course Mth 164 ڈِy{i|Wyassignment #008
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13:22:55 Query problem 7.1.B-10 c = 10 , `alpha = 40 deg, right triangle
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RESPONSE --> confidence assessment:
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13:24:35 what are a, b, and B?
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RESPONSE --> B = 90 deg - 40 deg = 50 deg sin 40 deg = a/10 =>10sin(40deg) = a => a is approx 6.4279 cos 40 deg = b/10 => 10cos(40deg) = b => b is approx 7.6604 confidence assessment: 3
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13:26:53 Explain how you obtained your solutions.
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RESPONSE --> Since a triangle has a total of 180 deg and 90 deg are used in the right angle this leaves the other two angles to equal 90 deg, a 40 deg angle is known so B = 90 deg - 40 deg = 50deg. a is the opposite side from angle 40 deg and the hypentuse is known to equal 10 so sin 40deg = a/10 which solves out to be about 6.4279. b is the adjacent side to angle 4o deg, with the known hypotenuse you use cos 40 deg = b/10 to solve to be about 7.6604. confidence assessment: 3
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13:27:04 Query problem 7.1.B-24 cliff height 100 feet, angle of elevation 25 deg. Dist of ship from shore.
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RESPONSE --> confidence assessment:
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13:27:32 what is the distance so shore?
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RESPONSE --> tan 25 deg = 100/b => 100/tan (25deg) = b => b is approx 214.4507ft. confidence assessment: 3
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13:28:56 Explain how you obtained your solution
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RESPONSE --> I positioned the triangle with a vertical leg of 100 ft, the cliff, which at the bottom, the horizon is the right angle. The horizontal leg, b, is what needs to be solved for. The angle of elevation at the right of the horizontal leg is 25 deg. I have the leg opposite and adjacent to the angle 25 deg so tangent is chosen, tan 25 deg = 100/b which solves out to be about 214.4507ft. confidence assessment: 3
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13:29:28 describe the triangle you used to find this distance and how you solved it
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RESPONSE --> The right triangle has legs 100 ft and b, the angle 25 deg is between leg b and the hypotenuse. confidence assessment: 3
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13:29:34 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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RESPONSE --> confidence assessment:
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13:30:05 What is the height of the tower?
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RESPONSE --> sin 25 deg = a/80 =>80sin(25deg) = a => a is approx 33.8095 ft. confidence assessment: 3
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13:31:17 Explain how you obtained your solution.
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RESPONSE --> The height, vertical leg a is what is being solved for. The hypotenuse, length of wire is 80ft and the angle between the hypotenuse and horizontal leg is 25 deg. So I have the opposite leg and hypotenuse from angle 25 deg so I use the sine function and set up sin 25 deg = a/80 which simplifies to about 33.8095ft. confidence assessment: 3
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13:31:31 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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RESPONSE --> confidence assessment:
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13:32:11 If the angle is `theta, as indicated, then how long is the ladder?
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RESPONSE --> The length of the ladder, expressed in terms of 'theta is sin^(-1)(a/L) where a is the opposite side from angle 'theta. confidence assessment: 2
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13:32:21 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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RESPONSE --> confidence assessment:
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13:33:44 how can we tell that the area of the triangle is a^2 sin(`theta) cos(`theta)?
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RESPONSE --> A = a^2 sin ('theta) cos ('theta) = a^2(opp/hyp)(adj/hyp) = a^2(h/a)(.5b/a) = a^2(.5bh/a^2) = .5bh confidence assessment: 2
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13:34:15 What is the altitude of the triangle?
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RESPONSE --> The altitude is unknown so I set it to be represented by h or it could be represented by sin 'theta = h/a confidence assessment: 3
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13:34:51 If the triangle is divided into two symmetric halves, then what is the length of a base of one of the halves? What therefore is the length of the base of the whole triangle?
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RESPONSE --> The base length of one of the halves would be half of the total base, so .5b. The length of the base of the whole triangle is b. confidence assessment: 3
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13:35:22 What therefore must be the area of the triangle?
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RESPONSE --> The area of the triangle is .5bh or as previously stated, a^2 sin('theta)cos('theta) confidence assessment: 3
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13:35:28 Query problems 7.2.12 `alpha = 70 deg; `beta = 60 deg, c = 4
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RESPONSE --> confidence assessment:
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13:37:32 specify the unknown sides and angles of your triangle.
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RESPONSE --> 'lamba = 180 deg - 70 deg - 60 deg = 50 deg sin (50)/4 = sin (70)/a =>(4sin(70))/sin(5) = a => a ~= 4.9067 sin (60)/b = sin(50)/4 =>(4sin(60))/sin(50) = b => b ~= 4.5221 confidence assessment: 3
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13:38:20 Explain how you obtained your results.
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RESPONSE --> The third angle is found from subtracting the known angles from 180 deg. Then a and b can be found because the 'lamda and c pair are known using the law of sines. confidence assessment: 3
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13:38:33 Query problems 7.2.28 b = 4, c = 5, `beta = 40 deg
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RESPONSE --> confidence assessment:
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13:40:34 specify the unknown sides and angles of your triangle.
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RESPONSE --> sin(40)/4 = sin 'lambda/5 => (5sin(40))/4 = sin 'lambda => 'lambda `+53.4641 'alpha = 180 - 40 - 53.4641 = 86.5359 deg sin(40)/4 = sin 86.5359/a => (4sin(86.5359))/sin (40) = a => a~6.2115 confidence assessment: 3
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13:41:56 Explain how you obtained your results.
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RESPONSE --> The beta and b pair are known so the law of sines can be set up to solve for lambda. Using the found value for lambda, alpha can be found by subtracting the beta dn lambda values from 180 deg. Set th elaw of sines up using the beta and b pair and the found alpha value to solve for a. confidence assessment: 3
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13:42:34 Query problems 7.2.40 line-of-sight angles 15 deg and 35 deg with line directly to shore
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RESPONSE --> I do not have this problem in the book, I have the seventh edition, and there is not enough information here to solve it. confidence assessment:
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13:42:38 how far is the ship from each Lighthouse and from the shore?
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RESPONSE --> confidence assessment:
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13:42:41 Explain a you obtained your results.
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RESPONSE --> confidence assessment:
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13:42:44 Comm on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> confidence assessment:
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