#$&* course Mth 164 3/22 4:30 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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10:55:51
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Given Solution: ** c is the hypotenuse. The angle a opposite the angle alpha satisfies hypotenuse * sin(alpha) = a so that a = 10 * sin(40 deg) = 6.43, approx.. We also have b = c * cos(alpha) = 10 * cos(40 deg) = 7.66 approx.. The remaining angle of the triangle is beta = 90 deg - alpha = 90 deg - 40 deg = 50 deg. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK **** Query problem 7.1.B-24 cliff height 100 feet, angle of elevation 25 deg. Dist of ship from shore.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The problem can be solved by observing that a right triangle is formed by the cliff height, distance from cliff and terminal side of the angle of elevation to the cliff height. The cliff height is 100 feet and the angle of elevation is 25 degrees, therefore, the distance to the ship is approximately 214.45 feet and figured as follows: tan(25 deg) = 100 ft/distance from cliff to ship distance from cliff to ship * tan(25 deg) = 100 ft distance from cliff to ship = 100 ft/ tan(25 deg) or approx. 21.45 ft confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The cliff height forms a leg of a right triangle, oppposite the 25 deg angle. The distance from ship to shore forms the other leg of the triangle, adjacent to the 25 deg angle. Cliff height / distance from ship to shore = opposite side / adjacent side = tan(25 deg) so adjacent side = opposite side / tan(25 deg) = 100 ft / tan(25 deg) = 214.5 ft. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK **** Query problem 7.1.B-36 guy wire 80 ft long makes an angle of 25 deg with a ground; ht of tower?
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11:07:20 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The problem can be solved by observing that a right triangle is formed by the tower height, distance from tower and terminal side of the angle of elevation to the tower height. The guy wire length is 80 feet and the angle of elevation is 25 degrees, therefore, the height of the tower is approximately 33.81 feet and figured as follows: sin(25 deg) = tower height/80 ft tower height = 80 ft * sin(25 deg) or approx. 33.81 feet confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The guy wire is the hypotenuse of a right triangle for which the altitude is opposite the 25 degree angle. Thus we have altitude = hypotenuse * sin(25 deg) = 80 ft * sin(25 deg) = 33.8 ft. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK **** query problem 7.1.A-72 length of ladder around corner hall widths 3 ft and 4 ft `theta relative to wall in 4' hall, ladder in contact with walls
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The longest ladder that will fit is about 9.86 feet. After studying the picture, we can see that we have two triangles that are similar. The sum of the two hypotenuses represents the length of the longest ladder that will fit around the corner. We know that sin(theta) = opposite side/ hypotenuse and cos(theta) = adjacent side / hypotenuse. Using the formulas we have: sin(theta) = 4/ hypotenuse hypotenuse *(sin(theta)) = 4 hypotenuse = 4/(sin(theta)) = the length of the ladder in the 4ft hallway The hypotenuse value above is added to the hypotenuse value below to make up the entire longest ladder. cos(theta) = 3/ hypotenuse hypotenuse *(cos(theta)) = 3 hypotenuse = 3/(cos(theta)) = the length of the ladder in the 3ft hallway The longest ladder to fit around the corner is 3* (cos(theta)) + 4* (sin(theta)) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** In the triangle formed by the ladder in the wider hall, `theta is the angle opposite the 4-foot leg of the triangle. If the length of the part of the ladder in that hall is c1, then c1 = 4 / sin(`theta). In the triangle formed in the narrower hall, the 3-foot leg of the triangle is parallel to the sides of the wall in the first hall so by corresponding angles `theta is the angle adjacent to that leg, and if c2 is the hypotenuse of that triangle we have c2 = 3 ft / cos(`theta). The length of the ladder is therefore 3 ft / cos(`theta) + 4 ft / sin(`theta) or 3 ft sec(`theta) + 4 ft csc(`theta). **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK **** query problem 7.1.A-78 area of isosceles triangle A = a^2 sin`theta cos`theta, a length of equal side
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have the problem: sketch an isosceles triangle and label the equal sides both a, indicating length a along each of the equal sides, and the two equal angles `theta at the base of each side. Show that the area of the triangle is A = a^2 sin(`theta) cos(`theta). Triangle area = ½ base * height. The triangle can be divided into two equal right triangles. Each triangle has hypotenuse = a, one side = the line of division = height , and one side = ½ original base therefore: one side = the line of division = height = a * sin(theta) and one side = ½ original base = a * cos(theta) so we have: a * sin(theta) * a * cos(theta) = a^2 * sin(theta)cos(theta). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If you divide the triangle by its axis of symmetry you get two congruent right triangles, each with angle `theta opposite the altitude and adjacent to the base. The side a makes up the hypotenuse of either of these triangles. The altitude of each is therefore a sin(`theta) and the base is a cos(`theta). The area of each triangle is thus 1/2 * base * height = 1/2 a sin(`theta) a cos(`theta) = 1/2 a^2 sin(`theta) cos(`theta). The areas of the two right triangles add up to the area of the isosceles triangle. This area is therefore 2 ( 1/2 a^2 sin(`theta) cos(`theta) ) = a^2 sin(`theta) cos(`theta). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK **** Query problems 7.2.12 alpha = 70 deg; `beta = 60 deg, c = 4
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11:48:52 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: C = 180 deg - 70 deg - 60 deg = 50 deg Using the Law of Sines we have: sin(50 deg)/4 = sin(60 deg)/b b = 4sin(60 deg)/sin(50 deg) or approx. 4.52 sin(50 deg)/4 = sin(70 deg)/a a = 4sin(70 deg)/sin(50 deg) or approx. 4.91 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** GOOD STUDENT SOLUTION: if alpha = 70 deg; and `beta = 60 deg, then `gamma = 50 deg alpha + `beta + `gamma = 180 deg 70 deg + 60 deg = `gamma = 180 deg. `gamma = 180 deg - 130 deg `gamma = 50 deg. Now for the sides - knowing what the three angles are and knowing that c = 4, a = : sin alpha / a = sin`gamma / c sin70 deg / a = sin 50 deg / 4 a = 4(sin70 deg) / sin50 deg a is approx. 4.91 b = : sin `beta / b = sin `gamma/ c sin 60 deg / b = sin 50 deg / 4 b = 4(sin 60 deg) / sin 50 deg b is approx. 4.52 **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK **** Query problems 7.2.28 b = 4, c = 5, `beta = 40 deg
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the Law of Cosines we have: a^2 = 4^2 + 5^2 - 2*4*5*cos(40 deg) a^2 = 16 +25 - 40cos(40 deg) a^2 = 41 - 40cos(40 deg) a = `sqrt(41 - 40cos(40 deg)) or approx. 3.22 5^2 = (`sqrt(41 - 40cos(40 deg)))^2 + 16 - 2*(`sqrt(41 - 40cos(40 deg))*4*cos(C) 2*(`sqrt(41 - 40cos(40 deg))*4*cos(C) = (`sqrt(41 - 40cos(40 deg)))^2 +16 - 25 cos(C) = (`sqrt(41 - 40cos(40 deg)))^2 +16 - 25/2*(`sqrt(41 - 40cos(40 deg))*4 C = cos^-1((`sqrt(41 - 40cos(40 deg)))^2 +16 - 25/2*(`sqrt(41 - 40cos(40 deg))*4) or approx 86.9 deg A = 180 deg - 87 deg - 40 deg = approx. 53 deg confidence rating #$&*: OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** sin(`gamma) = .80. Thus `gamma = arcsin(.80) = 53 deg, approx., or 180 deg - 53 deg = 117 deg. Note that we have to consider both angles because the sine doesn't distinguish between the first and second quadrant, whereas the cosine (which is negative in the second quadrant) would. If `gamma = 53 deg then alpha would be 87 deg. In this case the Law of Sines tells us that a = sin(87 deg) * 4 / sin(40 deg) = 6.2, approx.. If `gamma = 117 deg then alpha would be 23 deg so that a = sin(23 deg) * 4 / sin(40 deg) = 2.8 or so. You should draw both triangles to see that both of these solutions are possible. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Does not only one solution fit the triangle? Therefore, we should use the Law of cosines. ------------------------------------------------ Self-critique Rating: 2
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11:57:48 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the previous problem the sides and angles are specified as follow: angle A is approx. 53 deg angle B is approx. 87 deg angle C = 40 deg side a is approx. 3.22 side b = 4 side c = 5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: if it is possible to draw the triangle or even if it isn't we can solve for a. so we can say alpha+beta + gamma=180 deg. so alpha + 40+.80=180 so alpha= 139.2 we can then find the value of a by saying sin 139.2/a= sin 40/4 which is 4 sin 139.2= a sin 40 deg so a= 4.07 and alpha=139.2
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11:57:48
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK **** Query problems 7.2.40 line-of-sight angles 15 deg and 35 deg with line directly to shore points are 3 miles apart .
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The line going direct to shore forms a 90 deg angle therefore we can calculate the other two angles as follows: 180 deg - 15 deg - 90 deg = 75 deg and 180 deg - 35 deg - 90 deg = 55 deg Now that we have all the angles and one side, we can get the other sides using the Law of Sines as follows: angle A = 15d deg + 35 deg = 50 deg angle B = 75 deg angle C = 55 deg side a = 3 mi sin(50 deg) / 3 mi = sin(75 deg) / b b = (3mi * sin(75 deg))/ sin(50 deg) or approx. 3.78 mi sin(50 deg) / 3 mi = sin(55 deg) / c c = (3mi * sin(75 deg))/ sin(50 deg) or approx.3.21 mi confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** First form two right triangles. The first is from ship to shore to lighthouse A. Angles are 15 deg, 90 deg and 75 deg. The second is from ship to shore to lighthouse B. Angles are 35 deg, 90 deg and 55 deg. Now form the triangle from ship to lighthouse A to lighthouse B. Let alpha be the angle formed at the ship. Then 'alpha = 50deg 'beta = 55deg 'gamma = 75deg a = 3mi (the separation of the lighthouses). distance to lighthouse A is the side b: Law of sines tells us that sin(50deg)/3 = sin(55deg)/b so b = 3sin(55deg)/sin(50deg) b = 3.21 mi. distance to light house B is side c: By Law of Sines c = 3(sin75deg)/sin(50deg) c = 3.78 mi distance to shore: Using first right triangle Theta = 15 Hypotenuse = distance to light house A = 3.21mi cos`theta = dist to shore / hypotenuse so dist to short = hypotenuse * cos(`theta) = 3.21 mi * cos(15 deg) = 3.1 mi. The same distance would be confirmed by solving the other right triangle. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!