so the tree is 35 feet tall. 8. Ladder: There are two right triangles to consider here. One has the ladder as hypotenuse and a 5foot horizontal segment as a leg. The other has the lower feet of the ladder as hypotenuse and for a leg the horizontal segment from the base of the ladder to the point beneath the rung. We denote by d the length of this latter leg. Because these right triangles have the angle of the ladder with the horizontal in common, the triangles are similar. By similarity d 5 = 0. Thus d = 5 0 = 3. Hence the distance from the rung to the wall is 5 3 = feet.
17 SECTION 6.3 Right Angle Trigonometry An angle and a circle: By similarity we have AB AD = AE AC. Crossmultiplying gives AB AC = AD AE, as desired. 0. Area equals arc length: Let r denote the radius of the circle. We know that for every angle of radian measure a the corresponding area and arc length are the same, so ar = ar. This implies that and solving for r gives r =. r = r, 6.3 RIGHT TRIANGLE TRIGONOMETRY E. Other trigonometric functions of special angles: We know that sin 30 = and Hence tan 30 = cos 30 = sin 30 cos 30 = 3. ( ) 3 = 3, cot 30 = sec 30 = tan 30 = cos 30 = ( 3 ) = 3, ( ) 3 = 3, and csc 30 = sin 30 = ( ) =. E. Other trigonometric functions of special angles: We know that sin π 4 = cos π 4 =. Hence tan π 4 = sin π 4 cos π 4 = ( ) =,
18 506 Solution Guide for Chapter 6 cot π 4 = tan π 4 sec π 4 = cos π 4 = = =, ( ) =, and csc π 4 = sin π 4 ( ) = =. E3. The 5degree angle: (a) Because the angle marked s has degree measure 5, the angle of the big triangle that includes s has degree measure = 75. Now we show that the angle marked t has measure 75 degrees. In addition to the angle marked t, the big triangle has a 30degree angle, and also the angle of measure 75 degrees that we just considered. Because the angle sum of the triangle is 80 degrees, the angle marked t has measure = 75 degrees. Here is another way to see the same thing: The adjoined smaller right triangle has acute angles s and t. Because the angle sum of the triangle is 80 degrees, the angle marked t has measure = 75 degrees. (b) The sides opposite the 75degree angles in the big triangle are the same, so A+ 3 =, and thus A = 3. (c) We apply the Pythagorean theorem to the adjoined smaller right triangle: B = + A, so by Part (b) we have B = + ( 3) = = Thus B = (d) In the adjoined smaller right triangle, the angle marked s has degree measure 5, the side opposite s is A, and the hypotenuse is B. Hence as desired. sin 5 = Opposite Hypotenuse = A B = 3, 8 4 3
19 SECTION 6.3 Right Angle Trigonometry 507 E4. An angle of π 8 radian: (a) Because the angle marked s has radian measure π, the angle of the big triangle that 8 includes s has radian measure π 4 + π 8 = 3π 8. Now we show that the angle marked t has radian measure 3π 8. In addition to the angle marked t, the big triangle has an angle of radian measure π 4, and also the angle of radian measure 3π that we just 8 considered. Because the angle sum of the triangle is π radians, the angle marked t has radian measure π π 4 3π 8 = 3π. Here is another way to see the same thing: 8 The adjoined smaller right triangle has acute angles s and t. Because the angle sum of the triangle is π radians, the angle marked t has radian measure π π π 8 = 3π 8. (b) The sides opposite the angles in the big triangle of radian measure 3π 8 so A + =, and thus A =. are the same, (c) We apply the Pythagorean theorem to the adjoined smaller right triangle: B = + A, so by Part (b) we have B = + ( ) = + + = 4. Thus B = 4. (d) In the adjoined smaller right triangle, the angle marked s has radian measure π 8, the side opposite s is A, and the hypotenuse is B. Hence as desired. sin π 8 = Opposite Hypotenuse = A B =, 4 S. Calculating trigonometric functions: Now sin θ = Opposite Hypotenuse = 5 7 = 0.88, and cos θ = tan θ = Opposite = 5 8 Hypotenuse = 8 7 = 0.47, =.875, or about.88.
20 508 Solution Guide for Chapter 6 S. Calculating trigonometric functions: Now sin θ = Opposite Hypotenuse = = 0.88, and cos θ = Hypotenuse = 6 34 = 0.47, tan θ = Opposite = 30 =.875, or about S3. Calculating trigonometric functions: Now sin θ = Opposite Hypotenuse = 4 5 = 0.96, and cos θ = Hypotenuse = 7 5 = 0.8, tan θ = Opposite = 4 7 = S4. Calculating trigonometric functions: Now sin θ = Opposite Hypotenuse = = 0.95, and cos θ = Hypotenuse = 4 74 = 0.3, tan θ = Opposite = 70 4 =.9. S5. Calculating trigonometric functions: By the Pythagorean theorem the length of the hypotenuse is = 3. Now sin θ = Opposite Hypotenuse = 8 = 0.75, 3 and cos θ = Hypotenuse = 7 = 0.66, 3 tan θ = Opposite = 8 7 =.4.
21 SECTION 6.3 Right Angle Trigonometry 509 S6. Calculating trigonometric functions: By the Pythagorean theorem the length of the hypotenuse is 5 + = 69 = 3. Now sin θ = cos θ = Opposite Hypotenuse = 3 = 0.9, Hypotenuse = 5 3 = 0.38, and tan θ = Opposite = 5 =.40. S7. Calculating trigonometric functions: By the Pythagorean theorem the length of the opposite side is 9 = 63. Now sin θ = cos θ = Opposite 63 Hypotenuse = = 0.66, Hypotenuse = 9 = 0.75, and tan θ = Opposite 63 = = S8. Calculating trigonometric functions: By the Pythagorean theorem the length of the opposite side is 50 5 = 75. Now sin θ = cos θ = Opposite 75 Hypotenuse = = 0.95, 50 Hypotenuse = 5 50 = 0.30, and tan θ = Opposite 75 = = S9. Getting a length: If θ is the angle and H is the hypotenuse then 0. = sin θ = Opposite Hypotenuse = 8 H, so H = 8 0. = 7.73.
22 50 Solution Guide for Chapter 6 S0. Getting a length: If θ is the angle and H is the hypotenuse then 0.7 = cos θ = Hypotenuse = 8 H, so H = =.7. S. Getting an angle: We want to find θ so that sin θ = We use the crossinggraphs method in degree mode with a horizontal span from 0 to 90 degrees and a vertical span from 0 to. We see from the graph below that θ = 3.0 degrees. S. Getting an angle: We want to find θ so that tan θ = 4. We use the crossinggraphs method in radian mode with a horizontal span from 0 to π radians and a vertical span from 0 to 5. We see from the graph below that θ =.33 radians. S3. Other trigonometric functions: Now and sec 30 = csc 30 = cot 30 = cos 30 =.5, sin 30 =, tan 30 =.73. Here we used the calculator to find the cosine, sine, and tangent. The exact values can be found in the tables before the Enrichment Exercises.
23 SECTION 6.3 Right Angle Trigonometry 5 S4. Other trigonometric functions: Now sec π 4 = cos π 4 csc π 4 = sin π 4 =.4, =.4, and cot π 4 = tan π 4 =. Here we used the calculator to find the cosine, sine, and tangent. The exact values can be found in the tables before the Enrichment Exercises.. Calculating height: If the man sits 33 horizontal feet from the base of a wall and he must incline his eyes at an angle of 6. degrees to look at the top of the wall, then the ground and the wall form two legs of a right triangle with angle θ = 6. degrees, and the adjacent side has length 33 feet. Since we want to find the length of the opposite side, we use the tangent (as it is the ratio of opposite to adjacent): tan θ = Opposite tan 6. = Opposite 33 Opposite = 33 tan 6. = Thus the wall is 96.6 feet tall.. Calculating height: If the man sits horizontal feet from the base of a wall and he must incline his eyes at an angle of 6.4 degrees to look at the top of the wall, then the ground and the wall form two legs of a right triangle with angle θ = 6.4 degrees, and the adjacent side has length feet. Since we want to find the length of the opposite side, we use the tangent (as it is the ratio of opposite to adjacent): tan θ = Opposite tan 6.4 = Opposite Opposite = tan 6.4 = Thus the wall is 4.90 feet tall. 3. Calculating an angle: If the man sits 70 horizontal feet from the base of a wall which is 76 feet high, then the ground and the wall form two legs of a right triangle with adjacent side of length 70 feet and opposite side of length 76 feet. Since we want to
24 5 Solution Guide for Chapter 6 find the angle θ at which he must incline his eyes to view the top of the wall, we use the tangent (as it is the ratio of opposite to adjacent): tan θ = Opposite = We solve this using the crossinggraphs method. We use a horizontal span from 90 to 90 degrees and a vertical span from 0 to. We see from the graph below that θ = 5.7 degrees. In radians the solution is Calculating an angle: If the wall is 38 feet high, the man sits on the ground, and the distance from the man to the top of the wall is 83 feet, then the ground and the wall form two legs of a right triangle with opposite side of length 38 feet and a hypotenuse of length 83 feet. Since we want to find the angle θ at which he must incline his eyes to view the top of the wall, we use the sine (as it is the ratio of opposite to hypotenuse): Opposite sin θ = Hypotenuse = We solve this using the crossinggraphs method. We use a horizontal span from 0 to 90 degrees and a vertical span from 0 to. We see from the graph below that θ = 7.5 degrees. In radians the solution is 0.48.
25 SECTION 6.3 Right Angle Trigonometry Calculating distance: The horizontal distance of the man from the wall in Exercise 4 is the length of the adjacent side of the triangle. Since we already know the lengths of the other two sides, as well as the angle θ, there are many ways to calculate the horizontal distance. One way is to use cosine (as it is the ratio of adjacent to hypotenuse): cos θ = Hypotenuse cos 7.5 = 83 = 83 cos 7.5 = Thus the horizontal distance of the man from the wall is feet. 6. Calculating distance: If the man sits 30 horizontal feet from the base of a wall and he must incline his eyes at an angle of 3 degrees to look at the top of the wall, then the ground and the wall form two legs of a right triangle with angle θ = 3 degrees, and the adjacent side has length 30 feet. Since we want to find the distance from the man directly to the top of the wall, that is, the length of the hypotenuse, we use the cosine (as it is the ratio of adjacent to hypotenuse): cos θ = cos 3 = Hypotenuse 30 Hypotenuse. Hence Hypotenuse = 30 cos 3 = 33.4, so the distance from the man directly to the top of the wall is 33.4 feet. 7. Calculating distance: If the man sits 8 horizontal feet from the base of a wall and he must incline his eyes at an angle of degrees to look at the top of the wall, then the ground and the wall form two legs of a right triangle with angle θ = degrees, and the adjacent side has length 8 feet. Since we want to find the distance from the man directly to the top of the wall, that is, the length of the hypotenuse, we use the cosine (as it is the ratio of adjacent to hypotenuse): cos θ = cos = Hypotenuse 8 Hypotenuse. Hence Hypotenuse = 8 cos = 9.8, so the distance from the man directly to the top of the wall is 9.8 feet.
26 54 Solution Guide for Chapter 6 8. The right triangle: (a) In a right triangle, the legs have lengths 3 and 4, while the hypotenuse has length 5. One acute angle has an opposite side of length 3 and hypotenuse of length 5, so its sine is 3 5 = 0.6, while the other acute angle has an opposite side of length 4 and hypotenuse of length 5, so its sine is 4 5 = 0.8. (b) To find the two acute angles of the right triangle, by Part (a) we need only solve the equations sin θ = 0.6 and sin θ = 0.8. The corresponding solutions (found using crossing graphs, for example) are θ = degrees and θ = 53.3 degrees, respectively. In radians the angles are 0.64 and 0.93, respectively. 9. Finding the angle: We want to find θ so that cot θ = 5, or cos θ sin θ = 5. We solve this using the crossinggraphs method. We use a horizontal span from 0 to 90 degrees and a vertical span from 0 to 0. We see from the graph below that θ =.3 degrees. 0. A building: If you must incline your transit at an angle of 0 degrees to look at the top of the building, which is 50 feet taller than the transit, then the ground and the building form two legs of a right triangle with angle θ = 0 degrees, and the opposite side has length 50 feet. Since we want to find the horizontal distance from your transit to the building, that is, the length of the adjacent side, we use the tangent (as it is the ratio of opposite to adjacent): Hence tan θ = Opposite tan 0 50 =. = 50 tan 0 = 4., so the horizontal distance from your transit to the building is 4. feet.
27 SECTION 6.3 Right Angle Trigonometry 55. The width of a river: If you must rotate your transit through an angle of degrees to point toward the second tree, and the distance between the trees is 35 yards, then the two trees and the transit form the vertices of a right triangle, with the right angle at the first tree. This is a right angle because the transittofirsttree line is northsouth, while the line between the trees is eastwest. The transittofirsttree and the firsttreetosecondtree sides form two legs of a right triangle with angle θ = degrees, and the opposite side has length 35 yards. Since we want to find the distance from your transit to the first tree, that is, the length of the adjacent side, we use the tangent (as it is the ratio of opposite to adjacent): Hence tan θ = Opposite tan 35 =. = 35 tan = 64.66, so the distance from your transit to the first tree, that is, the width of the river, is yards, or about 65 yards.. A cannon: The distance downrange a cannonball strikes the ground is given by m sin (t), g where, in our case, g = 3 and m = 300, and we want the cannonball to strike the ground 000 feet downrange. Thus we want to find t such that 300 sin(t) = This equation can be solved using the crossinggraphs method. We use a horizontal span from 0 to 90 degrees and a vertical span from 0 to 000. We see from the graph below that one solution is an angle of t = 0.4 degrees. Another angle that could be used is t = degrees. In radians these are 0.8 and.39, respectively.
28 56 Solution Guide for Chapter 6 3. Dallas to Fort Smith: Since Dallas is 90 miles due south of Oklahoma City, and Fort Smith is 40 miles due east of Oklahoma City, the triangle Dallas to Oklahoma City to Fort Smith to Dallas is a right triangle with the right angle at Oklahoma City, and the airplane s flight path forms the hypotenuse. (a) The tangent of the angle that the flight path makes with Interstate 35 is given by Opposite = = (b) The angle θ that the flight path makes with Interstate 35 has the property that tan θ = 40. Solving, for example using crossing graphs, shows that θ = degrees. In radians the solution is (c) The distance that the airplane flies is the length of the hypotenuse. Using cosine, for example, we can calculate that length: cos θ = cos = Hypotenuse 90 Hypotenuse. Hence Hypotenuse = so the length of the flight path is 36 miles. 4. Intensity of sunlight: 90 cos = 36.00, (a) If the angle θ is 35 degrees, then the intensity is reduced by a factor of sin θ = sin 35 = (b) If the intensity is reduced by a factor of 0.3, then the angle θ of the incident rays satisfies the relation sin θ = 0.3. Solving (for example using crossing graphs), we find that θ = 7.46 degrees. 5. Grasping prey: As the diagram shows, the triangle ABC is a right triangle with the right angle at B. The diameter BC is the opposite side from the angle θ, while the length AC is the hypotenuse. Sine relates the lengths of the opposite side and the hypotenuse: Thus the formula is BC = AC sin θ. Opposite sin θ = Hypotenuse sin θ = BC AC sin θ AC = BC.
29 SECTION 6.3 Right Angle Trigonometry Getting the tangent from the sine: (a) We are given that Opposite = sin θ = 3 Hypotenuse, so one such triangle has a side of length opposite θ and a hypotenuse of length 3, as in the figure below. All such triangles are similar to this one. 3 0 (b) By the Pythagorean theorem, the length of the adjacent side is 3 = 5 =.4. (c) Now tan θ = Opposite = 5 = Getting the cosine from the tangent: (a) We are given that 5 Opposite = tan θ =, so one such triangle has a side of length 5 opposite θ and a side of length adjacent to θ, as in the figure below. All such triangles are similar to this one.
30 58 Solution Guide for Chapter (b) By the Pythagorean theorem, the length of the hypotenuse is 5 + = 6 = 5.0. (c) Now cos θ = Hypotenuse = = Getting the sine from the cotangent: (a) We are given that = cot θ = 3 Opposite, so one such triangle has a side of length 3 opposite θ and a side of length adjacent to θ, as in the figure below. All such triangles are similar to this one. 3 0 (b) By the Pythagorean theorem, the length of the hypotenuse is 3 + = 3 = 3.6.
31 SECTION 6.3 Right Angle Trigonometry 59 (c) Now sin θ = Opposite Hypotenuse = 3 = Dispersal method: (a) As indicated in the figure, v is the length of the side opposite the angle θ and d is the length of the hypotenuse. Thus sin θ = v and so v = d sin θ. d (b) If the distance the organisms move is 30 centimeters, then d = 30, and so the formula from Part (a) becomes v = 30 sin θ. Thus if the angle θ is 5 degrees, then v = 30 sin 5 = 7.76 centimeters, while if the angle is 30 degrees, then v = 30 sin 30 = 5 centimeters. (c) If the distance the organisms move is d = 30 centimeters and the change in elevation is v = 0 centimeters, then, by the formula from Part (a), sin θ = Solving for θ, for example using crossing graphs, we obtain θ = 4.8 degrees. 0. Jumping locust: We are given that g = 9.8, so the formula for the horizontal distance that an animal jumps is d = m sin θ. 9.8 (a) If a locust jumps at an angle of θ = 55 degrees for a horizontal distance d of 0.8 meter, then 0.8 = m sin( 55 ), so m sin( 55 ) = Hence m = sin( 55. Finally, the initial velocity is ) m = sin( 55 =.89 meters per second. ) (b) If the distance that an animal jumps is meter and the initial velocity is 3. meters per second, then = 3. sin(θ), and so 9.8 = 3. sin(θ), or 9.8 sin(θ) = Solving for θ, for example by using crossing graphs, we find that θ = degrees.
32 50 Solution Guide for Chapter 6 Chapter 6 Review Exercises. Area and perimeter: By the Pythagorean theorem a = 7 3 = 40 = 6.3 and b = 0 a = = 60 = Thus the area is and the perimeter is ( ) 60 = 33.98, = Maximizing area and perimeter: (a) By the Pythagorean theorem the length of the other leg is 30 a = 900 a. (b) The perimeter is a a. (c) The area is a 900 a. (d) We want to find when the graph of the perimeter function from Part (b) reaches a maximum. We made the graph using a horizontal span from 0 to 30 and a vertical span from 0 to 00. In the figure below we see that the maximum perimeter of 7.43 occurs at a =..
33 Chapter 6 Review Exercises 5 (e) We want to find when the graph of the area function from Part (c) reaches a maximum. We made the graph using a horizontal span from 0 to 30 and a vertical span from 0 to 300. In the figure below we see that the maximum area of 5 occurs at a =.. 3. Lengths of legs: (a) In general, the sum of the lengths of the legs must be larger than the hypotenuse because the shortest distance between two points is along a line. Here is another way to see this: If the hypotenuse has length c and the legs have lengths a and b, then (a + b) = a + b + ab > a + b = c, where the last equality holds by the Pythagorean theorem. Thus a + b > c. Of course, in this exercise c = 0. Hence the sum of the lengths of the legs must be at least 0. (b) In general, the hypotenuse is the longest side by the Pythagorean theorem: If the hypotenuse has length c and the legs have lengths a and b, then c = a + b > a, so c is larger than a. Of course, in this exercise c = 0. Hence the sum of the lengths of the legs is less than twice the length of the hypotenuse, so is less than 40. (c) Yes. By the Pythagorean theorem the length of the other leg is 0 = 399 = (d) No: By the reasoning in Part (b), the hypotenuse is longer than each leg. 4. Pythagorean triples: (a) Such a triangle is similar to the right triangle, so it has a right angle. (b) There are many possibilities, including (3,4,5), (5,,3), and (8,5,7). These can be found by trial and error, or by using Exercise 5 in Section 6..
34 5 Solution Guide for Chapter 6 (c) We would need to find a right triangle whose legs are whole numbers a and b and whose hypotenuse is 3. Then a and b would have to be less than 3, so each would have to be or. No such choice satisfies the equation a + b Pythagorean theorem. = 3 from the 5. Area and arc length: The arc length is 0 π 8 =.79 inches. The area is 80.7 square inches. 0 π = 6. Angle sum of a triangle: (a) The angle sum of a triangle is 80 degrees or π radians. (b) The sum of the two acute angles is = 90 degrees or π radians. (c) Now radian is 80 = degrees, and that is the degree measure of the angle π at vertex B. The degree measure of the angle at vertex C is π = (d) Now 60 degrees is 60 π 80 = π =.05 radians, and that is the radian measure of 3 the angle at vertex A. The radian measure of the angle at vertex C is π π 3 =.09. (This can also be found by converting the answer from Part (c) to radians.) 7. Similar triangles: We use vertical bars to denote the length of a segment. (a) These are right triangles, and they have the angle at E in common, so they are similar. (b) By similarity we have Thus so AE = 3 7 = 8.4. (c) Now AE CE = AB CD. AE 3 = 7, AC = AE CE = = 5.4. We find BD by first finding BE. By similarity we have BE DE = AB CD. Thus so BE = 5 7 = Hence BE 5 = 7, BD = BE DE = =.08.
35 Chapter 6 Review Exercises 53 The length of BE can also be found using by the Pythagorean theorem. There is small variation due to rounding. (d) The quadrilateral is a right trapezoid, and its area is ( AB + CD ) BD = (7 + ).08 = Here is another way to do this: The area of the quadrilateral is the difference between the areas of triangle ABE and triangle CDE. There is small variation due to rounding. 8. Ladder against a wall: (a) We consider the right triangle with the ladder as hypotenuse, a 7foot vertical segment as a leg, and a 6foot horizontal segment as a leg. By the Pythagorean theorem the length of the hypotenuse is = Thus the ladder is about 8 feet long. (b) In addition to the right triangle from Part (a), there is another right triangle to consider here. It has the lower 4 feet of the ladder as hypotenuse and for a leg the vertical segment from the rung to the ground. We denote by d the length in feet of this latter leg. Because these right triangles have the angle of the ladder with the horizontal in common, the triangles are similar. By similarity Thus d 7 = 4 8. d = = Hence the rung is about 3.78 feet above the ground. The answer will vary slightly if we use the length of the ladder from Part (a) without rounding. 9. Trigonometric functions: (a) By the Pythagorean theorem the length of the hypotenuse is = 49 =.. (b) Now and sin θ = cos θ = Opposite Hypotenuse = 7 = 0.57, 49 Hypotenuse = 0 = 0.8, 49 tan θ = Opposite = 7 0 = 0.70.
36 54 Solution Guide for Chapter 6 (c) Now and csc θ = 49 sin θ = =.74, 7 sec θ = 49 cos θ = =., 0 cot θ = tan θ = 0 7 =.43. Using the rounded values from Part (b) leads to some variation. 0. Calculating vertical distances: (a) The ground and the cliff form the two legs of a right triangle with angle θ = 3 degrees, and the adjacent side has length 60 feet. The tangent function is the ratio of opposite to adjacent, so it involves the height of the cliff and the distance from the base of the cliff. (b) Let c denote the height of the cliff in feet. By the reasoning in Part (a), so tan 3 = Thus the height of the cliff is feet. c 60, c = 60 tan 3 = (c) We want the length of the hypotenuse of the right triangle described in Part (a). Let H denote that length in feet. We use the cosine function because it involves the adjacent side and the hypotenuse. We have so cos 3 = Hypotenuse = 60 H, H = 60 cos 3 = Thus the distance from the person to the top of the cliff is 66.6 feet. This can also be found using the Pythagorean theorem and the answer from Part (b).. Calculating horizontal distances: (a) The height above the road of the mountain s peak is 4,60 50 = 9390 feet. (b) The road and the mountain form the two legs of a right triangle with angle θ = 9.5 degrees, and by Part (a) the opposite side has length 9390 feet. We use the tangent function because it is the ratio of opposite to adjacent, so it involves the height of
37 Chapter 6 Review Exercises 55 the peak above the road and the distance from the base of the mountain. Let d denote that distance in feet. Then so tan 9.5 = 9390 d, d = 9390 tan 9.5 = 56,.43. Thus the distance from the base of the mountain is about 56, feet, or 0.63 miles.. Finding an angle: We want to find θ so that tan θ =.3. We solve this using the crossinggraphs method. We use a horizontal span from 0 to 90 using degrees and a vertical span from 0 to 5. We see from the graph below that θ = degrees. 3. Getting the sine and cosine from the tangent: (a) We are given that Opposite = tan θ = 3, so one such triangle has a side of length opposite θ and a side of length 3 adjacent to θ, as in the figure below. All such triangles are similar to this one. 0 3 (b) By the Pythagorean theorem, the length of the hypotenuse is + 3 = 0 = 3.6. (c) Now sin θ = Opposite Hypotenuse = 0 = 0.3,
38 56 Solution Guide for Chapter 6 and cos θ = Hypotenuse = 3 = Jumping: (a) If an animal jumps at an angle of θ = 50 degrees and the horizontal distance d is. meters, then. = 9.8 m sin( 50 ), so m. 9.8 = sin( 50. Hence the initial ) velocity is. 9.8 m = sin( 50 = 3.3 meters per second. ) (b) If the distance that an animal jumps is meters and the initial velocity is 6 meters per second, then = sin(θ), and so sin(θ) = This equation can be solved using the crossinggraphs method. We use a horizontal span from 0 to 90 degrees and a vertical span from 0 to. We see from the graph below that there are two solutions and that one solution is an angle of θ = 6.49 degrees. The other solution is an angle of θ = 73.5 degrees. (c) If the distance that an animal jumps is meters and the initial velocity is 4 meters per second, then = sin(θ), and so sin(θ) = Now the righthand side of this equation is =.5, which is greater than. Because the sine function is never greater than, there is no solution to this equation. (This can also be seen from a graph.) Hence there are no such angles.
39 Chapter 6 Review Exercises The 53 right triangle: (a) Because 5 + = 3, the 53 triangle is a right triangle. (b) In a 53 right triangle, the legs have lengths 5 and, while the hypotenuse has length 3. One acute angle has an opposite side of length 5 and hypotenuse of length 3, so its sine is 5 = 0.38, and the other acute angle has an opposite side 3 of length and hypotenuse of length 3, so its sine is 3 = 0.9. (c) To find the two acute angles of the 53 right triangle, by Part (b) we need only solve the equations sin θ = 5 and sin θ =. The corresponding solutions (found 3 3 using crossing graphs, for example) are θ =.6 degrees and θ = degrees, respectively. (d) In radians the angles from Part (c) are 0.39 and.8.
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