answer：Okay, so I have this quadratic equation, right? It's x squared plus p times x plus q equals zero. And the problem is asking for the values of p and q such that the equation has two distinct solutions, which are 2p and p plus q. Hmm, that sounds a bit tricky, but let me see if I can figure it out step by step.First, I remember that for a quadratic equation of the form ax² + bx + c = 0, the solutions can be found using the quadratic formula, but I also recall something called Vieta's formulas. Vieta's formulas relate the coefficients of the polynomial to the sum and product of its roots. Maybe that can help here.So, if the equation is x² + px + q = 0, then the sum of the roots should be equal to -p (since the coefficient of x is p, and Vieta's formula says the sum is -b/a, and here a is 1). Similarly, the product of the roots should be equal to q (since the constant term is q, and Vieta's formula says the product is c/a, which is q here).Alright, so the roots are given as 2p and p + q. Let me write down the sum and product of these roots.Sum of the roots: 2p + (p + q) = 3p + q.Product of the roots: 2p * (p + q) = 2p² + 2pq.According to Vieta's formulas, the sum of the roots is -p, and the product is q. So, I can set up two equations:1. 3p + q = -p2. 2p² + 2pq = qLet me simplify the first equation. If I move all terms to one side:3p + q + p = 04p + q = 0So, q = -4p.Okay, that's straightforward. Now, let's substitute q = -4p into the second equation.So, equation 2 becomes:2p² + 2p*(-4p) = -4pSimplify the left side:2p² - 8p² = -6p²So, the equation is:-6p² = -4pLet me bring all terms to one side:-6p² + 4p = 0I can factor out a common term, which is -2p:-2p(3p - 2) = 0So, either -2p = 0 or 3p - 2 = 0.If -2p = 0, then p = 0.If 3p - 2 = 0, then 3p = 2, so p = 2/3.Now, I need to check these solutions to see if they make sense in the context of the problem.First, let's consider p = 0.If p = 0, then from q = -4p, q = 0.So, the quadratic equation becomes x² + 0x + 0 = x² = 0.This equation has a double root at x = 0.But the problem specifies that the equation should have two distinct solutions, 2p and p + q.If p = 0, then 2p = 0 and p + q = 0 + 0 = 0. So, both roots are 0, which are not distinct. Therefore, p = 0 is not a valid solution because it doesn't give us two distinct roots.Now, let's check p = 2/3.From q = -4p, substituting p = 2/3:q = -4*(2/3) = -8/3.So, the quadratic equation becomes x² + (2/3)x - 8/3 = 0.Let me check if the roots are indeed 2p and p + q.First, 2p = 2*(2/3) = 4/3.Second, p + q = (2/3) + (-8/3) = (2 - 8)/3 = (-6)/3 = -2.So, the roots should be 4/3 and -2.Let me verify if these are indeed the roots of the equation x² + (2/3)x - 8/3 = 0.First, plug in x = 4/3:(4/3)² + (2/3)*(4/3) - 8/3 = 16/9 + 8/9 - 24/9 = (16 + 8 - 24)/9 = 0/9 = 0. So, that works.Next, plug in x = -2:(-2)² + (2/3)*(-2) - 8/3 = 4 - 4/3 - 8/3 = 4 - (4 + 8)/3 = 4 - 12/3 = 4 - 4 = 0. That also works.Great, so both roots satisfy the equation. Additionally, 4/3 and -2 are distinct, so the condition of having two distinct solutions is satisfied.Therefore, the valid solution is p = 2/3 and q = -8/3.I think that's it. I don't see any mistakes in my calculations, and both roots check out. So, I'm confident that these are the correct values for p and q.

answer：First, I need to determine the total number of pages Liam read during the first four days. Since he read an average of 42 pages per day, I can calculate this by multiplying 42 by 4.Next, I'll calculate the total number of pages he read in the following two days. With an average of 50 pages per day, I'll multiply 50 by 2.After finding the pages read in the first six days, I'll add the 30 pages he read on the last day to get the total number of pages in the book.Finally, I'll sum all these values to determine the total number of pages in the book.

answer：Okay, so I have this problem: A regular polygon has interior angles of 135 degrees. How many sides does the polygon have? Hmm, I need to figure out the number of sides, which is 'n'. First, I remember that the sum of the interior angles of a polygon is given by the formula 180(n-2) degrees. Yeah, that sounds right. So, if each interior angle is 135 degrees, then the sum of all the interior angles should also be 135 multiplied by the number of sides, which is 135n. So, I can set up an equation: 180(n-2) equals 135n. Let me write that down:180(n - 2) = 135nNow, I need to solve for 'n'. Let me expand the left side first:180n - 360 = 135nOkay, so now I have 180n minus 360 equals 135n. I want to get all the terms with 'n' on one side and the constants on the other. So, I'll subtract 135n from both sides:180n - 135n - 360 = 0That simplifies to:45n - 360 = 0Now, I can add 360 to both sides to isolate the term with 'n':45n = 360To find 'n', I divide both sides by 45:n = 360 / 45Calculating that, 360 divided by 45 is 8. So, n equals 8.Wait, does that make sense? Let me double-check. If a polygon has 8 sides, it's an octagon. The formula for each interior angle is (n-2)*180/n. Plugging in 8:(8-2)*180 / 8 = 6*180 / 8 = 1080 / 8 = 135 degrees. Yep, that matches the given interior angle. So, it looks like 8 sides is the correct answer.I think I've got it right. The polygon has 8 sides.

answer：Okay, so I've got this integral to solve: the definite integral from 0 to -1 of (sqrt(1 - x²) + x) dx. Hmm, that seems a bit tricky at first glance, but let me try to break it down step by step.First off, I remember that definite integrals can represent areas under curves, but in this case, the limits of integration are from 0 to -1. That's a bit unusual because usually, the lower limit is less than the upper limit. I think when the upper limit is smaller than the lower limit, the integral is the negative of the integral from the upper limit to the lower limit. So, maybe I can rewrite this integral to make it easier to handle.Let me write it down: ∫₀^{-1} (sqrt(1 - x²) + x) dx. Since the upper limit is -1 and the lower limit is 0, which is greater than -1, I can reverse the limits and add a negative sign. So, that becomes -∫_{-1}^0 (sqrt(1 - x²) + x) dx. That might make things simpler because integrating from -1 to 0 is more straightforward.Now, I can split this integral into two separate integrals because of the linearity of integrals. So, it becomes - [∫_{-1}^0 sqrt(1 - x²) dx + ∫_{-1}^0 x dx]. That way, I can handle each part individually.Starting with the first integral: ∫_{-1}^0 sqrt(1 - x²) dx. I recall that the integral of sqrt(1 - x²) dx from -1 to 1 is the area of a semicircle with radius 1, which is (π * 1²)/2 = π/2. But here, we're only integrating from -1 to 0, which should be half of that semicircle, right? So, that would be a quarter of the full circle's area. Therefore, ∫_{-1}^0 sqrt(1 - x²) dx should be π/4.Wait, let me double-check that. If the integral from -1 to 1 is π/2, then from -1 to 0 should indeed be half of that, which is π/4. Yeah, that makes sense because the area from -1 to 0 is a quarter-circle in the second quadrant.Now, moving on to the second integral: ∫_{-1}^0 x dx. That's straightforward. The integral of x with respect to x is (x²)/2. So, evaluating from -1 to 0, it would be (0²)/2 - ((-1)²)/2 = 0 - (1)/2 = -1/2.Putting it all together, the original integral becomes - [π/4 + (-1/2)] = - [π/4 - 1/2]. Wait, that doesn't seem right. Let me check my steps again.Oh, I see where I might have messed up. When I split the integral, I had - [∫_{-1}^0 sqrt(1 - x²) dx + ∫_{-1}^0 x dx]. So, substituting the values I found: ∫_{-1}^0 sqrt(1 - x²) dx is π/4, and ∫_{-1}^0 x dx is -1/2. So, adding those together: π/4 + (-1/2) = π/4 - 1/2. Then, applying the negative sign outside: - (π/4 - 1/2) = -π/4 + 1/2.Wait, that would make the integral equal to 1/2 - π/4. But I'm not sure if that's correct. Let me think about it differently.Alternatively, maybe I should have kept the original limits and not reversed them. Let's try that approach. So, ∫₀^{-1} (sqrt(1 - x²) + x) dx. Since integrating from a higher limit to a lower limit is the negative of integrating from the lower to the higher, I can write it as -∫_{-1}^0 (sqrt(1 - x²) + x) dx, which is what I did before.But perhaps another way is to recognize that sqrt(1 - x²) is an even function, meaning it's symmetric about the y-axis. So, integrating from -1 to 0 is the same as integrating from 0 to 1. Therefore, ∫_{-1}^0 sqrt(1 - x²) dx = ∫₀^1 sqrt(1 - x²) dx = π/4, which matches what I found earlier.And for the integral of x from -1 to 0, since x is an odd function, integrating from -a to a would give zero, but here we're only going from -1 to 0, so it's half of that. Wait, no, that's not quite right. The integral of x from -1 to 0 is indeed -1/2, as I calculated before.So, putting it all together: - [π/4 + (-1/2)] = -π/4 + 1/2, which simplifies to 1/2 - π/4.But wait, is that the final answer? Let me verify by considering the original integral without reversing the limits. So, ∫₀^{-1} (sqrt(1 - x²) + x) dx. If I integrate from 0 to -1, which is moving to the left on the x-axis, the integral would represent the area under the curve from 0 to -1, but since we're moving in the negative direction, it's the negative of the area from -1 to 0.So, the integral is negative of the area from -1 to 0, which is - [π/4 - 1/2] = -π/4 + 1/2, which is the same as 1/2 - π/4.Alternatively, I can think of it as integrating from 0 to -1, which is the same as integrating from 0 to 1 with a substitution. Let me try that substitution method to confirm.Let me set u = -x. Then, when x = 0, u = 0, and when x = -1, u = 1. Also, dx = -du. So, the integral becomes ∫₀^1 (sqrt(1 - (-u)²) + (-u)) (-du). Simplifying, sqrt(1 - u²) remains the same, and -u becomes -u. The negative sign from dx = -du flips the integral limits back to 0 to 1, so we have ∫₀^1 (sqrt(1 - u²) - u) du.Now, integrating term by term: ∫₀^1 sqrt(1 - u²) du is π/4, as before, and ∫₀^1 (-u) du is - (u²)/2 evaluated from 0 to 1, which is - (1/2 - 0) = -1/2.So, adding those together: π/4 - 1/2. But wait, that's the same as before, but without the negative sign. Hmm, that's confusing. Let me see.Wait, no, because when I did the substitution, I ended up with ∫₀^1 (sqrt(1 - u²) - u) du, which is π/4 - 1/2. But the original integral was ∫₀^{-1} (sqrt(1 - x²) + x) dx, which after substitution became ∫₀^1 (sqrt(1 - u²) - u) du. So, the result is π/4 - 1/2.But earlier, when I reversed the limits, I got 1/2 - π/4. These two results are negatives of each other. That can't be right. I must have made a mistake in one of the approaches.Let me go back to the substitution method. When I set u = -x, then x = -u, and dx = -du. So, the integral becomes ∫₀^{-1} (sqrt(1 - x²) + x) dx = ∫₀^1 (sqrt(1 - (-u)²) + (-u)) (-du).Simplifying inside the integral: sqrt(1 - u²) remains, and (-u) is just -u. The (-du) flips the integral limits from 0 to 1 to 1 to 0, but since we're integrating from 0 to 1, we can write it as ∫₀^1 (sqrt(1 - u²) - u) du.So, that integral is π/4 - 1/2. But wait, the original integral was from 0 to -1, which is a negative interval. So, does that mean the result is negative?Wait, no, because when I did the substitution, I accounted for the negative by flipping the limits and introducing the negative sign, which cancels out. So, the result of the substitution is ∫₀^{-1} (sqrt(1 - x²) + x) dx = π/4 - 1/2.But earlier, when I reversed the limits, I got - [π/4 - 1/2] = -π/4 + 1/2, which is 1/2 - π/4. Now I'm confused because two different methods are giving me two different results.Let me check the substitution method again. When I set u = -x, then x = -u, dx = -du. So, when x = 0, u = 0; when x = -1, u = 1. So, the integral becomes ∫₀^1 (sqrt(1 - (-u)²) + (-u)) (-du). That simplifies to ∫₀^1 (sqrt(1 - u²) - u) (-du) = -∫₀^1 (sqrt(1 - u²) - u) du.Wait, I think I missed the negative sign in front of the integral. So, it's -∫₀^1 (sqrt(1 - u²) - u) du. Which is - [π/4 - 1/2] = -π/4 + 1/2, which is 1/2 - π/4.Ah, there we go. So, the substitution method also gives me 1/2 - π/4, which matches the earlier result after reversing the limits. So, both methods agree that the integral is 1/2 - π/4.Therefore, the value of the definite integral ∫₀^{-1} (sqrt(1 - x²) + x) dx is 1/2 - π/4.Wait, but earlier I thought of it as - [π/4 - 1/2] = 1/2 - π/4, which is the same. So, both approaches confirm that the result is 1/2 - π/4.Just to be thorough, let me compute the integral directly without reversing the limits or substitution.So, ∫₀^{-1} (sqrt(1 - x²) + x) dx. Let's integrate term by term.First, ∫ sqrt(1 - x²) dx. The antiderivative of sqrt(1 - x²) is (x/2) sqrt(1 - x²) + (1/2) arcsin(x) + C. Let me verify that by differentiation:d/dx [ (x/2) sqrt(1 - x²) + (1/2) arcsin(x) ] = (1/2) sqrt(1 - x²) + (x/2)(-2x)/(2 sqrt(1 - x²)) + (1/2)(1/sqrt(1 - x²)).Simplifying: (1/2) sqrt(1 - x²) - (x²)/(2 sqrt(1 - x²)) + 1/(2 sqrt(1 - x²)).Combine terms: [ (1 - x²) - x² + 1 ] / (2 sqrt(1 - x²)) = [2 - 2x²]/(2 sqrt(1 - x²)) = (1 - x²)/sqrt(1 - x²) = sqrt(1 - x²). So, yes, that's correct.Now, the integral of x dx is (x²)/2.So, putting it all together, the antiderivative F(x) is (x/2) sqrt(1 - x²) + (1/2) arcsin(x) + (x²)/2.Now, evaluate F(x) from 0 to -1.First, at x = -1:F(-1) = (-1/2) sqrt(1 - (-1)²) + (1/2) arcsin(-1) + ((-1)²)/2.Simplify:sqrt(1 - 1) = 0, so first term is 0.arcsin(-1) = -π/2, so second term is (1/2)(-π/2) = -π/4.Third term: (1)/2 = 1/2.So, F(-1) = 0 - π/4 + 1/2.Now, at x = 0:F(0) = (0/2) sqrt(1 - 0) + (1/2) arcsin(0) + (0²)/2 = 0 + 0 + 0 = 0.Therefore, the integral ∫₀^{-1} (sqrt(1 - x²) + x) dx = F(-1) - F(0) = ( -π/4 + 1/2 ) - 0 = 1/2 - π/4.So, that confirms it again. The value is 1/2 - π/4.Just to make sure, let me approximate the numerical value to see if it makes sense. π is approximately 3.1416, so π/4 is about 0.7854. Then, 1/2 is 0.5. So, 0.5 - 0.7854 ≈ -0.2854.Wait, that's a negative number. Is that possible? Let me think about the function we're integrating.From x = 0 to x = -1, the function sqrt(1 - x²) is the upper half of the unit circle, which is positive, and x is negative in that interval. So, the integrand is sqrt(1 - x²) + x, which is positive (since sqrt is positive and x is negative, but sqrt is larger in magnitude near x=0, but as x approaches -1, x becomes -1, and sqrt(1 - (-1)^2) is 0. So, at x = -1, the integrand is 0 + (-1) = -1.Wait, so the integrand starts at x=0: sqrt(1 - 0) + 0 = 1, and goes down to x=-1: 0 + (-1) = -1. So, the function starts at 1, decreases, crosses zero somewhere, and ends at -1.Therefore, the area under the curve from 0 to -1 would have both positive and negative parts. But since we're integrating from 0 to -1, which is moving to the left, the integral would be the net area, which could be negative.But in our calculation, we got 1/2 - π/4 ≈ -0.2854, which is negative. That makes sense because the area below the curve (where the function is negative) might outweigh the area above the curve (where the function is positive) in this interval.Alternatively, if we consider the integral as the net signed area, it's possible to have a negative result.So, all in all, after multiple methods and checks, I'm confident that the value of the integral is 1/2 - π/4.

answer：Okay, so I have this problem here: the equation is a cubic, ax³ + bx² + cx + d = 0, and the roots are given as 2, 4, and 5. I need to find the ratio c/d. Hmm, alright, let me think about how to approach this.First, I remember that for polynomials, especially cubic ones, Vieta's formulas are really useful. Vieta's formulas relate the coefficients of the polynomial to sums and products of its roots. So, I should probably use those to find expressions for c and d in terms of the roots.Let me recall Vieta's formulas for a cubic equation. For a general cubic equation ax³ + bx² + cx + d = 0, the sum of the roots is -b/a, the sum of the products of the roots taken two at a time is c/a, and the product of the roots is -d/a. So, if the roots are r1, r2, r3, then:1. r1 + r2 + r3 = -b/a2. r1r2 + r1r3 + r2r3 = c/a3. r1r2r3 = -d/aAlright, so in this problem, the roots are 2, 4, and 5. Let me assign them as r1=2, r2=4, r3=5.So, first, let me compute the sum of the roots: 2 + 4 + 5. That's 11. According to Vieta, this should be equal to -b/a. But I don't need b for this problem, so maybe I can skip that part.Next, the sum of the products of the roots taken two at a time. That would be (2*4) + (2*5) + (4*5). Let me compute each term:2*4 is 8,2*5 is 10,4*5 is 20.Adding those together: 8 + 10 + 20 = 38. So, according to Vieta, this sum is equal to c/a. So, c/a = 38.Then, the product of the roots is 2*4*5. Let me compute that: 2*4 is 8, and 8*5 is 40. So, the product is 40. According to Vieta, this is equal to -d/a. So, -d/a = 40, which means d/a = -40.Now, I need to find c/d. Let me see. I have c/a = 38 and d/a = -40. So, if I take c/a divided by d/a, that would be (c/a) / (d/a) = c/d. Because the a's would cancel out.So, let me compute that: (38) / (-40) = -38/40. I can simplify that fraction by dividing numerator and denominator by 2: -19/20.Wait, but the question asks for c/d, not necessarily a positive number. So, is it okay that it's negative? Let me check my steps.First, c/a is 38, which is positive. Then, d/a is -40, which is negative. So, c/d is (c/a)/(d/a) = 38/(-40) = -19/20. So, that seems correct.But let me double-check my calculations because sometimes signs can be tricky.Sum of products two at a time: 2*4 + 2*5 + 4*5 = 8 + 10 + 20 = 38. That seems right.Product of roots: 2*4*5 = 40. Correct.Vieta's formula says that the sum of products two at a time is c/a, so c/a = 38.Product of roots is -d/a, so 40 = -d/a, which implies d/a = -40.Therefore, c/d = (c/a)/(d/a) = 38/(-40) = -19/20.Wait, but in the original problem, the equation is ax³ + bx² + cx + d = 0. So, the leading coefficient is a, and the constant term is d. So, the product of the roots is -d/a, which is 40, so d = -40a. Similarly, the sum of products two at a time is c/a = 38, so c = 38a.Therefore, c/d is (38a)/(-40a) = 38/(-40) = -19/20.So, that seems consistent. Therefore, c/d is -19/20.But wait, the problem didn't specify whether a is positive or not. But since a is just a coefficient, unless specified otherwise, we can assume a ≠ 0, but its sign doesn't affect the ratio c/d because both c and d are scaled by a, so the a cancels out.Therefore, the ratio c/d is indeed -19/20.Wait, but in the initial problem statement, the equation is given as ax³ + bx² + cx + d = 0, so the coefficients are a, b, c, d. So, the leading coefficient is a, and the constant term is d. So, Vieta's formula applies as I used.Alternatively, sometimes people write cubic equations as x³ + px² + qx + r = 0, where the leading coefficient is 1. In that case, the formulas are similar but without the a in the denominator. But in this problem, the leading coefficient is a, so I have to keep that in mind.So, to recap:Given roots 2, 4, 5,Sum of roots: 11 = -b/a,Sum of products two at a time: 38 = c/a,Product of roots: 40 = -d/a.Therefore, c = 38a,d = -40a,Thus, c/d = 38a / (-40a) = -38/40 = -19/20.So, yeah, that's the answer.I think that's solid. I don't see any mistakes in my reasoning. So, I'm confident that c/d is -19/20.**Final Answer**The value of frac{c}{d} is boxed{-dfrac{19}{20}}.

answer：Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem gives me an equation involving angles B and C: cos(B - C) - 2 sin B sin C = -1/2. I need to find the measure of angle A first, and then, given that a = 5 and b = 4, calculate the area of the triangle.Starting with part (1): Find the measure of angle A.Hmm, the equation given is cos(B - C) - 2 sin B sin C = -1/2. I remember some trigonometric identities that might help here. Let me think... There's a formula for cos(B - C), which is cos B cos C + sin B sin C. So, if I substitute that into the equation, it becomes:cos B cos C + sin B sin C - 2 sin B sin C = -1/2.Simplifying that, the sin B sin C terms combine:cos B cos C - sin B sin C = -1/2.Wait, that looks familiar. Isn't that the formula for cos(B + C)? Yes, because cos(B + C) = cos B cos C - sin B sin C. So, substituting that in:cos(B + C) = -1/2.Alright, so cos(B + C) = -1/2. Now, in a triangle, the sum of angles B and C is equal to 180 degrees minus angle A, because the total sum of angles in a triangle is 180 degrees. So, B + C = 180° - A. Therefore, cos(180° - A) = -1/2.I also remember that cos(180° - x) = -cos x. So, substituting that in:-cos A = -1/2.Multiplying both sides by -1:cos A = 1/2.So, cos A = 1/2. What angle A between 0° and 180° has a cosine of 1/2? That's 60°, because cos 60° = 0.5. So, angle A is 60 degrees.Wait, let me double-check. If cos A = 1/2, then A could be 60° or 300°, but since in a triangle angles are less than 180°, 300° is not possible. So, yes, A is 60°. That seems right.So, part (1) is solved: angle A is 60 degrees.Moving on to part (2): Given that a = 5 and b = 4, calculate the area of triangle ABC.Alright, so sides a and b are given as 5 and 4 respectively. Side a is opposite angle A, which we found to be 60°, and side b is opposite angle B. So, angle B is opposite side b = 4.I need to find the area. The formula for the area of a triangle is (1/2)ab sin C, but I don't know angle C or side c. Alternatively, I can use the formula (1/2)bc sin A, since I know angle A. But I don't know side c yet. So, maybe I should find side c first.To find side c, I can use the Law of Cosines. The Law of Cosines states that a² = b² + c² - 2bc cos A. Plugging in the known values:5² = 4² + c² - 2 * 4 * c * cos 60°.Calculating each term:25 = 16 + c² - 8c * (1/2).Because cos 60° is 0.5, so 8c * 0.5 is 4c.So, simplifying:25 = 16 + c² - 4c.Subtracting 25 from both sides:0 = 16 + c² - 4c - 25.Simplifying further:0 = c² - 4c - 9.So, the quadratic equation is c² - 4c - 9 = 0.To solve for c, I can use the quadratic formula: c = [4 ± sqrt(16 + 36)] / 2.Because the quadratic is c² - 4c - 9 = 0, so a = 1, b = -4, c = -9.So discriminant D = (-4)^2 - 4*1*(-9) = 16 + 36 = 52.Therefore, c = [4 ± sqrt(52)] / 2.Simplify sqrt(52): sqrt(4*13) = 2 sqrt(13). So,c = [4 ± 2 sqrt(13)] / 2 = 2 ± sqrt(13).Since side lengths can't be negative, we take the positive solution: c = 2 + sqrt(13).So, side c is 2 + sqrt(13).Now, to find the area, I can use the formula (1/2)ab sin C, but I don't know angle C. Alternatively, since I know sides a, b, and c, I can use Heron's formula, but that might be more complicated. Alternatively, since I know angle A, I can use the formula (1/2)bc sin A.Wait, actually, let's see: if I use (1/2)ab sin C, I don't know angle C. If I use (1/2)bc sin A, I know angle A is 60°, and I know sides b and c.Yes, that seems better. So, area = (1/2) * b * c * sin A.Plugging in the values: (1/2) * 4 * (2 + sqrt(13)) * sin 60°.Sin 60° is sqrt(3)/2. So,Area = (1/2) * 4 * (2 + sqrt(13)) * (sqrt(3)/2).Simplify step by step:First, (1/2) * 4 = 2.Then, 2 * (2 + sqrt(13)) = 4 + 2 sqrt(13).Then, multiply by (sqrt(3)/2):(4 + 2 sqrt(13)) * (sqrt(3)/2) = [4 * sqrt(3) + 2 sqrt(13) * sqrt(3)] / 2.Simplify each term:4 sqrt(3)/2 = 2 sqrt(3).2 sqrt(13) sqrt(3)/2 = sqrt(13) sqrt(3) = sqrt(39).So, adding them together: 2 sqrt(3) + sqrt(39).Therefore, the area is 2 sqrt(3) + sqrt(39).Wait, let me double-check the calculations:Starting with area = (1/2) * 4 * (2 + sqrt(13)) * (sqrt(3)/2).Multiply (1/2) and 4: that's 2.Multiply 2 and (2 + sqrt(13)): that's 4 + 2 sqrt(13).Multiply that by sqrt(3)/2: (4 + 2 sqrt(13)) * sqrt(3)/2.Which is (4 sqrt(3) + 2 sqrt(13) sqrt(3)) / 2.Divide each term by 2: 2 sqrt(3) + sqrt(13) sqrt(3).Which is 2 sqrt(3) + sqrt(39). Yes, that seems correct.Alternatively, I can compute the area using Heron's formula to verify.First, compute the semi-perimeter: s = (a + b + c)/2 = (5 + 4 + 2 + sqrt(13))/2 = (11 + sqrt(13))/2.Then, area = sqrt[s(s - a)(s - b)(s - c)].Plugging in:sqrt[ (11 + sqrt(13))/2 * ( (11 + sqrt(13))/2 - 5 ) * ( (11 + sqrt(13))/2 - 4 ) * ( (11 + sqrt(13))/2 - (2 + sqrt(13)) ) ].Simplify each term:s = (11 + sqrt(13))/2.s - a = (11 + sqrt(13))/2 - 5 = (11 + sqrt(13) - 10)/2 = (1 + sqrt(13))/2.s - b = (11 + sqrt(13))/2 - 4 = (11 + sqrt(13) - 8)/2 = (3 + sqrt(13))/2.s - c = (11 + sqrt(13))/2 - (2 + sqrt(13)) = (11 + sqrt(13) - 4 - 2 sqrt(13))/2 = (7 - sqrt(13))/2.So, the area is sqrt[ (11 + sqrt(13))/2 * (1 + sqrt(13))/2 * (3 + sqrt(13))/2 * (7 - sqrt(13))/2 ].That's a bit complicated, but let's compute it step by step.First, multiply the numerators:(11 + sqrt(13))(1 + sqrt(13))(3 + sqrt(13))(7 - sqrt(13)).This seems messy, but maybe we can pair them:First pair: (11 + sqrt(13))(7 - sqrt(13)).Second pair: (1 + sqrt(13))(3 + sqrt(13)).Compute first pair:(11)(7) + 11(-sqrt(13)) + sqrt(13)(7) + sqrt(13)(-sqrt(13)).= 77 - 11 sqrt(13) + 7 sqrt(13) - 13.Simplify:77 - 13 = 64.-11 sqrt(13) + 7 sqrt(13) = -4 sqrt(13).So, first pair is 64 - 4 sqrt(13).Second pair:(1)(3) + 1(sqrt(13)) + sqrt(13)(3) + sqrt(13)(sqrt(13)).= 3 + sqrt(13) + 3 sqrt(13) + 13.Simplify:3 + 13 = 16.sqrt(13) + 3 sqrt(13) = 4 sqrt(13).So, second pair is 16 + 4 sqrt(13).Now, multiply these two results:(64 - 4 sqrt(13))(16 + 4 sqrt(13)).Let me compute this:64*16 + 64*4 sqrt(13) - 4 sqrt(13)*16 - 4 sqrt(13)*4 sqrt(13).Compute each term:64*16 = 1024.64*4 sqrt(13) = 256 sqrt(13).-4 sqrt(13)*16 = -64 sqrt(13).-4 sqrt(13)*4 sqrt(13) = -16 * 13 = -208.Now, add all terms together:1024 + 256 sqrt(13) - 64 sqrt(13) - 208.Simplify:1024 - 208 = 816.256 sqrt(13) - 64 sqrt(13) = 192 sqrt(13).So, total is 816 + 192 sqrt(13).Therefore, the product of the numerators is 816 + 192 sqrt(13).Now, the denominator is 2^4 = 16, since each term was divided by 2, and there are four terms.So, the area squared is (816 + 192 sqrt(13))/16.Simplify:816 / 16 = 51.192 / 16 = 12.So, area squared is 51 + 12 sqrt(13).Therefore, area is sqrt(51 + 12 sqrt(13)).Hmm, let's see if this is equal to 2 sqrt(3) + sqrt(39).Compute (2 sqrt(3) + sqrt(39))²:= (2 sqrt(3))² + 2*(2 sqrt(3))*(sqrt(39)) + (sqrt(39))².= 4*3 + 4 sqrt(117) + 39.= 12 + 4 sqrt(117) + 39.= 51 + 4 sqrt(117).But sqrt(117) = sqrt(9*13) = 3 sqrt(13).So, 4 sqrt(117) = 12 sqrt(13).Therefore, (2 sqrt(3) + sqrt(39))² = 51 + 12 sqrt(13), which matches the Heron's formula result.So, sqrt(51 + 12 sqrt(13)) is indeed equal to 2 sqrt(3) + sqrt(39). Therefore, the area is 2 sqrt(3) + sqrt(39).So, that confirms the area calculation.Therefore, the answers are:(1) Angle A is 60 degrees.(2) The area of triangle ABC is 2 sqrt(3) + sqrt(39).**Final Answer**(1) The measure of angle ( A ) is (boxed{60^circ}).(2) The area of triangle ( ABC ) is (boxed{2sqrt{3} + sqrt{39}}).