Find: αb wave direction breaking αo wave contours αb beach A) 8.9

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Find: αb wave direction breaking αo wave contours αb beach A) 8.9 Ho = 20 [ft] Find alpha b, the apporach angle of the wave when the wave breaks. [pause] In this problem we are provided, --- o T = 15 [s] o αo = 30 o o

Find: αb wave direction breaking αo wave contours αb beach A) 8.9 Ho = 20 [ft] the deepwater waveheight, H knot, the period of the wave, --- o T = 15 [s] o αo = 30 o o

Find: αb wave direction breaking αo wave contours αb beach A) 8.9 Ho = 20 [ft] T, and alpha knot, which is, --- o T = 15 [s] o αo = 30 o o

Find: αb wave direction breaking αo wave contours αb beach A) 8.9 Ho = 20 [ft] the deepwater approach angle. [pause] The problem asks to find alpha b, --- o T = 15 [s] o αo = 30 o o

Find: αb wave direction breaking αo wave contours αb beach A) 8.9 Ho = 20 [ft] which is the angle between the crestline of the wave and the ocean bottom contours, when the wave breaks. To find alpha b, --- o T = 15 [s] o αo = 30 o o

Find: αb = wave direction breaking αo wave contours αb beach sin (αb) sin (αo) = we’ll use the relationship which relates alpha b to alpha knot, L b and L knot. Where L b and L knot --- Lb Lo

Find: αb = wave direction breaking αo wave contours αb beach sin (αb) sin (αo) deep water = correspond to the wavelength for the breaking condition and deepwater condition, respectively. [pause] After solving for alpha b, --- wavelength Lb Lo wavelength at the breaking depth

Find: αb = wave direction breaking αo wave contours αb beach sin (αb) sin (αo) deep water = we can plug in the value of alpha knot, which is 30 degrees, --- wavelength Lb Lo Lb wavelength at the αb = sin-1 sin(αo) * Lo breaking depth

Find: αb = wave direction breaking αo wave contours αb beach sin (αb) sin (αo) deep water = and was provided in the problem statement. [pause] But we still have to find --- wavelength Lb Lo o 30 Lb wavelength at the αb = sin-1 sin(αo) * Lo breaking depth

Find: αb = wave direction breaking αo wave contours αb beach sin (αb) sin (αo) deep water = L b, and, L knot. [pause] L knot, equals, --- wavelength Lb Lo o 30 Lb wavelength at the αb = sin-1 sin(αo) * Lo breaking depth

Find: αb = wave direction breaking αo wave contours αb beach sin (αb) sin (αo) g * T2 = g times T squared, divided by 2 PI. [pause] Variable g represents the --- Lo = Lb Lo 2 * π o 30 Lb wavelength at the αb = sin-1 sin(αo) * Lo breaking depth

Find: αb = wave direction breaking αo wave contours αb beach 32.16 [ft/s2] sin (αb) sin (αo) g * T2 = gravitational acceleration constant, and variable T is the wave period, --- Lo = Lb Lo 2 * π o 30 Lb wavelength at the αb = sin-1 sin(αo) * Lo breaking depth

Find: αb = wave direction breaking αo wave contours αb beach 32.16 [ft/s2] T = 15 [s] sin (αb) sin (αo) g * T2 = which was provided as 15 seconds. [pause] This makes L knot, equal to, --- Lo = Lb Lo 2 * π o 30 Lb wavelength at the αb = sin-1 sin(αo) * Lo breaking depth

Find: αb = =1,151.645 [ft] wave direction breaking αo wave contours αb beach 32.16 [ft/s2] T = 15 [s] sin (αb) sin (αo) g * T2 = 1,151.645 feet. [pause] Lastly, we need to determine, L b, --- Lo = =1,151.645 [ft] Lb Lo 2 * π o 30 Lb wavelength at the αb = sin-1 sin(αo) * Lo breaking depth

Find: αb ? = =1,151.645 [ft] wave direction breaking αo wave contours beach 32.16 [ft/s2] T = 15 [s] sin (αb) sin (αo) g * T2 = the wavelength of the wave, when the wave breaks. [pause] We know we have found L b, when we select a wave depth, d, --- Lo = =1,151.645 [ft] Lb Lo 2 * π o 30 ? Lb wavelength at the αb = sin-1 sin(αo) * Lo breaking depth

Find: αb ? =1,151.645 [ft] wave direction breaking αo wave contours αb beach 32.16 [ft/s2] T = 15 [s] d=? g * T2 which results in a wave condition, where the actual height of the wave, H A, --- Lo = =1,151.645 [ft] 2 * π wave o 30 ? depth Lb αb = sin-1 sin(αo) * Lo

Find: αb ? = Lo=1,151.645 [ft] wave direction breaking αo wave contours actual wave wave height at height breaking depth Ha = Hb d=? equals the breaking height of the wave, H b. [pause] This is the same comparison, as if we happen to divide --- Lo=1,151.645 [ft] o wave 30 ? depth Lb αb = sin-1 sin(αo) * Lo

Find: αb ? = Lo=1,151.645 [ft] wave direction breaking αo wave contours actual wave wave height at height breaking depth Ha Hb d=? = both sides of the equation by the wavelength, L. [pause] To begin, we’ll arbitrarily choose a wave depth of --- Lo=1,151.645 [ft] L L o wave 30 ? wavelength depth Lb αb = sin-1 sin(αo) * Lo

Find: αb ? = Lo=1,151.645 [ft] wave direction breaking αo wave contours actual wave wave height at height breaking depth Ha Hb d=20 [ft] = 20 feet for our first iteration. [pause] Next, we’ll compute d over L knot, --- Lo=1,151.645 [ft] L L o wave 30 ? wavelength depth Lb αb = sin-1 sin(αo) * Lo

Find: αb ? = Lo=1,151.645 [ft] wave direction αo αb beach d Lo d=20 [ft] and after plugging the appropriate variables, d over L knot, --- Lo=1,151.645 [ft] o wave 30 ? depth Lb αb = sin-1 sin(αo) * Lo

Find: αb ? =0.01736 Lo=1,151.645 [ft] wave direction αo αb beach d Lo d=20 [ft] equals, 0.01736. [pause] This value of d over L knot corresponds to a d over L value of, --- Lo=1,151.645 [ft] o wave 30 ? depth Lb αb = sin-1 sin(αo) * Lo

Find: αb ? =0.05353 =0.01736 Lo=1,151.645 [ft] wave direction αo d L beach d =0.01736 Lo d=20 [ft] 0.05353. Therefore, the wavelength for this wave, --- Lo=1,151.645 [ft] o wave 30 ? depth Lb αb = sin-1 sin(αo) * Lo

Find: αb ? = 0.05353 =0.01736 Lo=1,151.645 [ft] wave direction αo d L beach d =0.01736 Lo d=20 [ft] at a depth of 20 feet, equals, -- Lo=1,151.645 [ft] o wave 30 ? depth Lb αb = sin-1 sin(αo) * Lo

Find: αb ? = 0.05353 = 373.622 [ft] =0.01736 Lo=1,151.645 [ft] wave direction αo d = 0.05353 L αb d L= = 373.622 [ft] d/L beach d =0.01736 Lo d=20 [ft] 373.622 feet. [pause] Also, by knowing the value of d over L knot, --- Lo=1,151.645 [ft] o wave 30 ? depth Lb αb = sin-1 sin(αo) * Lo

Find: αb ? = 0.05353 =0.01736 Lo=1,151.645 [ft] wave direction αo d L L= 373.622 [ft] beach d=20 [ft] d =0.01736 Lo we can interpolate to find the hyperbolic tangent of k d, and, --- Lo=1,151.645 [ft] o 30 ? Lb αb = sin-1 sin(αo) * Lo

Find: αb ? = 0.05353 =0.01736 Lo=1,151.645 [ft] wave direction αo d L L= 373.622 [ft] beach d=20 [ft] d =0.01736 Lo the shoaling coeffcient, H over H knot prime, where the variable H, --- Lo=1,151.645 [ft] o tanh (k*d) = 0.3241 30 ? Lb H/Ho = 1.2652 ‘ αb = sin-1 sin(αo) * Lo

Find: αb ? = 0.05353 =0.01736 Lo=1,151.645 [ft] wave direction αo d L L= 373.622 [ft] beach d=20 [ft] d =0.01736 Lo represents the actual height of the wave, H A, at the tested depth of, --- Lo=1,151.645 [ft] o tanh (k*d) = 0.3241 30 ? Lb H/Ho = 1.2652 ‘ αb = sin-1 sin(αo) * Lo

Find: αb ? = 0.05353 =0.01736 Lo=1,151.645 [ft] wave direction αo d L L= 373.622 [ft] beach d=20 [ft] d =0.01736 Lo 20 feet, and that variable H, is what we’ll use to compare the actual wave, H a, --- Lo=1,151.645 [ft] o tanh (k*d) = 0.3241 30 ? Lb H/Ho = 1.2652 ‘ αb = sin-1 sin(αo) * Lo

Find: αb ? = 0.05353 = =0.01736 Lo=1,151.645 [ft] actual wave d height Ha Hb L= 373.622 [ft] = L L d=20 [ft] d =0.01736 Lo with the wave at its breaking condition, H b. [pause] The right hand side of this equation, H b over L, --- Lo=1,151.645 [ft] o tanh (k*d) = 0.3241 30 ? Lb H/Ho = 1.2652 ‘ αb = sin-1 sin(αo) * Lo

Find: αb = 0.05353 = =0.01736 = 0.142 * tanh (k*d) actual wave d height L Ha Hb L= 373.622 [ft] = L L d=20 [ft] d =0.01736 Hb Lo equals, 0.142 times the hyperbolic tangent of k d, --- = 0.142 * tanh (k*d) L tanh (k*d) = 0.3241 H/Ho = 1.2652 ‘

Find: αb = 0.05353 = = 0.0460 =0.01736 = 0.142 * tanh (k*d) actual wave d = 0.05353 height L Ha Hb L= 373.622 [ft] = L L d=20 [ft] Hb = 0.0460 d L =0.01736 Hb Lo which equals, 0.0460. [pause] Solving for the actual wave height, H a, --- = 0.142 * tanh (k*d) L tanh (k*d) = 0.3241 H/Ho = 1.2652 ‘

Find: αb = 0.05353 = = 0.0460 =0.01736 = 0.142 * tanh (k*d) actual wave d = 0.05353 height L Ha Hb L= 373.622 [ft] = L L d=20 [ft] Hb = 0.0460 d L =0.01736 Hb Lo we must first compute the approach angle the wave would make, --- = 0.142 * tanh (k*d) L tanh (k*d) = 0.3241 H/Ho = 1.2652 ‘

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L d=20 [ft] L Lo=1,151.645 [ft] α = sin-1 sin(αo) * Lo tanh (k*d) = 0.3241 at a depth of 20 feet. And after plugging in the known variables of, --- H/Ho = 1.2652 ‘

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L o 30 d=20 [ft] L Lo=1,151.645 [ft] α = sin-1 sin(αo) * Lo tanh (k*d) = 0.3241 L, L knot, and alpha knot, the crestline of the wave would be ---- H/Ho = 1.2652 ‘

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L o 30 d=20 [ft] L Lo=1,151.645 [ft] α = sin-1 sin(αo) * Lo tanh (k*d) = 0.3241 9.319 degrees offset from the ocean bottom contours. With the value of alpha, we can compute --- α = 9.319 o H/Ho = 1.2652 ‘

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L o 30 d=20 [ft] L Lo=1,151.645 [ft] α = sin-1 sin(αo) * Lo tanh (k*d) = 0.3241 the wave refraction index, K r, which equals, --- α = 9.319 o H/Ho = 1.2652 ‘ cos (αo) KR = cos (α)

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L o 30 d=20 [ft] L Lo=1,151.645 [ft] α = sin-1 sin(αo) * Lo tanh (k*d) = 0.3241 0.9368. [pause] After that, we’ll compute the unrefracted deepwater wave height, --- α = 9.319 o H/Ho = 1.2652 ‘ o cos (αo) 30 KR = cos (α) KR =0.9368

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L d=20 [ft] Lo=1,151.645 [ft] Ho = KR * Ho ‘ tanh (k*d) = 0.3241 H knot prime, by multipliying the refraction coefficient by the deepwater wave height. We just computed --- unrefracted H/Ho = 1.2652 ‘ deepwater α = 9.319 o wave height KR =0.9368

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L d=20 [ft] Lo=1,151.645 [ft] Ho = KR * Ho ‘ tanh (k*d) = 0.3241 the refraction coefficient, K r, and the problem statement provided the deepwater --- unrefracted H/Ho = 1.2652 ‘ deepwater α = 9.319 o wave height KR =0.9368

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L d=20 [ft] Ho = 20 [ft] Lo=1,151.645 [ft] Ho = KR * Ho ‘ tanh (k*d) = 0.3241 wave height as 20 feet. This makes the unrefracted deepwater wave height, equal to, --- unrefracted H/Ho = 1.2652 ‘ deepwater α = 9.319 o wave height KR =0.9368

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L d=20 [ft] Ho = 20 [ft] Lo=1,151.645 [ft] Ho = KR * Ho ‘ tanh (k*d) = 0.3241 18.736 feet. [pause] Now, to find the actual height of the wave at this depth, --- Ho = 18.736 [ft] ‘ H/Ho = 1.2652 ‘ α = 9.319 o KR =0.9368

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L d=20 [ft] Ho = 20 [ft] Lo=1,151.645 [ft] Ho = KR * Ho ‘ tanh (k*d) = 0.3241 we’ll multiply H knot prime by the shoaling coefficient, and the actual height, equals, --- Ho = 18.736 [ft] ‘ H/Ho = 1.2652 ‘ α = 9.319 o H = Ho * (H/Ho ) ‘ ‘ KR =0.9368

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L d=20 [ft] Ho = 20 [ft] Lo=1,151.645 [ft] Ho = KR * Ho ‘ tanh (k*d) = 0.3241 23.705 feet. [pause] When divided by the wavelength, L, --- Ho = 18.736 [ft] ‘ H/Ho = 1.2652 ‘ α = 9.319 o H = Ho * (H/Ho ) ‘ ‘ KR =0.9368 H = 23.705 [ft]

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L d=20 [ft] Ha Lo=1,151.645 [ft] L tanh (k*d) = 0.3241 the quotient, H a over L, equals, -- Ho = 18.736 [ft] ‘ H/Ho = 1.2652 ‘ α = 9.319 o H = Ho * (H/Ho ) ‘ ‘ KR =0.9368 H = 23.705 [ft]

Find: αb ? = 0.05353 = Lo=1,151.645 [ft] = 0.0634 actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] = L L d=20 [ft] Ha Lo=1,151.645 [ft] = 0.0634 L tanh (k*d) = 0.3241 0.0634. [pause] Since the actual height of the wave H a, --- Ho = 18.736 [ft] ‘ H/Ho = 1.2652 ‘ α = 9.319 o H = Ho * (H/Ho ) ‘ ‘ KR =0.9368 H = 23.705 [ft]

Find: αb = 0.05353 > Lo=1,151.645 [ft] = 0.0634 actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] > L L d=20 [ft] Ha Lo=1,151.645 [ft] = 0.0634 L tanh (k*d) = 0.3241 is greater than the breaking height of the wave, then, this wave does not exist, because, it broke, at a depth, --- Ho = 18.736 [ft] ‘ H/Ho = 1.2652 ‘ α = 9.319 o H = Ho * (H/Ho ) ‘ ‘ KR =0.9368 H = 23.705 [ft]

Find: αb = 0.05353 > Lo=1,151.645 [ft] = 0.0634 actual wave d height 0.0460 L Ha Hb L= 373.622 [ft] > L L d=20 [ft] d < db Ha Lo=1,151.645 [ft] = 0.0634 L tanh (k*d) = 0.3241 deeper than 20 feet. [pause] For our second iteration, we’ll use a depth based on --- Ho = 18.736 [ft] ‘ H/Ho = 1.2652 ‘ α = 9.319 o H = Ho * (H/Ho ) ‘ ‘ KR =0.9368 H = 23.705 [ft]

Find: αb = 0.05353 > Lo=1,151.645 [ft] d 0.0634 0.0460 L Ha Hb L= 373.622 [ft] > L L d=20 [ft] d < db ( HA,i / Li ) di+1= di Lo=1,151.645 [ft] * ( Hb,i / Li ) tanh (k*d) = 0.3241 our previous depth, and variables H a, and, H b. [pause] After plugging in --- Ho = 18.736 [ft] ‘ H/Ho = 1.2652 ‘ α = 9.319 o H = Ho * (H/Ho ) ‘ ‘ KR =0.9368 H = 23.705 [ft]

Find: αb = 0.05353 > Lo=1,151.645 [ft] d 0.0634 0.0460 L Ha Hb L= 373.622 [ft] > L L d=20 [ft] ( HA,i / Li ) di+1= di Lo=1,151.645 [ft] * ( Hb,i / Li ) tanh (k*d) = 0.3241 the appropriate variables, our next depth to test will be, --- Ho = 18.736 [ft] ‘ H/Ho = 1.2652 ‘ α = 9.319 o H = Ho * (H/Ho ) ‘ ‘ KR =0.9368 H = 23.705 [ft]

Find: αb = 0.05353 > Lo=1,151.645 [ft] d 0.0634 0.0460 L Ha Hb L= 373.622 [ft] > L L d=20 [ft] ( HA,i / Li ) d2= d1 Lo=1,151.645 [ft] * ( Hb,i / Li ) tanh (k*d) = 0.3241 27.58 feet. [pause] When using the depth of 27.58 feet, --- d2= 27.58 [ft] H/Ho = 1.2652 ‘ α = 9.319 o KR =0.9368

Find: αb ? = ? = Lo=1,151.645 [ft] d ? ? L Ha Hb L= ? [ft] L L d=27.58 [ft] ( HA,i / Li ) Lo=1,151.645 [ft] di+1= di * ( Hb,i / Li ) tanh (k*d) = ? and performing the same calculations as the first iteration, --- H/Ho = ? ‘ α = ? o KR =?

Find: αb = 0.06333 < Lo=1,151.645 [ft] d 0.0508 0.0537 L Ha Hb L= 435.497 [ft] < L L d=27.58 [ft] ( HA,i / Li ) Lo=1,151.645 [ft] di+1= di * ( Hb,i / Li ) tanh (k*d) = 0.3782 we compute the values shown here. At this depth, we notice the actual height of the wave, --- H/Ho = 1.1786 ‘ α = 10.899 o KR =0.9391

Find: αb = 0.06333 < Lo=1,151.645 [ft] d 0.0508 0.0537 L Ha Hb L= 435.497 [ft] < L L d=27.58 [ft] ( HA,i / Li ) Lo=1,151.645 [ft] di+1= di * ( Hb,i / Li ) tanh (k*d) = 0.3782 H a, is less than the breaking height of the wave, H b. Therefore, our subsequent depth, --- H/Ho = 1.1786 ‘ α = 10.899 o KR =0.9391

Find: αb = 0.06333 < Lo=1,151.645 [ft] d 0.0508 0.0537 L Ha Hb L= 435.497 [ft] < L L d=27.58 [ft] ( HA,i / Li ) Lo=1,151.645 [ft] d3= d2 * ( Hb,i / Li ) tanh (k*d) = 0.3782 d 3, will be shallower than 27.58 feet. For our next depth value we’ll try --- H/Ho = 1.1786 ‘ α = 10.899 o KR =0.9391

Find: αb = 0.06333 < Lo=1,151.645 [ft] d 0.0508 0.0537 L Ha Hb L= 435.497 [ft] < L L d=27.58 [ft] ( HA,i / Li ) Lo=1,151.645 [ft] d3= d2 * ( Hb,i / Li ) tanh (k*d) = 0.3782 26.09 feet. [pause] After running through a third iteration, --- d3= 26.09 [ft] H/Ho = 1.1786 ‘ α = 10.899 o KR =0.9391

Find: αb ? = ? = Lo=1,151.645 [ft] d ? ? L Ha Hb L= ? [ft] L L d=26.09 [ft] ( HA,i / Li ) Lo=1,151.645 [ft] di+1= di * ( Hb,i / Li ) tanh (k*d) = ? for this new depth value, we again calculate out all the values, ---- H/Ho = ? ‘ α = ? o KR =?

Find: αb = 0.06150 < Lo=1,151.645 [ft] d 0.05280 0.05338 L Ha Hb L= 424.288 [ft] < L L d=26.09 [ft] ( HA,i / Li ) Lo=1,151.645 [ft] di+1= di * ( Hb,i / Li ) tanh (k*d) = 0.3759 as shown. [pause] As the actual wave height approaches the breaking wave height, --- H/Ho = 1.1932 ‘ α = 10.613 o KR =0.9387

Find: αb = 0.06150 < Lo=1,151.645 [ft] d 0.05280 0.05338 L Ha Hb L= 424.288 [ft] < L L d=26.09 [ft] ( HA,i / Li ) Lo=1,151.645 [ft] di+1= di * ( Hb,i / Li ) tanh (k*d) = 0.3759 the actual approach angle converges on the approach angle for the breaking wave, alpha b. [pause] The approach angle for the first 3 iterations are, --- H/Ho = 1.1932 ‘ α = 10.613 o KR =0.9387

Find: αb = 0.06150 < Lo=1,151.645 [ft] d 0.05280 0.05338 L Ha Hb L= 424.288 [ft] < L L d=26.09 [ft] ( HA,i / Li ) Lo=1,151.645 [ft] di+1= di * ( Hb,i / Li ) tanh (k*d) = 0.3759 9.319 degrees, 10.899 degrees, and 10.613 degrees. Where 10.613 degrees --- α1 = 9.319 o H/Ho = 1.1932 ‘ α2 = 10.899 o α = 10.613 o α3 = 10.613 o KR =0.9387

Find: αb = 0.06150 < Lo=1,151.645 [ft] d 0.05280 0.05338 L Ha Hb L= 424.288 [ft] < L L d=26.09 [ft] ( HA,i / Li ) Lo=1,151.645 [ft] di+1= di * ( Hb,i / Li ) tanh (k*d) = 0.3759 is the most accurate approximation for alpha b, because this value in our algorithm converges to alpha b. [pause] α1 = 9.319 o H/Ho = 1.1932 ‘ α2 = 10.899 o α = 10.613 o α3 = 10.613 o KR =0.9387

Find: αb = 0.06150 Lo=1,151.645 [ft] d A) 8.9 B) 10.5 L C) 12.1 L= 424.288 [ft] o d=26.09 [ft] Lo=1,151.645 [ft] tanh (k*d) = 0.3759 When reviewing the possible solutions, --- α1 = 9.319 o H/Ho = 1.1932 ‘ α2 = 10.899 o α = 10.613 o α3 = 10.613 o KR =0.9387

Find: αb = 0.06150 Lo=1,151.645 [ft] d A) 8.9 B) 10.5 L C) 12.1 L= 424.288 [ft] o d=26.09 [ft] Lo=1,151.645 [ft] answerB tanh (k*d) = 0.3759 the answer is B. [fin] α1 = 9.319 o H/Ho = 1.1932 ‘ α2 = 10.899 o α = 10.613 o α3 = 10.613 o KR =0.9387