💡Okay, so I have this problem here where the motion equation of an object is given as s = 3 + t². I need to find the average velocity of the object during the time interval [2, 2.1]. Alright, let me think about how to approach this.First, I remember that average velocity is different from average speed. Velocity is a vector quantity, which means it has both magnitude and direction, while speed is just the scalar quantity representing how fast something is moving. But in this case, since we're dealing with one-dimensional motion (I assume because the equation is in terms of t² and not something more complex), average velocity should be straightforward.I think average velocity is calculated by the change in position divided by the change in time. So, if I can find the position of the object at the start and end of the time interval, I can subtract them to get the change in position, and then divide by the change in time to get the average velocity.Let me write that down to make it clearer. The formula for average velocity (v_avg) should be:v_avg = (s_final - s_initial) / (t_final - t_initial)In this problem, the time interval is from t = 2 to t = 2.1. So, t_initial is 2 and t_final is 2.1. I need to find the position of the object at these two times.The position function is given by s = 3 + t². So, let me calculate s at t = 2 and t = 2.1.First, at t = 2:s_initial = 3 + (2)² = 3 + 4 = 7Okay, so at t = 2, the position s is 7 units.Next, at t = 2.1:s_final = 3 + (2.1)²Hmm, I need to calculate (2.1)². Let me do that step by step. 2.1 times 2.1 is... Well, 2 times 2 is 4, 2 times 0.1 is 0.2, 0.1 times 2 is another 0.2, and 0.1 times 0.1 is 0.01. Adding those up: 4 + 0.2 + 0.2 + 0.01 = 4.41.So, (2.1)² = 4.41. Therefore, s_final = 3 + 4.41 = 7.41Alright, so at t = 2.1, the position s is 7.41 units.Now, I can plug these values into the average velocity formula:v_avg = (7.41 - 7) / (2.1 - 2)Calculating the numerator first: 7.41 - 7 = 0.41Denominator: 2.1 - 2 = 0.1So, v_avg = 0.41 / 0.1Dividing 0.41 by 0.1 is the same as multiplying 0.41 by 10, which gives 4.1.Therefore, the average velocity of the object during the time interval [2, 2.1] is 4.1 units per time unit.Wait, let me double-check my calculations to make sure I didn't make a mistake. First, s at t=2: 3 + 4 = 7. That seems right.s at t=2.1: 3 + (2.1)^2. I calculated (2.1)^2 as 4.41, so 3 + 4.41 is 7.41. That looks correct.Then, the change in position is 7.41 - 7 = 0.41, and the change in time is 0.1. Dividing 0.41 by 0.1 gives 4.1. Yep, that seems correct.I think I did everything right. So, the average velocity is 4.1.

💡First, identify the initial number of apples Ruth has, which is 89.Next, determine how many apples Ruth shares with Peter, which is 5.To find out how many apples Ruth has left after sharing, subtract the number of apples she gives away from her initial amount.Perform the subtraction: 89 minus 5 equals 84.Therefore, Ruth ends with 84 apples.

💡First, I need to determine how far the rat has traveled during its 6-hour head start. Since the rat's speed is 36 km/h, the distance covered in 6 hours is 36 multiplied by 6, which equals 216 km.Next, I'll calculate the relative speed between the cat and the rat. The cat's speed is 90 km/h, and the rat's speed is 36 km/h. The relative speed is the difference between these two speeds, so 90 km/h minus 36 km/h equals 54 km/h.Finally, to find out how long it takes for the cat to catch up to the rat, I'll divide the distance the rat has traveled by the relative speed. That is, 216 km divided by 54 km/h equals 4 hours.

💡First, we note the initial composition of the box: 4 red balls and 3 white balls, making a total of 7 balls.Given that the first ball drawn is red, we update the composition of the box. After removing one red ball, there are now 3 red balls and 3 white balls left, totaling 6 balls.To find the probability that the second ball is also red, we consider the remaining red balls and the total remaining balls:[text{Probability} = frac{text{Number of remaining red balls}}{text{Total number of remaining balls}} = frac{3}{6} = frac{1}{2}]

💡First, I need to determine the measure of the angle that is 30% larger than a right angle. A right angle is 90 degrees, so 30% of 90 degrees is 27 degrees. Adding this to 90 degrees gives an angle of 117 degrees.Next, since the triangle is isosceles and obtuse, it has two equal angles. The sum of all angles in a triangle is 180 degrees. Subtracting the obtuse angle of 117 degrees from 180 degrees leaves 63 degrees for the two equal angles.Finally, to find the measure of one of the two smallest angles, I divide 63 degrees by 2, which equals 31.5 degrees.

💡Alright, so I have this triangle ABC where angle B is 45 degrees and angle C is 60 degrees. I need to find the ratio of sin(angle BAD) to sin(angle CAD), where point D divides BC in the ratio 3:1. Hmm, okay, let me try to visualize this.First, I know that in any triangle, the sum of the angles is 180 degrees. So, angle A must be 180 - 45 - 60, which is 75 degrees. So, angle BAC is 75 degrees. That makes sense.Now, point D is on BC such that BD:DC is 3:1. So, if I let BD be 3x, then DC would be x, making the entire length of BC equal to 4x. I don't know the actual lengths, but maybe I can assign some variables or use the Law of Sines or Cosines to find relationships between the sides.Since I need to find the ratio of sines of two angles, BAD and CAD, maybe I can use the Law of Sines in triangles ABD and ACD. Let me recall the Law of Sines: in any triangle, the ratio of a side to the sine of its opposite angle is constant. So, in triangle ABD, I can write something like BD/sin(angle BAD) = AD/sin(angle ABD). Similarly, in triangle ACD, CD/sin(angle CAD) = AD/sin(angle ACD).Wait, angle ABD is angle at B, which is 45 degrees, and angle ACD is angle at C, which is 60 degrees. So, in triangle ABD, BD/sin(angle BAD) = AD/sin(45°), and in triangle ACD, CD/sin(angle CAD) = AD/sin(60°). That seems right.Let me write that down:For triangle ABD:BD / sin(angle BAD) = AD / sin(45°)So, sin(angle BAD) = (BD * sin(45°)) / ADSimilarly, for triangle ACD:CD / sin(angle CAD) = AD / sin(60°)So, sin(angle CAD) = (CD * sin(60°)) / ADNow, I can take the ratio of sin(angle BAD) to sin(angle CAD):[sin(angle BAD)] / [sin(angle CAD)] = [(BD * sin(45°)) / AD] / [(CD * sin(60°)) / AD]Oh, the AD cancels out, which is nice. So, it simplifies to:(BD / CD) * (sin(45°) / sin(60°))We know BD:CD is 3:1, so BD/CD is 3. Now, sin(45°) is √2/2 and sin(60°) is √3/2. So, plugging those in:3 * (√2/2) / (√3/2) = 3 * (√2/2) * (2/√3) = 3 * (√2 / √3)Simplify that, √2/√3 is the same as √(2/3), which can be rationalized as √6/3. So, 3 * (√6/3) = √6.Wait, that doesn't seem right. Let me check my steps again.Wait, when I took the ratio, I had (BD/CD) * (sin45/sin60). BD/CD is 3, sin45 is √2/2, sin60 is √3/2. So, 3 * (√2/2) / (√3/2) = 3 * (√2/2) * (2/√3) = 3 * √2 / √3. Then, √2/√3 is √6/3, so 3 * √6/3 is √6.Hmm, but in the initial problem, the angles at B and C are 45 and 60, so maybe I need to consider the sides opposite these angles. Wait, maybe I should use the Law of Sines on the entire triangle ABC first to find the ratios of the sides.In triangle ABC, by Law of Sines:AB / sin(C) = BC / sin(A) = AC / sin(B)So, AB / sin(60°) = BC / sin(75°) = AC / sin(45°)Let me assign some lengths. Let me let BC = 4 units, so BD = 3 and DC = 1. Then, using the Law of Sines:AB / sin(60°) = 4 / sin(75°)So, AB = (4 * sin(60°)) / sin(75°)Similarly, AC = (4 * sin(45°)) / sin(75°)Compute sin(75°), which is sin(45° + 30°) = sin45*cos30 + cos45*sin30 = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4So, sin75° = (√6 + √2)/4Therefore, AB = (4 * sin60°) / sin75° = (4 * √3/2) / [(√6 + √2)/4] = (2√3) / [(√6 + √2)/4] = 2√3 * (4 / (√6 + √2)) = 8√3 / (√6 + √2)Similarly, AC = (4 * sin45°) / sin75° = (4 * √2/2) / [(√6 + √2)/4] = (2√2) / [(√6 + √2)/4] = 2√2 * (4 / (√6 + √2)) = 8√2 / (√6 + √2)Hmm, okay, so AB and AC are expressed in terms of √6 and √2. Maybe I can rationalize the denominators.Let me rationalize AB:8√3 / (√6 + √2) = 8√3 * (√6 - √2) / [(√6 + √2)(√6 - √2)] = 8√3*(√6 - √2) / (6 - 2) = 8√3*(√6 - √2)/4 = 2√3*(√6 - √2)Simplify √3*√6 = √18 = 3√2, and √3*√2 = √6So, AB = 2*(3√2 - √6) = 6√2 - 2√6Similarly, AC = 8√2 / (√6 + √2) = 8√2*(√6 - √2)/4 = 2√2*(√6 - √2) = 2*(√12 - √4) = 2*(2√3 - 2) = 4√3 - 4Wait, let me double-check that:AC = 8√2*(√6 - √2)/4 = 2√2*(√6 - √2) = 2*(√12 - √4) = 2*(2√3 - 2) = 4√3 - 4. Yeah, that's correct.So, AB = 6√2 - 2√6 and AC = 4√3 - 4.Now, going back to triangles ABD and ACD.In triangle ABD, sides are AB, BD, and AD.In triangle ACD, sides are AC, CD, and AD.We can use the Law of Sines in both triangles to relate the angles.In triangle ABD:BD / sin(angle BAD) = AD / sin(angle ABD)Angle ABD is angle at B, which is 45 degrees.So, BD / sin(angle BAD) = AD / sin(45°)Similarly, in triangle ACD:CD / sin(angle CAD) = AD / sin(angle ACD)Angle ACD is angle at C, which is 60 degrees.So, CD / sin(angle CAD) = AD / sin(60°)We can write:sin(angle BAD) = (BD * sin(45°)) / ADsin(angle CAD) = (CD * sin(60°)) / ADTherefore, the ratio sin(angle BAD)/sin(angle CAD) = (BD/CD) * (sin45/sin60)We know BD/CD = 3/1 = 3sin45 = √2/2, sin60 = √3/2So, ratio = 3 * (√2/2) / (√3/2) = 3 * √2 / √3 = 3√6 / 3 = √6Wait, that's different from my initial thought. Hmm, but earlier I thought it was 3√3/2, but now I'm getting √6.Wait, maybe I made a mistake in the initial approach. Let me check again.Wait, in the initial approach, I didn't consider the entire triangle ABC and just used the Law of Sines on ABD and ACD directly, which gave me √6. But in the second approach, I went through the entire triangle and still got √6.Wait, but in the initial problem statement, the user had a different approach which led to 3√3/2. So, which one is correct?Wait, maybe I need to use the Law of Sines in a different way or consider the areas or use Ceva's theorem or something else.Alternatively, maybe using coordinates would help. Let me try that.Let me place point B at (0,0), point C at (4,0), since BC is 4 units. Then, point D is at (3,0).Now, I need to find coordinates of point A such that angle at B is 45 degrees and angle at C is 60 degrees.Wait, but if I place B at (0,0) and C at (4,0), then the coordinates of A can be found using the angles.Let me denote point A as (x,y). Then, the angle at B is 45 degrees, so the slope of BA is tan(45) = 1, so y = x.Similarly, the angle at C is 60 degrees, so the slope of CA is tan(180 - 60) = tan(120) = -√3. So, the line from C to A has slope -√3.So, the line from C(4,0) to A(x,y) has slope (y - 0)/(x - 4) = -√3, so y = -√3(x - 4)But we also have y = x from the angle at B.So, setting x = -√3(x - 4)x = -√3 x + 4√3x + √3 x = 4√3x(1 + √3) = 4√3x = (4√3)/(1 + √3)Rationalizing the denominator:x = (4√3)(1 - √3)/(1 - 3) = (4√3 - 12)/(-2) = (12 - 4√3)/2 = 6 - 2√3So, x = 6 - 2√3, and y = x = 6 - 2√3So, point A is at (6 - 2√3, 6 - 2√3)Now, point D is at (3,0). So, we can find vectors BA and DA to find angles BAD and CAD.Wait, maybe it's easier to compute the coordinates and then find the angles using vectors or slopes.Alternatively, compute the lengths AD, BD, CD, and use the Law of Sines.Wait, let's compute AD.Point A is at (6 - 2√3, 6 - 2√3), point D is at (3,0).So, AD distance is sqrt[(6 - 2√3 - 3)^2 + (6 - 2√3 - 0)^2] = sqrt[(3 - 2√3)^2 + (6 - 2√3)^2]Compute (3 - 2√3)^2 = 9 - 12√3 + 12 = 21 - 12√3(6 - 2√3)^2 = 36 - 24√3 + 12 = 48 - 24√3So, AD^2 = (21 - 12√3) + (48 - 24√3) = 69 - 36√3So, AD = sqrt(69 - 36√3)Hmm, that's a bit messy, but maybe we can simplify it.Wait, 69 - 36√3 can be written as (a - b√3)^2 = a^2 - 2ab√3 + 3b^2So, set a^2 + 3b^2 = 69 and 2ab = 36, so ab = 18Looking for integers a and b such that ab=18 and a^2 + 3b^2=69Possible pairs: a=6, b=3: 6*3=18, 6^2 + 3*3^2=36 + 27=63≠69a=9, b=2: 9*2=18, 81 + 12=93≠69a=3, b=6: same as firsta=18, b=1: 324 + 3=327≠69Hmm, maybe not integers. Maybe a=√(something). Alternatively, maybe it's not a perfect square, so we can leave it as sqrt(69 - 36√3).Alternatively, maybe I can compute sin(angle BAD) and sin(angle CAD) using coordinates.Point A is (6 - 2√3, 6 - 2√3), point B is (0,0), point D is (3,0), point C is (4,0).So, vector BA is from B to A: (6 - 2√3, 6 - 2√3)Vector DA is from D to A: (6 - 2√3 - 3, 6 - 2√3 - 0) = (3 - 2√3, 6 - 2√3)Similarly, vector CA is from C to A: (6 - 2√3 - 4, 6 - 2√3 - 0) = (2 - 2√3, 6 - 2√3)Now, angle BAD is the angle between BA and DA.Similarly, angle CAD is the angle between CA and DA.We can use the dot product formula to find the cosines of these angles, and then find the sines.But since we need the ratio of sines, maybe we can find the ratio of the sines using the areas or something else.Alternatively, since we have coordinates, maybe we can compute the slopes and then the angles.Wait, let's compute the slopes.Slope of BA: from B(0,0) to A(6 - 2√3, 6 - 2√3). The slope is (6 - 2√3 - 0)/(6 - 2√3 - 0) = 1. So, angle of BA with x-axis is 45 degrees, which makes sense since angle at B is 45 degrees.Slope of DA: from D(3,0) to A(6 - 2√3, 6 - 2√3). The slope is (6 - 2√3 - 0)/(6 - 2√3 - 3) = (6 - 2√3)/(3 - 2√3)Let me simplify that:(6 - 2√3)/(3 - 2√3) = [2*(3 - √3)] / (3 - 2√3)Hmm, maybe multiply numerator and denominator by (3 + 2√3):[2*(3 - √3)(3 + 2√3)] / [(3 - 2√3)(3 + 2√3)] = [2*(9 + 6√3 - 3√3 - 2*3)] / (9 - 12) = [2*(9 + 3√3 - 6)] / (-3) = [2*(3 + 3√3)] / (-3) = [6 + 6√3]/(-3) = -2 - 2√3So, the slope of DA is -2 - 2√3. Therefore, the angle of DA with the x-axis is arctangent of that, which is a negative angle, but since it's in the second quadrant (because the slope is negative and point A is above the x-axis), the angle is 180 - arctan(2 + 2√3).Similarly, slope of CA: from C(4,0) to A(6 - 2√3, 6 - 2√3). The slope is (6 - 2√3 - 0)/(6 - 2√3 - 4) = (6 - 2√3)/(2 - 2√3)Simplify:(6 - 2√3)/(2 - 2√3) = [2*(3 - √3)] / [2*(1 - √3)] = (3 - √3)/(1 - √3)Multiply numerator and denominator by (1 + √3):(3 - √3)(1 + √3) / (1 - 3) = [3(1) + 3√3 - √3 - 3] / (-2) = [3 + 2√3 - 3]/(-2) = (2√3)/(-2) = -√3So, the slope of CA is -√3, which makes sense because angle at C is 60 degrees, so the line from C to A makes 60 degrees with the vertical, hence slope is -√3.Now, angle CAD is the angle between CA and DA. Since CA has slope -√3 and DA has slope -2 - 2√3, we can find the angle between them using the formula:tan(theta) = |(m2 - m1)/(1 + m1*m2)|So, tan(angle CAD) = |(-2 - 2√3 - (-√3))/(1 + (-√3)(-2 - 2√3))| = |(-2 - 2√3 + √3)/(1 + 2√3 + 6)| = |(-2 - √3)/(7 + 2√3)|Hmm, that's a bit messy. Let me compute it step by step.Numerator: -2 - √3Denominator: 1 + ( -√3)(-2 - 2√3) = 1 + 2√3 + 6 = 7 + 2√3So, tan(angle CAD) = |(-2 - √3)/(7 + 2√3)|Let me rationalize the denominator:Multiply numerator and denominator by (7 - 2√3):[(-2 - √3)(7 - 2√3)] / [(7 + 2√3)(7 - 2√3)] = [(-14 + 4√3 - 7√3 + 2*3)] / (49 - 12) = [(-14 - 3√3 + 6)] / 37 = (-8 - 3√3)/37Since we're taking absolute value, it's (8 + 3√3)/37So, tan(angle CAD) = (8 + 3√3)/37Similarly, angle BAD is the angle between BA and DA. BA has slope 1, DA has slope -2 - 2√3.So, tan(angle BAD) = |(m2 - m1)/(1 + m1*m2)| = |(-2 - 2√3 - 1)/(1 + (1)(-2 - 2√3))| = |(-3 - 2√3)/(1 - 2 - 2√3)| = |(-3 - 2√3)/(-1 - 2√3)|Simplify numerator and denominator:= |(3 + 2√3)/(1 + 2√3)|Multiply numerator and denominator by (1 - 2√3):[(3 + 2√3)(1 - 2√3)] / [(1 + 2√3)(1 - 2√3)] = [3(1) - 6√3 + 2√3 - 4*3] / (1 - 12) = [3 - 4√3 - 12]/(-11) = (-9 - 4√3)/(-11) = (9 + 4√3)/11So, tan(angle BAD) = (9 + 4√3)/11Now, we have tan(angle BAD) and tan(angle CAD). We need sin(angle BAD)/sin(angle CAD).Recall that sin(theta) = tan(theta)/sqrt(1 + tan^2(theta))So, sin(angle BAD) = tan(angle BAD)/sqrt(1 + tan^2(angle BAD)) = [(9 + 4√3)/11] / sqrt(1 + [(9 + 4√3)/11]^2)Similarly, sin(angle CAD) = tan(angle CAD)/sqrt(1 + tan^2(angle CAD)) = [(8 + 3√3)/37] / sqrt(1 + [(8 + 3√3)/37]^2)This seems very complicated, but maybe we can find a ratio that simplifies.Alternatively, maybe there's a better approach. Let me think.Wait, earlier when I used the Law of Sines directly on triangles ABD and ACD, I got the ratio as √6. But when I tried using coordinates, I ended up with a complicated expression. Maybe I made a mistake in the coordinate approach.Wait, let me go back to the Law of Sines approach. In triangle ABD:BD / sin(angle BAD) = AD / sin(45°)Similarly, in triangle ACD:CD / sin(angle CAD) = AD / sin(60°)So, sin(angle BAD) = (BD * sin45)/ADsin(angle CAD) = (CD * sin60)/ADTherefore, ratio = (BD/CD) * (sin45/sin60) = 3 * (√2/2)/(√3/2) = 3√2/√3 = 3√6/3 = √6So, the ratio is √6.Wait, but in the initial problem, the user had a different approach and got 3√3/2. Maybe I need to check that.Wait, the user's approach was:In triangle ABD:sin(angle BAD) = (BD / AD) * sin45In triangle ACD:sin(angle CAD) = (CD / AD) * sin60Then ratio = (3x / AD * sin45) / (x / AD * sin60) = 3 * sin45 / sin60 = 3*(√2/2)/(√3/2) = 3√2/√3 = 3√6/3 = √6Wait, that's the same as my approach. So, why did the user get 3√3/2? Maybe they made a mistake.Wait, looking back at the user's initial problem, they wrote:"In triangle ABC, angle B = 45°, angle C = 60°. Point D divides BC in ratio 3:1. Find sin(angle BAD)/sin(angle CAD)."Then, in their solution, they wrote:"By the Law of Sines in triangle ABC, angle A = 75°. BD:DC = 3:1, so BC = 4x. In triangle ABD: BD/sin(angle BAD) = AD/sin45° => sin(angle BAD) = BD/(AD√2) = 3x/(AD√2). In triangle ACD: CD/sin(angle CAD) = AD/sin60° => sin(angle CAD) = x/(AD√3). Therefore, ratio = (3x/(AD√2))/(x/(AD√3)) = 3√3/2."Wait, so in their solution, they have sin(angle BAD) = 3x/(AD√2) and sin(angle CAD) = x/(AD√3). Therefore, ratio = 3√3/2.But in my approach, I have ratio = √6.So, which one is correct? Let me check the calculations.Wait, in the user's solution, they wrote:In triangle ABD: BD/sin(angle BAD) = AD/sin45° => sin(angle BAD) = BD/(AD√2) = 3x/(AD√2)Similarly, in triangle ACD: CD/sin(angle CAD) = AD/sin60° => sin(angle CAD) = CD/(AD√3) = x/(AD√3)Therefore, ratio = (3x/(AD√2))/(x/(AD√3)) = 3√3/2Wait, but in my approach, I used the Law of Sines correctly, and got ratio = √6.Wait, maybe the user made a mistake in their calculation.Wait, let me re-examine the Law of Sines in triangle ABD:BD / sin(angle BAD) = AD / sin(angle ABD)Angle ABD is 45°, so sin(angle ABD) = sin45 = √2/2Therefore, BD / sin(angle BAD) = AD / (√2/2) => BD / sin(angle BAD) = 2AD / √2 = AD√2Therefore, sin(angle BAD) = BD / (AD√2)Similarly, in triangle ACD:CD / sin(angle CAD) = AD / sin(angle ACD) = AD / sin60 = AD / (√3/2) = 2AD / √3Therefore, sin(angle CAD) = CD / (2AD / √3) = CD * √3 / (2AD)So, sin(angle CAD) = (CD * √3)/(2AD)Therefore, ratio = [BD / (AD√2)] / [ (CD * √3)/(2AD) ] = [BD / (AD√2)] * [2AD / (CD√3)] = (BD * 2) / (CD * √2 * √3) = (2BD)/(CD * √6)Since BD/CD = 3, ratio = (2*3)/√6 = 6/√6 = √6Ah, so the user made a mistake in their calculation. They wrote sin(angle CAD) = x/(AD√3), but actually it's (x * √3)/(2AD). Therefore, their ratio was incorrect.So, the correct ratio is √6.Therefore, the answer should be √6.But wait, in the initial problem, the user had a different approach and got 3√3/2, which is approximately 2.598, while √6 is approximately 2.449. So, they are different.Wait, but in my coordinate approach, I ended up with tan(angle BAD) = (9 + 4√3)/11 and tan(angle CAD) = (8 + 3√3)/37. If I compute sin(angle BAD)/sin(angle CAD), it should equal √6.Let me compute sin(angle BAD):sin(angle BAD) = tan(angle BAD) / sqrt(1 + tan^2(angle BAD)) = [(9 + 4√3)/11] / sqrt(1 + [(9 + 4√3)/11]^2)Similarly, sin(angle CAD) = [(8 + 3√3)/37] / sqrt(1 + [(8 + 3√3)/37]^2)This seems complicated, but maybe I can compute the ratio:[sin(angle BAD)/sin(angle CAD)] = [ (9 + 4√3)/11 / sqrt(1 + [(9 + 4√3)/11]^2 ) ] / [ (8 + 3√3)/37 / sqrt(1 + [(8 + 3√3)/37]^2 ) ]This is very messy, but perhaps if I compute numerically:Compute tan(angle BAD) = (9 + 4√3)/11 ≈ (9 + 6.928)/11 ≈ 15.928/11 ≈ 1.448So, angle BAD ≈ arctan(1.448) ≈ 55.2 degreesSimilarly, tan(angle CAD) = (8 + 3√3)/37 ≈ (8 + 5.196)/37 ≈ 13.196/37 ≈ 0.3566So, angle CAD ≈ arctan(0.3566) ≈ 19.6 degreesThen, sin(55.2)/sin(19.6) ≈ 0.821/0.335 ≈ 2.448, which is approximately √6 ≈ 2.449. So, that matches.Therefore, the correct ratio is √6.So, the user's initial solution was incorrect because they miscalculated sin(angle CAD). The correct ratio is √6.Therefore, the answer is √6.