answer：Alright, so I have this triangle ABC, and I need to find its area and then the lengths of its sides given some conditions. Let me try to break this down step by step.First, part (1) asks for the area of triangle ABC. I know that the area of a triangle can be found using the formula:[text{Area} = frac{1}{2}absin C]But in this case, I don't have all the sides or angles. Instead, I'm given that ( bc = 5 ) and ( cos frac{A}{2} = frac{3sqrt{10}}{10} ). Hmm, okay. So I need to figure out how to use this information.I remember that ( cos frac{A}{2} ) is related to the half-angle formulas. Maybe I can find ( cos A ) from ( cos frac{A}{2} ). Let me recall the double-angle formula for cosine:[cos A = 2cos^2 frac{A}{2} - 1]Plugging in the given value:[cos A = 2left(frac{3sqrt{10}}{10}right)^2 - 1]Calculating the square:[left(frac{3sqrt{10}}{10}right)^2 = frac{9 times 10}{100} = frac{90}{100} = frac{9}{10}]So,[cos A = 2 times frac{9}{10} - 1 = frac{18}{10} - 1 = frac{18}{10} - frac{10}{10} = frac{8}{10} = frac{4}{5}]Okay, so ( cos A = frac{4}{5} ). Now, to find the area, I need ( sin A ). Since ( sin^2 A + cos^2 A = 1 ), I can find ( sin A ):[sin A = sqrt{1 - cos^2 A} = sqrt{1 - left(frac{4}{5}right)^2} = sqrt{1 - frac{16}{25}} = sqrt{frac{9}{25}} = frac{3}{5}]Great, so ( sin A = frac{3}{5} ). Now, going back to the area formula. I have ( bc = 5 ) and ( sin A = frac{3}{5} ). So,[text{Area} = frac{1}{2} times b times c times sin A = frac{1}{2} times 5 times frac{3}{5}]Simplifying this:[text{Area} = frac{1}{2} times 5 times frac{3}{5} = frac{1}{2} times 3 = frac{3}{2}]So, the area is ( frac{3}{2} ). That seems straightforward. I think that's part (1) done.Moving on to part (2), it says if ( sin B = 5 sin C ), find the values of ( a ), ( b ), and ( c ). Hmm, okay. So I have a relationship between ( sin B ) and ( sin C ). I remember the Law of Sines, which states:[frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} = 2R]Where ( R ) is the radius of the circumcircle. So, from this, I can write:[frac{b}{sin B} = frac{c}{sin C}]Given that ( sin B = 5 sin C ), substituting into the equation:[frac{b}{5 sin C} = frac{c}{sin C}]Simplifying this, since ( sin C ) is not zero (as it's a triangle angle), we can cancel ( sin C ):[frac{b}{5} = c implies b = 5c]So, ( b = 5c ). But we also know from the problem statement that ( bc = 5 ). So, substituting ( b = 5c ) into ( bc = 5 ):[5c times c = 5 implies 5c^2 = 5 implies c^2 = 1 implies c = 1]Since lengths are positive, ( c = 1 ). Then, ( b = 5c = 5 times 1 = 5 ).So, now we have ( b = 5 ) and ( c = 1 ). We need to find ( a ). For this, I can use the Law of Cosines, which relates the sides of a triangle to one of its angles. The formula is:[a^2 = b^2 + c^2 - 2bc cos A]We already know ( b = 5 ), ( c = 1 ), and ( cos A = frac{4}{5} ). Plugging these values in:[a^2 = 5^2 + 1^2 - 2 times 5 times 1 times frac{4}{5}]Calculating each term:- ( 5^2 = 25 )- ( 1^2 = 1 )- ( 2 times 5 times 1 = 10 )- ( 10 times frac{4}{5} = 8 )So,[a^2 = 25 + 1 - 8 = 18]Therefore, ( a = sqrt{18} = 3sqrt{2} ).Let me just recap what I did:1. For part (1), I used the half-angle formula to find ( cos A ), then used the Pythagorean identity to find ( sin A ). With ( bc = 5 ) and ( sin A ), I applied the area formula.2. For part (2), I used the given relationship ( sin B = 5 sin C ) along with the Law of Sines to find that ( b = 5c ). Then, using ( bc = 5 ), I solved for ( b ) and ( c ). Finally, I applied the Law of Cosines to find ( a ).I think that covers everything. Let me just double-check my calculations to make sure I didn't make any arithmetic errors.For part (1):- ( cos frac{A}{2} = frac{3sqrt{10}}{10} )- ( cos A = 2 times left(frac{9 times 10}{100}right) - 1 = 2 times frac{9}{10} - 1 = frac{18}{10} - 1 = frac{8}{10} = frac{4}{5} ) ✔️- ( sin A = sqrt{1 - left(frac{16}{25}right)} = sqrt{frac{9}{25}} = frac{3}{5} ) ✔️- Area = ( frac{1}{2} times 5 times frac{3}{5} = frac{3}{2} ) ✔️For part (2):- ( sin B = 5 sin C ) implies ( frac{b}{sin B} = frac{c}{sin C} ) so ( frac{b}{5 sin C} = frac{c}{sin C} ) leading to ( b = 5c ) ✔️- ( bc = 5 ) with ( b = 5c ) gives ( 5c^2 = 5 ) so ( c = 1 ), ( b = 5 ) ✔️- Law of Cosines: ( a^2 = 25 + 1 - 8 = 18 ) so ( a = 3sqrt{2} ) ✔️Everything seems to check out. I think I'm confident with these answers.**Final Answer**(1) The area of triangle ABC is boxed{dfrac{3}{2}}.(2) The lengths of the sides are a = boxed{3sqrt{2}}, b = boxed{5}, and c = boxed{1}.

answer：Okay, so I have this problem where I'm given three integers a, b, and c, and their sum a + b + c is divisible by six. I need to prove that a⁵ + b³ + c is also divisible by six. Hmm, interesting. Let me try to break this down step by step.First, I know that if a number is divisible by six, it must be divisible by both 2 and 3 because 6 is the product of these two primes. So, maybe I can approach this problem by checking divisibility by 2 and 3 separately. That might make things easier.Starting with divisibility by 2: If a + b + c is divisible by 2, then the sum of a, b, and c is even. So, I need to see if a⁵ + b³ + c is also even. Let me think about how exponents affect the parity of a number. If a is even, then any power of a, like a⁵, will also be even. Similarly, if a is odd, a⁵ will be odd. The same goes for b³: if b is even, b³ is even; if b is odd, b³ is odd. So, the parity of a⁵ and b³ depends on the parity of a and b, respectively.Now, c's parity is directly given by the sum a + b + c being even. So, if a + b is even, then c must be even, and if a + b is odd, then c must be odd. Wait, but since a + b + c is even, the sum of a and b must have the same parity as c. So, if a + b is even, c is even; if a + b is odd, c is odd. Let me try to express a⁵ + b³ + c in terms of a + b + c. Maybe I can manipulate the expression to include a + b + c somehow. Let's see:a⁵ + b³ + c = (a⁵ - a) + (b³ - b) + (a + b + c)Hmm, that seems a bit forced, but let's go with it. So, if I can show that (a⁵ - a) and (b³ - b) are both divisible by 6, and since a + b + c is divisible by 6, then the entire expression would be divisible by 6.Wait, but I need to check if (a⁵ - a) and (b³ - b) are divisible by 6. Let me think about that. I remember that for any integer n, n³ ≡ n mod 6. Is that true? Let me test it with some numbers. If n = 2, 2³ = 8, and 8 mod 6 is 2, which is equal to n mod 6. If n = 3, 3³ = 27, and 27 mod 6 is 3, which is equal to n mod 6. Hmm, seems like it works. So, n³ ≡ n mod 6. That means b³ - b is divisible by 6. That's good.What about a⁵? Does a similar property hold? Let me check. For n = 2, 2⁵ = 32, and 32 mod 6 is 2, which is equal to n mod 6. For n = 3, 3⁵ = 243, and 243 mod 6 is 3, which is equal to n mod 6. For n = 4, 4⁵ = 1024, and 1024 mod 6 is 4, which is equal to n mod 6. Hmm, interesting. It seems like a⁵ ≡ a mod 6 as well. So, a⁵ - a is divisible by 6.Therefore, both (a⁵ - a) and (b³ - b) are divisible by 6. And since a + b + c is also divisible by 6, the entire expression a⁵ + b³ + c is the sum of three numbers each divisible by 6, which means the whole thing is divisible by 6.Wait, but let me make sure I'm not missing something. Is it always true that a⁵ ≡ a mod 6 and b³ ≡ b mod 6? Let me test more numbers to confirm.Take n = 5: 5³ = 125, 125 mod 6 is 5, which is equal to n mod 6. 5⁵ = 3125, 3125 mod 6 is 5, which is equal to n mod 6. Okay, seems consistent.What about negative numbers? Let's try n = -1. (-1)³ = -1, which mod 6 is 5, and -1 mod 6 is also 5. (-1)⁵ = -1, which mod 6 is 5, same as n mod 6. Okay, that works too.So, it seems like for any integer n, n³ ≡ n mod 6 and n⁵ ≡ n mod 6. Therefore, a⁵ - a and b³ - b are both divisible by 6.Since a + b + c is divisible by 6, and both (a⁵ - a) and (b³ - b) are divisible by 6, their sum a⁵ + b³ + c must also be divisible by 6.I think that makes sense. Let me recap:1. We know a + b + c is divisible by 6.2. We showed that a⁵ - a is divisible by 6.3. We showed that b³ - b is divisible by 6.4. Therefore, a⁵ + b³ + c = (a⁵ - a) + (b³ - b) + (a + b + c) is the sum of three numbers each divisible by 6, so the whole expression is divisible by 6.Yeah, that seems solid. I don't see any gaps in the logic. Maybe I should also consider modular arithmetic properties to reinforce this.In modular arithmetic, if we have two numbers congruent modulo 6, their sum is also congruent modulo 6. So, since a⁵ ≡ a mod 6 and b³ ≡ b mod 6, then a⁵ + b³ ≡ a + b mod 6. Adding c to both sides, we get a⁵ + b³ + c ≡ a + b + c mod 6. But since a + b + c is divisible by 6, it's congruent to 0 mod 6. Therefore, a⁵ + b³ + c ≡ 0 mod 6, meaning it's divisible by 6.That's another way to look at it, using congruences. It reinforces the earlier conclusion.I think I've covered all bases here. I've checked specific cases, considered negative numbers, used both expression manipulation and modular arithmetic. Everything points to the conclusion that a⁵ + b³ + c is divisible by 6 given that a + b + c is divisible by 6.Just to be thorough, let me try plugging in some actual numbers to test this.Let's say a = 1, b = 2, c = 3. Then a + b + c = 6, which is divisible by 6. Now, a⁵ = 1, b³ = 8, c = 3. So, a⁵ + b³ + c = 1 + 8 + 3 = 12, which is divisible by 6. Good.Another example: a = 4, b = 5, c = -3. Then a + b + c = 4 + 5 - 3 = 6, which is divisible by 6. Now, a⁵ = 1024, b³ = 125, c = -3. So, a⁵ + b³ + c = 1024 + 125 - 3 = 1146. Dividing 1146 by 6 gives 191, which is an integer. So, it works.One more example with negative numbers: a = -2, b = -3, c = 5. Then a + b + c = -2 - 3 + 5 = 0, which is divisible by 6. Now, a⁵ = (-2)^5 = -32, b³ = (-3)^3 = -27, c = 5. So, a⁵ + b³ + c = -32 - 27 + 5 = -54. Dividing -54 by 6 gives -9, which is an integer. So, it works here too.Alright, I'm pretty confident now that the proof holds. The key was recognizing that a⁵ ≡ a mod 6 and b³ ≡ b mod 6, which allows us to express a⁵ + b³ + c in terms of a + b + c plus terms that are each divisible by 6. Since a + b + c is given to be divisible by 6, the entire expression follows suit.I think I've thought this through thoroughly. Time to put it all together in a clear, step-by-step explanation.

answer：Okay, so I need to solve the equation (5 + 2iz = 4 - 6iz) for (z). Hmm, let me think about how to approach this. I remember that when solving equations with complex numbers, the goal is to isolate the variable, which in this case is (z). First, I should probably get all the terms containing (z) on one side of the equation and the constant terms on the other side. So, let me see, on the left side, I have (5 + 2iz), and on the right side, I have (4 - 6iz). I think I should move the (2iz) term to the right side and the (4) term to the left side. To do that, I can subtract (2iz) from both sides and also subtract (4) from both sides. Let me write that down:Starting with:[5 + 2iz = 4 - 6iz]Subtract (2iz) from both sides:[5 = 4 - 6iz - 2iz]And subtract (4) from both sides:[5 - 4 = -6iz - 2iz]Simplifying both sides:[1 = -8iz]Wait, let me check that again. When I subtract (2iz) from both sides, the left side becomes (5), and the right side becomes (4 - 6iz - 2iz), which is (4 - 8iz). Then, subtracting (4) from both sides gives me (1 = -8iz). Yeah, that seems right.So now, I have:[1 = -8iz]I need to solve for (z), so I should isolate it. To do that, I can divide both sides of the equation by (-8i). Let me write that:[z = frac{1}{-8i}]Hmm, but having an (i) in the denominator isn't usually preferred. I remember that to simplify expressions with (i) in the denominator, I can multiply the numerator and the denominator by (i) to rationalize it. Let me try that:Multiply numerator and denominator by (i):[z = frac{1 times i}{-8i times i}]Simplifying the denominator:[-8i times i = -8i^2]Since (i^2 = -1), this becomes:[-8(-1) = 8]So now, the equation is:[z = frac{i}{8}]Wait, but originally, I had a negative sign in the denominator. Let me double-check that step. Starting from:[z = frac{1}{-8i}]Multiplying numerator and denominator by (i):[z = frac{1 times i}{-8i times i} = frac{i}{-8i^2}]Since (i^2 = -1), this becomes:[frac{i}{-8(-1)} = frac{i}{8}]So, that simplifies to:[z = frac{i}{8}]But wait, that seems different from my initial thought. Let me verify my steps again. Starting equation:[5 + 2iz = 4 - 6iz]Subtract (2iz) from both sides:[5 = 4 - 8iz]Subtract (4) from both sides:[1 = -8iz]Divide both sides by (-8i):[z = frac{1}{-8i}]Multiply numerator and denominator by (i):[z = frac{i}{-8i^2} = frac{i}{-8(-1)} = frac{i}{8}]So, yes, that's correct. Therefore, the solution is (z = frac{i}{8}).Wait a second, but earlier I thought it was (-frac{1}{8}i). Did I make a mistake somewhere? Let me go back.Starting equation:[5 + 2iz = 4 - 6iz]Subtract (2iz) from both sides:[5 = 4 - 8iz]Subtract (4) from both sides:[1 = -8iz]Divide both sides by (-8i):[z = frac{1}{-8i}]Multiply numerator and denominator by (i):[z = frac{i}{-8i^2} = frac{i}{-8(-1)} = frac{i}{8}]So, it seems that the correct answer is (z = frac{i}{8}). But initially, I thought it was negative. Maybe I confused the signs somewhere. Let me check the division step again.Starting from:[1 = -8iz]Dividing both sides by (-8i):[z = frac{1}{-8i}]Yes, that's correct. So, (z = frac{1}{-8i}). Then, multiplying numerator and denominator by (i), we get:[z = frac{i}{-8i^2} = frac{i}{-8(-1)} = frac{i}{8}]So, indeed, (z = frac{i}{8}). Therefore, my initial thought of (-frac{1}{8}i) was incorrect. The correct solution is (z = frac{i}{8}).Wait, but let me verify by plugging the value back into the original equation to ensure it's correct.Original equation:[5 + 2iz = 4 - 6iz]Let (z = frac{i}{8}). Substitute into the equation:Left side:[5 + 2i times frac{i}{8} = 5 + frac{2i^2}{8} = 5 + frac{2(-1)}{8} = 5 - frac{2}{8} = 5 - frac{1}{4} = frac{20}{4} - frac{1}{4} = frac{19}{4}]Right side:[4 - 6i times frac{i}{8} = 4 - frac{6i^2}{8} = 4 - frac{6(-1)}{8} = 4 + frac{6}{8} = 4 + frac{3}{4} = frac{16}{4} + frac{3}{4} = frac{19}{4}]Both sides equal (frac{19}{4}), so the solution (z = frac{i}{8}) is correct.Wait, but earlier I thought the answer was (-frac{1}{8}i). Did I make a mistake in the sign somewhere? Let me re-examine the steps.Starting from:[1 = -8iz]Dividing both sides by (-8i):[z = frac{1}{-8i}]Yes, that's correct. So, (z = frac{1}{-8i}). Then, multiplying numerator and denominator by (i):[z = frac{i}{-8i^2} = frac{i}{-8(-1)} = frac{i}{8}]So, the negative sign in the denominator cancels out because (i^2 = -1), resulting in a positive denominator. Therefore, (z = frac{i}{8}) is indeed the correct solution.I think I initially made a mistake by not carefully following the sign changes when multiplying by (i). It's important to keep track of the negative signs to avoid errors. So, the correct answer is (z = frac{i}{8}).But just to be thorough, let me try solving the equation again step by step.Starting equation:[5 + 2iz = 4 - 6iz]Bring all (z) terms to the left and constants to the right:[2iz + 6iz = 4 - 5]Combine like terms:[8iz = -1]Divide both sides by (8i):[z = frac{-1}{8i}]Multiply numerator and denominator by (i):[z = frac{-1 times i}{8i times i} = frac{-i}{8i^2} = frac{-i}{8(-1)} = frac{-i}{-8} = frac{i}{8}]Yes, that confirms it again. So, (z = frac{i}{8}) is the correct solution.I think I confused myself earlier by misapplying the negative sign when multiplying by (i). It's crucial to handle each step carefully, especially when dealing with complex numbers and their properties. Taking it step by step and verifying each operation helps prevent mistakes.In summary, the solution to the equation (5 + 2iz = 4 - 6iz) is (z = frac{i}{8}).

answer：First, I need to find the value of ( x ) given that its reciprocal is (-1). The reciprocal of a number ( x ) is defined as ( frac{1}{x} ). Therefore, if ( frac{1}{x} = -1 ), I can solve for ( x ) by taking the reciprocal of both sides. This gives me ( x = -1 ).Next, I need to evaluate the expression ( |-x - 1| ) using the value of ( x ) I just found. Substituting ( x = -1 ) into the expression, it becomes ( |-(-1) - 1| ). Simplifying inside the absolute value, ( -(-1) ) is equal to ( 1 ), so the expression becomes ( |1 - 1| ).Finally, calculating the absolute value of ( 1 - 1 ) gives ( |0| = 0 ). Therefore, the value of ( |-x - 1| ) is ( 0 ).

answer：Okay, so I've got this problem about a workshop with 50 workers who need to process two types of car parts, A and B. The first part asks how many workers should be assigned to process type A parts if each car set needs 7 parts of type A and 2 parts of type B every day. The second part is about calculating the total processing cost for these 50 workers in one day, given the costs per part.Alright, let's tackle part (1) first. I need to figure out how many workers should work on type A parts. Each worker can process either 30 parts of type A or 20 parts of type B per day. So, if I assign x workers to type A, then the remaining (50 - x) workers will be processing type B.Now, each car set requires 7 type A parts and 2 type B parts. I think this means that the number of type A parts produced should be 7 times the number of car sets, and similarly, the number of type B parts should be 2 times the number of car sets. But wait, actually, the problem says "to produce a car set every day," so maybe it's per car set per day? Hmm, I'm not entirely sure, but I think it's safer to assume that each day, the workshop needs to produce enough parts to make as many car sets as possible, given the workers' capacities.So, if we let x be the number of workers on type A, they can produce 30x type A parts per day. Similarly, the remaining workers on type B can produce 20*(50 - x) type B parts per day.Now, since each car set needs 7 type A and 2 type B parts, the number of car sets that can be produced is limited by the part that runs out first. So, the number of car sets based on type A parts would be (30x)/7, and based on type B parts would be (20*(50 - x))/2. These two should be equal to ensure that we're producing the maximum number of car sets without any excess parts.So, setting them equal:(30x)/7 = (20*(50 - x))/2Let me solve this equation step by step.First, simplify the right side:(20*(50 - x))/2 = 10*(50 - x) = 500 - 10xSo, now the equation is:(30x)/7 = 500 - 10xTo eliminate the fraction, multiply both sides by 7:30x = 7*(500 - 10x)30x = 3500 - 70xNow, bring all terms to one side:30x + 70x = 3500100x = 3500x = 3500 / 100x = 35So, 35 workers should be assigned to process type A parts. That seems straightforward.Now, moving on to part (2). We need to find the total processing cost for these 50 workers in one day, given that processing one type A part costs 10 and one type B part costs 12.From part (1), we have 35 workers on type A and 15 workers on type B.First, let's find out how many parts each group processes in a day.Type A workers: 35 workers * 30 parts/day = 1050 type A parts.Type B workers: 15 workers * 20 parts/day = 300 type B parts.Now, the cost for type A parts is 1050 parts * 10/part = 10,500.The cost for type B parts is 300 parts * 12/part = 3,600.Adding these together gives the total processing cost:10,500 + 3,600 = 14,100.So, the total processing cost for these 50 workers in one day is 14,100.Wait, let me double-check my calculations to make sure I didn't make any mistakes.For part (1):(30x)/7 = (20*(50 - x))/2Simplify right side: 10*(50 - x) = 500 - 10xMultiply both sides by 7: 30x = 3500 - 70xCombine like terms: 100x = 3500x = 35. That seems correct.For part (2):35 workers * 30 = 1050 type A parts.15 workers * 20 = 300 type B parts.Cost A: 1050 * 10 = 10,500Cost B: 300 * 12 = 3,600Total cost: 10,500 + 3,600 = 14,100Yes, that all checks out. I think I've got it right.

answer：Okay, so I have this problem where I'm given two vectors, vector a and vector b. Vector a is (sin(α + π/6), 1) and vector b is (4, 4cosα - √3). It says that vector a is perpendicular to vector b, and I need to find the value of sin(α + 4π/3). Hmm, okay, let me think about how to approach this.First, I remember that if two vectors are perpendicular, their dot product is zero. So, I should calculate the dot product of vector a and vector b and set it equal to zero. That should give me an equation to solve for α.Let me write down the vectors again:Vector a = (sin(α + π/6), 1)Vector b = (4, 4cosα - √3)So, the dot product of a and b is:sin(α + π/6) * 4 + 1 * (4cosα - √3) = 0Let me write that equation out:4 sin(α + π/6) + (4 cosα - √3) = 0Okay, so I need to simplify this equation. Maybe I can expand sin(α + π/6) using the sine addition formula. The sine addition formula is sin(A + B) = sinA cosB + cosA sinB.So, sin(α + π/6) = sinα cos(π/6) + cosα sin(π/6)I know that cos(π/6) is √3/2 and sin(π/6) is 1/2. So substituting those in:sin(α + π/6) = sinα*(√3/2) + cosα*(1/2)So, sin(α + π/6) = (√3/2) sinα + (1/2) cosαNow, substitute this back into the equation:4[(√3/2) sinα + (1/2) cosα] + 4 cosα - √3 = 0Let me distribute the 4 into the brackets:4*(√3/2) sinα + 4*(1/2) cosα + 4 cosα - √3 = 0Simplify each term:(4*(√3)/2) sinα = 2√3 sinα(4*(1/2)) cosα = 2 cosαSo, the equation becomes:2√3 sinα + 2 cosα + 4 cosα - √3 = 0Combine like terms. The cosα terms are 2 cosα and 4 cosα, which add up to 6 cosα. The sinα term is 2√3 sinα. So:2√3 sinα + 6 cosα - √3 = 0Hmm, okay, so now I have:2√3 sinα + 6 cosα = √3I can factor out a common factor from the left side. Let me see, both terms have a factor of 2, but 2√3 and 6. Maybe I can factor out 2√3?Wait, 2√3 sinα + 6 cosα can be written as 2√3 sinα + 6 cosα. Let me factor out 2√3:2√3 (sinα + (6)/(2√3) cosα) = √3Simplify 6/(2√3):6/(2√3) = 3/√3 = √3So, that becomes:2√3 (sinα + √3 cosα) = √3Divide both sides by √3:2 (sinα + √3 cosα) = 1So, sinα + √3 cosα = 1/2Hmm, okay, so now I have:sinα + √3 cosα = 1/2This looks like a linear combination of sine and cosine. I remember that expressions like A sinα + B cosα can be written as C sin(α + φ) or C cos(α + φ), where C is the amplitude and φ is the phase shift.Let me try to write sinα + √3 cosα as a single sine function. The formula is:A sinα + B cosα = C sin(α + φ)Where C = √(A² + B²) and tanφ = B/AIn this case, A = 1 and B = √3. So,C = √(1² + (√3)²) = √(1 + 3) = √4 = 2And tanφ = B/A = √3/1 = √3, so φ = π/3Therefore, sinα + √3 cosα = 2 sin(α + π/3)So, substituting back into our equation:2 sin(α + π/3) = 1/2Divide both sides by 2:sin(α + π/3) = 1/4Okay, so sin(α + π/3) = 1/4. Now, the problem asks for sin(α + 4π/3). Hmm, let me think about how to relate these two.I know that 4π/3 is equal to π + π/3. So, α + 4π/3 = α + π + π/3 = (α + π/3) + πSo, sin(α + 4π/3) = sin[(α + π/3) + π]I remember that sin(θ + π) = -sinθ. Because adding π radians to an angle reflects it across the origin, flipping the sine value.So, sin[(α + π/3) + π] = -sin(α + π/3)Therefore, sin(α + 4π/3) = -sin(α + π/3)But we already found that sin(α + π/3) = 1/4. So,sin(α + 4π/3) = -1/4So, the value is -1/4.Let me just recap to make sure I didn't make any mistakes.1. Vectors a and b are perpendicular, so their dot product is zero.2. Calculated the dot product, expanded sin(α + π/6) using the sine addition formula.3. Simplified the equation to 2√3 sinα + 6 cosα = √3.4. Factored out 2√3 and simplified to sinα + √3 cosα = 1/2.5. Expressed sinα + √3 cosα as 2 sin(α + π/3) using the amplitude-phase form.6. Solved for sin(α + π/3) = 1/4.7. Recognized that sin(α + 4π/3) is sin[(α + π/3) + π], which equals -sin(α + π/3).8. Therefore, sin(α + 4π/3) = -1/4.Everything seems to check out. I don't see any calculation errors, and the logic flows correctly. So, I think the answer is -1/4.**Final Answer**The value of sinleft(alpha + frac{4pi}{3}right) is boxed{-frac{1}{4}}.