#$&* course mth 164 10/29 5:45pm SOLUTIONS/COMMENTARY ON QUERY 8**** Query problem 7.1.B-10 c = 10 , alpha = 40 deg, right triangle
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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): ------------------------------------------------ Self-critique Rating: **** 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: a = tan 65 deg. a/100 which means 100tan(65 deg) = 214.45 ft.
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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): The way I got my answer is different than the way you have it, will the way I did it work ? ------------------------------------------------ Self-critique Rating:2 **** 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: b = sin 25 deg = b/80, 80sin(25 deg) = 33.81 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): ------------------------------------------------ Self-critique Rating: **** 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: a^2 + b^2 = c^2 Sqrt4^2 + sqrt3^2 = sqrt25 = 5, c = 5 B = arctan(3/4) = 36.87 deg.
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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): I don’t understand what the answer means, where did I go wrong.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t know how to answer this question. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If you divide the triangle by its axis of symmetry you get two congruent right triangles, each with angle `theta opposite the altitude and adjacent to the base. The side a makes up the hypotenuse of either of these triangles. The altitude of each is therefore a sin(`theta) and the base is a cos(`theta). The area of each triangle is thus 1/2 * base * height = 1/2 a sin(`theta) a cos(`theta) = 1/2 a^2 sin(`theta) cos(`theta). The areas of the two right triangles add up to the area of the isosceles triangle. This area is therefore 2 ( 1/2 a^2 sin(`theta) cos(`theta) ) = a^2 sin(`theta) cos(`theta). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I still don’t understand how you arrive at this answer.
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11:48:52 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A+B+C = 180 70deg + 60deg = 130deg that makes C = 50deg a = sin 50deg / 4 = sin 70deg /a a = 4sin(70deg) / sin(50deg) a = 4.91 b = sin 50deg/4 = sin 60deg/b b = 4sin(60deg) / sin(50deg) b = 4.52 confidence rating #$&*: 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): ------------------------------------------------ Self-critique Rating: **** Query problems 7.2.28 b = 4, c = 5, `beta = 40 deg
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5sin(40deg)/4 = .8 so C = .8 arcsin(.8) = 53deg, so C = 53deg 53 + 40 = 93, 180-93 = 87 A= 87deg confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** sin(`gamma) = .80. Thus `gamma = arcsin(.80) = 53 deg, approx., or 180 deg - 53 deg = 117 deg. Note that we have to consider both angles because the sine doesn't distinguish between the first and second quadrant, whereas the cosine (which is negative in the second quadrant) would. If `gamma = 53 deg then alpha would be 87 deg. In this case the Law of Sines tells us that a = sin(87 deg) * 4 / sin(40 deg) = 6.2, approx.. If `gamma = 117 deg then alpha would be 23 deg so that a = sin(23 deg) * 4 / sin(40 deg) = 2.8 or so. You should draw both triangles to see that both of these solutions are possible. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** specify the unknown sides and angles of your triangle.
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11:57:48 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: **** 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: Use two triangles, the 1st being from the first lighthouse to the ship then back to shore then up to the lighthouse. Which would have the angles 90deg + 15deg + 75deg = 180deg. The 2nd being from the 2nd lighthouse to the ship then back to shore then to the lighthouse. Which would have the angles 90deg + 35deg + 55deg = 180deg. Then combine the two triangles together putting the two 90deg sides against each other. The angle at the ship would be 50deg which will be A. The angle at the 1st lighthouse would be 75deg which will be C. The angle at the 2nd lighthouse would be 55deg which will be B. The distance between lighthouses is 3 miles. So a = 3 So to find the distance between the ship and the 1st lighthouse we have to solve for b. Sin A/ a = sin B/ b, sin(50deg)/3 = sin(55deg)/b, 3sin(55deg)/sin(50deg) = 3.21 So the distance between the two is 3.21 miles. To find the distance between the ship and the 2nd lighthouse we have to solve for c. Sin A/ a = sin C/c, sin(50deg)/3 = sin(75deg)/c, 3sin(75deg)/sin(50deg) = 3.78 So the distance between the two is 3.78 miles. Kind of stuck on the distance from shore. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Did not understand how to get the distance from shore answer. ------------------------------------------------ Self-critique Rating:1 "