answer：Okay, so I have this problem where I need to find the lengths of the sides and the measures of the angles of a triangle. The given information is the area, which is 234 cm², the perimeter, which is 108 cm, and one of the angles, which is 130 degrees, 26 minutes, and 59 seconds. Hmm, that's quite a bit to work with, but I think I can figure it out step by step.First, I remember that the semi-perimeter (s) is half of the perimeter. So, if the perimeter is 108 cm, then the semi-perimeter would be 54 cm. That seems straightforward. So, s = 54 cm.Now, I need to find the lengths of the sides. I know the area, so maybe I can use some area formulas. The area of a triangle can be calculated using the formula:[ t = frac{1}{2}ab sin C ]But I don't know which sides correspond to which angles yet. Wait, I have one angle given, which is 130 degrees, 26 minutes, and 59 seconds. Let me convert that to decimal degrees to make calculations easier. So, 26 minutes is 26/60 degrees, which is approximately 0.4333 degrees, and 59 seconds is 59/3600 degrees, which is approximately 0.0164 degrees. Adding those together with the 130 degrees gives me:130 + 0.4333 + 0.0164 ≈ 130.4497 degrees.Okay, so angle α is approximately 130.4497 degrees. That's a pretty big angle, so it must be the largest angle in the triangle, meaning the side opposite to it will be the longest side.Since I have the semi-perimeter, maybe I can use Heron's formula, which relates the area, semi-perimeter, and the sides of the triangle. Heron's formula is:[ t = sqrt{s(s - a)(s - b)(s - c)} ]But I have three unknowns here: a, b, and c. I also know the perimeter, so a + b + c = 108 cm. That gives me one equation, but I still need more information.Wait, I also have one angle. Maybe I can use the Law of Cosines or the Law of Sines. The Law of Cosines relates the sides of a triangle with one of its angles. It states:[ c^2 = a^2 + b^2 - 2ab cos C ]But again, I have multiple unknowns here. Maybe I can combine this with the area formula.Let me think. If I can express two sides in terms of the third, I might be able to solve for them. Let's denote the sides opposite angles α, β, and γ as a, b, and c respectively. So, angle α is opposite side a.Given that, I can use the Law of Sines, which states:[ frac{a}{sin alpha} = frac{b}{sin beta} = frac{c}{sin gamma} = 2R ]Where R is the radius of the circumscribed circle. But I don't know R, so maybe that's not directly helpful yet.Alternatively, I can use the area formula with the given angle. The area is given by:[ t = frac{1}{2}ab sin C ]But if I take angle α, then the area can also be expressed as:[ t = frac{1}{2}bc sin alpha ]Wait, that might be useful. Let me write that down:[ 234 = frac{1}{2}bc sin 130.4497^circ ]First, let me compute sin(130.4497°). Using a calculator, sin(130.4497°) is approximately sin(180° - 49.5503°) = sin(49.5503°) ≈ 0.7595.So, plugging that in:[ 234 = frac{1}{2}bc times 0.7595 ][ 234 = 0.37975 bc ][ bc = frac{234}{0.37975} approx 616.3 ]So, bc ≈ 616.3 cm². That's one equation involving sides b and c.I also know that the perimeter is 108 cm, so:[ a + b + c = 108 ]But I don't know a yet. Maybe I can express a in terms of b and c. If I can find another equation involving a, b, and c, I can solve the system.Wait, I can use the Law of Cosines for angle α. Since angle α is opposite side a, the Law of Cosines gives:[ a^2 = b^2 + c^2 - 2bc cos alpha ]We already have bc ≈ 616.3, and we know cos(130.4497°). Let me compute that. Cos(130.4497°) is negative because it's in the second quadrant. Using a calculator, cos(130.4497°) ≈ -0.6494.So, plugging into the Law of Cosines:[ a^2 = b^2 + c^2 - 2 times 616.3 times (-0.6494) ][ a^2 = b^2 + c^2 + 2 times 616.3 times 0.6494 ][ a^2 = b^2 + c^2 + 808.3 ]Hmm, that's another equation. But I still have three variables: a, b, c. Maybe I can express a in terms of b and c from the perimeter equation and substitute.From the perimeter:[ a = 108 - b - c ]So, plugging that into the equation from the Law of Cosines:[ (108 - b - c)^2 = b^2 + c^2 + 808.3 ]Expanding the left side:[ 108^2 - 2 times 108 times (b + c) + (b + c)^2 = b^2 + c^2 + 808.3 ][ 11664 - 216(b + c) + b^2 + 2bc + c^2 = b^2 + c^2 + 808.3 ]Simplify by subtracting b² + c² from both sides:[ 11664 - 216(b + c) + 2bc = 808.3 ]We know bc ≈ 616.3, so 2bc ≈ 1232.6. Plugging that in:[ 11664 - 216(b + c) + 1232.6 = 808.3 ][ 11664 + 1232.6 - 216(b + c) = 808.3 ][ 12896.6 - 216(b + c) = 808.3 ][ -216(b + c) = 808.3 - 12896.6 ][ -216(b + c) = -12088.3 ][ b + c = frac{12088.3}{216} approx 56 ]So, b + c ≈ 56 cm. That's helpful because from the perimeter, a = 108 - (b + c) ≈ 108 - 56 = 52 cm.So, now I have a ≈ 52 cm, and b + c ≈ 56 cm. Also, bc ≈ 616.3 cm².Now, I have two equations:1. b + c = 562. bc = 616.3This is a system of equations that can be solved for b and c. Let me denote b and c as the roots of the quadratic equation:[ x^2 - (b + c)x + bc = 0 ][ x^2 - 56x + 616.3 = 0 ]Let me solve this quadratic equation using the quadratic formula:[ x = frac{56 pm sqrt{56^2 - 4 times 1 times 616.3}}{2} ][ x = frac{56 pm sqrt{3136 - 2465.2}}{2} ][ x = frac{56 pm sqrt{670.8}}{2} ][ x = frac{56 pm 25.9}{2} ]So, the two solutions are:1. x = (56 + 25.9)/2 ≈ 81.9/2 ≈ 40.95 cm2. x = (56 - 25.9)/2 ≈ 30.1/2 ≈ 15.05 cmTherefore, the sides b and c are approximately 40.95 cm and 15.05 cm. Let me round these to two decimal places: b ≈ 41.00 cm and c ≈ 15.00 cm.So, now I have all three sides:- a ≈ 52.00 cm- b ≈ 41.00 cm- c ≈ 15.00 cmLet me double-check if these add up to the perimeter: 52 + 41 + 15 = 108 cm. Yes, that's correct.Now, I need to find the other two angles, β and γ. I can use the Law of Sines for this. The Law of Sines states:[ frac{a}{sin alpha} = frac{b}{sin beta} = frac{c}{sin gamma} ]We already know a, b, c, and α, so we can solve for β and γ.First, let's compute the ratio:[ frac{a}{sin alpha} = frac{52}{sin 130.4497^circ} ]We already calculated sin(130.4497°) ≈ 0.7595, so:[ frac{52}{0.7595} ≈ 68.43 ]So, this ratio is approximately 68.43. Therefore:For side b:[ frac{41}{sin beta} = 68.43 ][ sin beta = frac{41}{68.43} ≈ 0.599 ]So, β ≈ arcsin(0.599) ≈ 36.7 degrees.Similarly, for side c:[ frac{15}{sin gamma} = 68.43 ][ sin gamma = frac{15}{68.43} ≈ 0.219 ]So, γ ≈ arcsin(0.219) ≈ 12.6 degrees.Let me convert these decimal degrees back to degrees, minutes, and seconds.For β ≈ 36.7 degrees:0.7 degrees is 0.7 × 60 = 42 minutes. So, β ≈ 36 degrees 42 minutes.For γ ≈ 12.6 degrees:0.6 degrees is 0.6 × 60 = 36 minutes. So, γ ≈ 12 degrees 36 minutes.Wait, let me check if these angles add up to 180 degrees with angle α.α ≈ 130.4497°, β ≈ 36.7°, γ ≈ 12.6°.Adding them up: 130.4497 + 36.7 + 12.6 ≈ 179.75°, which is approximately 180°, considering rounding errors. That seems reasonable.But let me check the exact values without rounding to see if they add up to 180 degrees.Wait, actually, when I calculated β and γ, I used approximate sine values, so there might be some error. Let me compute more accurately.First, let's compute sin β:[ sin beta = frac{41}{68.43} ≈ 0.599 ]Using a calculator, arcsin(0.599) is approximately 36.7 degrees, as I had before.Similarly, sin γ ≈ 0.219, so arcsin(0.219) ≈ 12.6 degrees.So, adding up the angles:130.4497 + 36.7 + 12.6 ≈ 179.75°, which is very close to 180°, so that seems okay.But just to be thorough, let me use more precise calculations.First, let's compute angle β:[ sin beta = frac{41}{68.43} ≈ 0.599 ]Using a calculator, arcsin(0.599) is approximately 36.7 degrees, but let's get more precise.Using a calculator, arcsin(0.599) ≈ 36.7 degrees, which is 36 degrees and 0.7 × 60 = 42 minutes. So, 36°42'.Similarly, for angle γ:[ sin gamma = frac{15}{68.43} ≈ 0.219 ]Using a calculator, arcsin(0.219) ≈ 12.6 degrees, which is 12 degrees and 0.6 × 60 = 36 minutes. So, 12°36'.Adding up the angles:130°26'59'' + 36°42' + 12°36' = ?First, convert all to degrees:130°26'59'' ≈ 130 + 26/60 + 59/3600 ≈ 130.4497°36°42' ≈ 36 + 42/60 ≈ 36.7°12°36' ≈ 12 + 36/60 ≈ 12.6°Adding them: 130.4497 + 36.7 + 12.6 ≈ 179.7497°, which is approximately 180°, considering the rounding.So, the angles are approximately:- α ≈ 130°26'59''- β ≈ 36°42'- γ ≈ 12°36'But let me check if these angles make sense with the sides.Using the Law of Sines:[ frac{a}{sin alpha} = frac{52}{sin 130.4497°} ≈ frac{52}{0.7595} ≈ 68.43 ][ frac{b}{sin beta} = frac{41}{sin 36.7°} ≈ frac{41}{0.6018} ≈ 68.13 ][ frac{c}{sin gamma} = frac{15}{sin 12.6°} ≈ frac{15}{0.2179} ≈ 68.75 ]These are all approximately equal to 68.43, considering rounding errors. So, that checks out.Therefore, the sides are approximately:- a ≈ 52 cm- b ≈ 41 cm- c ≈ 15 cmAnd the angles are approximately:- α ≈ 130°26'59''- β ≈ 36°42'- γ ≈ 12°36'Wait, but in the initial problem, angle α was given as 130°26'59''. My calculated angles for β and γ are 36°42' and 12°36', which seems a bit off because 36°42' is roughly 36.7°, and 12°36' is roughly 12.6°, which adds up with α to almost 180°, so that seems okay.But let me check if the area with these sides and angles gives the correct area.Using the formula:[ t = frac{1}{2}ab sin C ]Taking sides a = 52, b = 41, and angle γ = 12.6°, which is between sides a and b.Wait, actually, angle α is between sides b and c, right? Because in the standard notation, angle α is opposite side a, so it's between sides b and c.So, to compute the area, it's better to use:[ t = frac{1}{2}bc sin alpha ]We have b = 41, c = 15, and sin α ≈ 0.7595.So,[ t = frac{1}{2} times 41 times 15 times 0.7595 ][ t = frac{1}{2} times 615 times 0.7595 ][ t = 307.5 times 0.7595 ≈ 234 , text{cm}^2 ]Yes, that matches the given area. So, that confirms that the sides and angles are correct.Therefore, the lengths of the sides are approximately 52 cm, 41 cm, and 15 cm, and the measures of the angles are approximately 130°26'59'', 36°42', and 12°36'.**Final Answer**The lengths of the sides are ( boxed{52 , text{cm}} ), ( boxed{41 , text{cm}} ), and ( boxed{15 , text{cm}} ). The measures of the angles are ( boxed{130^circ 26' 59''} ), ( boxed{36^circ 42'} ), and ( boxed{12^circ 36'} ).

answer：Okay, so I have this problem where I need to prove that ( a = y + z ), ( b = x + z ), and ( c = x + y ) where ( x, y ), and ( z ) are positive numbers. Hmm, let's see. I think this might be related to some kind of system of equations, maybe something like solving for ( x, y, z ) in terms of ( a, b, c ). But I'm not entirely sure. Let me try to break it down step by step.First, I have three equations:1. ( a = y + z )2. ( b = x + z )3. ( c = x + y )I need to show that these relationships hold given that ( x, y, z ) are positive. Maybe I should try to solve for ( x, y, z ) in terms of ( a, b, c ). That could help me see if the relationships make sense.Starting with the first equation, ( a = y + z ). If I solve for ( y ), I get ( y = a - z ). Similarly, from the second equation, ( b = x + z ), solving for ( x ) gives ( x = b - z ). Now, I can substitute these expressions for ( x ) and ( y ) into the third equation, which is ( c = x + y ).Substituting, I get:( c = (b - z) + (a - z) )Simplifying that:( c = b + a - 2z )Now, solving for ( z ):( 2z = a + b - c )So,( z = frac{a + b - c}{2} )Okay, so I have ( z ) in terms of ( a, b, c ). Now, I can plug this back into the expressions for ( x ) and ( y ).Starting with ( x = b - z ):( x = b - frac{a + b - c}{2} )Simplifying:( x = frac{2b - (a + b - c)}{2} )( x = frac{2b - a - b + c}{2} )( x = frac{-a + b + c}{2} )Similarly, for ( y = a - z ):( y = a - frac{a + b - c}{2} )Simplifying:( y = frac{2a - (a + b - c)}{2} )( y = frac{2a - a - b + c}{2} )( y = frac{a - b + c}{2} )So now I have expressions for ( x, y, z ) in terms of ( a, b, c ):( x = frac{-a + b + c}{2} )( y = frac{a - b + c}{2} )( z = frac{a + b - c}{2} )Now, since ( x, y, z ) are positive numbers, the numerators in these fractions must be positive. That gives me the following conditions:1. ( -a + b + c > 0 ) → ( b + c > a )2. ( a - b + c > 0 ) → ( a + c > b )3. ( a + b - c > 0 ) → ( a + b > c )These are the triangle inequalities! So, as long as ( a, b, c ) satisfy the triangle inequalities, ( x, y, z ) will be positive. That makes sense because if ( a, b, c ) are sides of a triangle, then these inequalities hold.Wait, but the problem just says ( x, y, z ) are positive numbers. It doesn't explicitly say that ( a, b, c ) are sides of a triangle. So, maybe I need to assume that ( a, b, c ) satisfy these inequalities to ensure ( x, y, z ) are positive.Alternatively, if ( a, b, c ) are given such that these inequalities hold, then ( x, y, z ) will indeed be positive. So, the relationships ( a = y + z ), ( b = x + z ), and ( c = x + y ) hold under the condition that ( a, b, c ) satisfy the triangle inequalities, ensuring ( x, y, z ) are positive.I think I've got it. By expressing ( x, y, z ) in terms of ( a, b, c ) and ensuring their positivity through the triangle inequalities, I've shown that the given relationships hold when ( x, y, z ) are positive. Maybe I should double-check my algebra to make sure I didn't make any mistakes.Starting from the substitutions:1. ( y = a - z )2. ( x = b - z )3. Substituting into ( c = x + y ):( c = (b - z) + (a - z) = a + b - 2z )Solving for ( z ):( 2z = a + b - c ) → ( z = frac{a + b - c}{2} )Then, substituting back:( x = b - frac{a + b - c}{2} = frac{2b - a - b + c}{2} = frac{-a + b + c}{2} )( y = a - frac{a + b - c}{2} = frac{2a - a - b + c}{2} = frac{a - b + c}{2} )Looks correct. And ensuring positivity leads to the triangle inequalities. So, yes, the proof seems solid.

answer：Okay, so I have this problem where I need to prove that ( a^3 - 4a ) is divisible by 120, given that ( a - 1 ) and ( a + 1 ) are twin primes greater than 10. Hmm, twin primes are primes that are two units apart, like 11 and 13, right? So, ( a ) must be the number in between them, which would be 12 in the case of 11 and 13.First, I need to understand what it means for a number to be divisible by 120. Divisibility by 120 means that the number is divisible by all the prime factors of 120. Let me factorize 120: ( 120 = 2^3 times 3 times 5 ). So, to prove that ( a^3 - 4a ) is divisible by 120, I need to show that it's divisible by 8, 3, and 5.Let me start by looking at the expression ( a^3 - 4a ). I can factor this expression:( a^3 - 4a = a(a^2 - 4) = a(a - 2)(a + 2) ).So, ( a^3 - 4a ) is the product of ( a ), ( a - 2 ), and ( a + 2 ). That means it's the product of three consecutive even numbers if ( a ) is even. Wait, are ( a - 2 ), ( a ), and ( a + 2 ) consecutive even numbers? Let me check.If ( a ) is even, then ( a - 2 ) and ( a + 2 ) are also even. So, yes, they are three consecutive even numbers. That means among them, there must be a multiple of 4 and another multiple of 2, which together make a multiple of 8. So, ( a(a - 2)(a + 2) ) is divisible by 8. That takes care of the 8 part of 120.Next, I need to check divisibility by 3. Since ( a - 1 ) and ( a + 1 ) are primes greater than 10, they can't be multiples of 3. So, ( a - 1 ) and ( a + 1 ) are not divisible by 3. That means ( a ) must be divisible by 3 because in any set of three consecutive numbers, one of them is divisible by 3. Here, since ( a - 1 ) and ( a + 1 ) are not divisible by 3, ( a ) must be. So, ( a ) is divisible by 3, which means ( a(a - 2)(a + 2) ) is divisible by 3.Now, for divisibility by 5. Similar to the 3 case, ( a - 1 ) and ( a + 1 ) are primes greater than 10, so they can't be multiples of 5. Therefore, ( a ) must be divisible by 5 because in any set of five consecutive numbers, one is divisible by 5. Since ( a - 1 ) and ( a + 1 ) aren't, ( a ) must be. So, ( a ) is divisible by 5, making ( a(a - 2)(a + 2) ) divisible by 5.Putting it all together, since ( a^3 - 4a ) is divisible by 8, 3, and 5, it's divisible by their least common multiple, which is 120. Therefore, ( a^3 - 4a ) is divisible by 120.Wait, let me double-check if I missed anything. I considered the factors 8, 3, and 5. For 8, I used the fact that ( a ) is even, so ( a ), ( a - 2 ), and ( a + 2 ) are three consecutive even numbers, which must include a multiple of 4 and another multiple of 2, giving 8. For 3 and 5, I used the fact that since ( a - 1 ) and ( a + 1 ) are primes greater than 10, they can't be multiples of 3 or 5, so ( a ) must be a multiple of both 3 and 5. That seems solid.I think that covers all the necessary parts. So, yeah, ( a^3 - 4a ) should indeed be divisible by 120 under the given conditions.

answer：First, I need to determine the cost of each type of ice cream cone that Mrs. Hilt is purchasing.For the vanilla cones:She is buying 2 vanilla cones at 0.99 each.2 × 0.99 = 1.98For the chocolate cones:She is buying 3 chocolate cones at 1.29 each.3 × 1.29 = 3.87For the strawberry cone:She is buying 1 strawberry cone at 1.49.1 × 1.49 = 1.49Next, I will add up the costs of all the cones to find the total cost.1.98 (vanilla) + 3.87 (chocolate) + 1.49 (strawberry) = 7.34Therefore, the total cost of her purchase of 6 ice cream cones is 7.34.

answer：First, I recognize that each subsequent triangle is formed by joining the midpoints of the previous triangle, which means each new triangle has sides that are half the length of the previous one.Given that the side length of the first triangle ( T_1 ) is 50 cm, the perimeter of ( T_1 ) is ( 3 times 50 = 150 ) cm.The side length of ( T_2 ) will be ( frac{50}{2} = 25 ) cm, and its perimeter will be ( 3 times 25 = 75 ) cm.Similarly, the side length of ( T_3 ) will be ( frac{25}{2} = 12.5 ) cm, and its perimeter will be ( 3 times 12.5 = 37.5 ) cm.This pattern continues indefinitely, with each subsequent perimeter being half of the previous one. Therefore, the perimeters form a geometric series: 150 cm, 75 cm, 37.5 cm, and so on.The sum of an infinite geometric series can be calculated using the formula:[text{Sum} = frac{a}{1 - r}]where ( a ) is the first term and ( r ) is the common ratio.In this case, ( a = 150 ) cm and ( r = frac{1}{2} ).Plugging these values into the formula:[text{Sum} = frac{150}{1 - frac{1}{2}} = frac{150}{frac{1}{2}} = 300 text{ cm}]Therefore, the sum of the perimeters of all the triangles is 300 cm.

answer：Okay, so I have this problem about a factory that produces three different models of products: A, B, and C. The ratio of their quantities is 2:3:4. They're using a stratified sampling method to draw a sample of size n, and in this sample, there are 16 units of model A. I need to find out what the total sample size n is.Alright, let me start by understanding what stratified sampling is. From what I remember, stratified sampling is a method where the population is divided into subgroups or strata, and then samples are taken from each stratum in a way that's proportional to their size in the population. So, if the ratio of A:B:C is 2:3:4, then in the sample, the number of units from each model should also be in the same ratio.Given that, the ratio of A:B:C is 2:3:4. That means for every 2 units of A, there are 3 units of B and 4 units of C. So, if I think of it in terms of parts, A is 2 parts, B is 3 parts, and C is 4 parts. The total number of parts would be 2 + 3 + 4, which is 9 parts.Now, in the sample, there are 16 units of model A. Since the ratio is 2:3:4, I can think of these 16 units as representing the 2 parts of A in the ratio. So, each part would be equal to 16 divided by 2, which is 8 units. That means each part is 8 units.If each part is 8 units, then model B, which is 3 parts, would be 3 times 8, which is 24 units. Similarly, model C, which is 4 parts, would be 4 times 8, which is 32 units.So, in the sample, we have 16 units of A, 24 units of B, and 32 units of C. To find the total sample size n, I just need to add these up: 16 + 24 + 32.Let me do that calculation: 16 plus 24 is 40, and 40 plus 32 is 72. So, the total sample size n should be 72.Wait, let me double-check that. If the ratio is 2:3:4, and each part is 8 units, then A is 2 parts, so 16, B is 3 parts, so 24, and C is 4 parts, so 32. Adding them up: 16 + 24 is 40, and 40 + 32 is indeed 72. That seems correct.I think I've got it. The total sample size n is 72.