answer：First, I need to determine the total amount of rice in ounces. Since there are 4 containers and each contains 25 ounces, I multiply 25 by 4 to get the total ounces.Next, I convert the total ounces to pounds. Knowing that 1 pound equals 16 ounces, I divide the total ounces by 16 to find the total weight in pounds.

answer：First, I need to determine the total number of subjects Ashok has. I'll denote the total number of subjects as ( n ).Ashok's average marks across all subjects is 78, so the total marks he secured in all subjects is ( 78n ).The average of marks in the first ( n-1 ) subjects is 74, which means the total marks in these subjects is ( 74(n-1) ).Ashok scored 98 marks in the last subject.The total marks in all subjects can also be expressed as the sum of the total marks in the first ( n-1 ) subjects and the marks in the last subject:[78n = 74(n-1) + 98]Expanding and simplifying the equation:[78n = 74n - 74 + 98][78n - 74n = 98 - 74][4n = 24][n = frac{24}{4} = 6]Therefore, there are 6 subjects in total.

answer：To calculate (i^{2011}), I recognize that the powers of (i) follow a repeating cycle every four exponents: (i^1 = i), (i^2 = -1), (i^3 = -i), and (i^4 = 1). This cyclical pattern continues indefinitely.First, I determine how many complete cycles of four are in 2011. Dividing 2011 by 4 gives 502 complete cycles with a remainder of 3. This means (i^{2011}) can be expressed as (i^{4 times 502 + 3}).Next, I simplify this expression by separating the exponent into two parts: ((i^4)^{502} times i^3). Since (i^4 = 1), the term ((i^4)^{502}) simplifies to (1^{502}), which is just 1. Therefore, the expression reduces to (1 times i^3).Finally, I calculate (i^3), which is (-i). Combining these results, (i^{2011} = -i).

answer：First, I need to evaluate each option to determine which operation is correct.For option A: ( m^2 + m^3 = m^5 ). Adding ( m^2 ) and ( m^3 ) does not result in ( m^5 ), so this is incorrect.For option B: ( m^2 cdot m^3 = m^5 ). When multiplying like bases with exponents, I add the exponents: ( m^{2+3} = m^5 ). This is correct.For option C: ( (m^2)^3 = m^5 ). When raising a power to another power, I multiply the exponents: ( m^{2 times 3} = m^6 ). This is incorrect.For option D: ( (m^2)^3 = m^8 ). Again, multiplying the exponents gives ( m^6 ), not ( m^8 ). This is also incorrect.After evaluating all options, only option B is correct.

answer：First, I need to determine the production rate per person per hour. Given that ( x ) men working ( x ) hours a day for ( x ) days produce ( x ) articles, I can set up the equation:[x times x times x times k = x]Simplifying this, I find:[x^3 k = x]Solving for ( k ), the production rate per person per hour, I get:[k = frac{1}{x^2}]Next, I apply this production rate to the new scenario where ( x ) men work ( x ) hours a day for ( y ) days. Using the same rate ( k ), the number of articles produced is:[x times x times y times frac{1}{x^2} = y]Therefore, the number of articles produced in ( y ) days is ( y ).

answer：Okay, so I have this problem about a right triangle with an acute angle α, and there's a circle that touches the hypotenuse and the extensions of the two legs, and the radius of that circle is R. I need to find the length of the hypotenuse. Hmm, let me try to visualize this.First, I imagine a right triangle ABC with the right angle at C. So, sides AC and BC are the legs, and AB is the hypotenuse. The circle touches the hypotenuse AB and the extensions of the legs AC and BC. That means the circle is outside the triangle, touching the hypotenuse and the extensions of the legs beyond the triangle. So, it's like an ex-circle or something?Wait, actually, in a right triangle, there are ex-circles opposite each vertex. Maybe this circle is one of those ex-circles. But I'm not entirely sure. Let me think.The circle touches the hypotenuse and the extensions of the two legs. So, it's tangent to AB and the lines extending from AC and BC beyond the triangle. That makes me think it's an ex-circle opposite to the right angle. In a right triangle, the ex-circle opposite the right angle touches the hypotenuse and the extensions of the legs. So, maybe that's what this circle is.If that's the case, then the radius R of this ex-circle can be related to the sides of the triangle. I remember that for ex-circles, the radius can be expressed in terms of the triangle's sides and its semiperimeter. Let me recall the formula.For an ex-circle opposite to angle C, the radius is given by R = Δ / (s - c), where Δ is the area of the triangle, s is the semiperimeter, and c is the side opposite angle C. In this case, since the ex-circle is opposite the right angle, which is at C, the side opposite is AB, which is the hypotenuse.So, let me denote the sides as follows: let’s say AC = b, BC = a, and AB = c. Then, the semiperimeter s = (a + b + c)/2. The area Δ of the triangle is (a*b)/2.So, plugging into the formula for the ex-radius opposite C: R = Δ / (s - c). Let's compute s - c: s - c = (a + b + c)/2 - c = (a + b - c)/2.So, R = (a*b/2) / ((a + b - c)/2) = (a*b) / (a + b - c). So, R = (a*b) / (a + b - c).But I need to find c, the hypotenuse, in terms of R and α. So, I need to express a, b, and c in terms of α and maybe R.Since it's a right triangle with angle α at A, then angle at B is 90° - α. So, sides can be expressed in terms of c and α.So, AC = b = c*cos α, and BC = a = c*sin α.So, let me substitute a and b in terms of c and α into the expression for R.So, R = (a*b) / (a + b - c) = (c*sin α * c*cos α) / (c*sin α + c*cos α - c).Simplify numerator and denominator:Numerator: c² sin α cos αDenominator: c(sin α + cos α - 1)So, R = (c² sin α cos α) / [c(sin α + cos α - 1)] = (c sin α cos α) / (sin α + cos α - 1)So, R = [c sin α cos α] / (sin α + cos α - 1)I need to solve for c. So, let's write:c = R * (sin α + cos α - 1) / (sin α cos α)Hmm, that seems a bit complicated. Maybe I can simplify this expression.Let me factor the numerator:sin α + cos α - 1. Hmm, not sure if that factors nicely. Maybe I can write it as (sin α + cos α) - 1.Alternatively, perhaps I can express sin α + cos α in terms of a single trigonometric function. I know that sin α + cos α = √2 sin(α + 45°), but I'm not sure if that helps here.Alternatively, let's see if we can manipulate the expression:c = R * (sin α + cos α - 1) / (sin α cos α)Let me split the fraction:c = R * [ (sin α + cos α) / (sin α cos α) - 1 / (sin α cos α) ]Simplify each term:(sin α + cos α) / (sin α cos α) = (sin α)/(sin α cos α) + (cos α)/(sin α cos α) = 1/cos α + 1/sin α = sec α + csc αSimilarly, 1 / (sin α cos α) = 2 / sin(2α), since sin(2α) = 2 sin α cos α.So, putting it back:c = R * [ sec α + csc α - 2 / sin(2α) ]Hmm, not sure if that's helpful. Maybe another approach.Wait, let's go back to the formula R = [c sin α cos α] / (sin α + cos α - 1). Maybe instead of trying to solve for c directly, I can manipulate this equation.Let me denote t = sin α + cos α. Then, t = sin α + cos α, and t² = sin² α + 2 sin α cos α + cos² α = 1 + 2 sin α cos α. So, sin α cos α = (t² - 1)/2.So, substituting back into R:R = [c * (t² - 1)/2 ] / (t - 1) = [c (t - 1)(t + 1)/2 ] / (t - 1) = [c (t + 1)/2 ]So, R = c (t + 1)/2But t = sin α + cos α, so:R = c (sin α + cos α + 1)/2Therefore, solving for c:c = 2R / (sin α + cos α + 1)Hmm, that seems simpler. Let me check my steps to make sure I didn't make a mistake.Starting from R = [c sin α cos α] / (sin α + cos α - 1)Let t = sin α + cos αThen, sin α cos α = (t² - 1)/2So, R = [c * (t² - 1)/2 ] / (t - 1) = [c (t - 1)(t + 1)/2 ] / (t - 1) = c (t + 1)/2Yes, that cancels out (t - 1) in numerator and denominator, assuming t ≠ 1, which it isn't because sin α + cos α = 1 only for specific α, but generally not.So, R = c (t + 1)/2, so c = 2R / (t + 1) = 2R / (sin α + cos α + 1)Okay, that seems correct. So, c = 2R / (1 + sin α + cos α)Alternatively, we can rationalize the denominator or express it differently.Let me see if I can write 1 + sin α + cos α in another form.We know that 1 + sin α = 2 sin(α/2 + 45°)^2 or something like that, but maybe it's better to use the identity:1 + sin α + cos α = 2 cos(α/2) (sin(α/2) + cos(α/2))Wait, let me recall that:1 + sin α = 2 sin(α/2 + 45°)^2, but I'm not sure. Alternatively, perhaps factor it as:1 + sin α + cos α = 1 + √2 sin(α + 45°)But I don't know if that helps.Alternatively, let's multiply numerator and denominator by (1 + sin α - cos α):c = 2R / (1 + sin α + cos α) * (1 + sin α - cos α)/(1 + sin α - cos α)So, numerator becomes 2R (1 + sin α - cos α)Denominator becomes (1 + sin α)^2 - cos² α = 1 + 2 sin α + sin² α - cos² αSimplify denominator:1 + 2 sin α + sin² α - cos² αWe know that sin² α - cos² α = -cos(2α), but maybe better to express in terms of sin α.Alternatively, 1 - cos² α = sin² α, so:Denominator = sin² α + 2 sin α + sin² α = 2 sin² α + 2 sin α = 2 sin α (sin α + 1)Wait, let me check:Wait, denominator is (1 + sin α)^2 - cos² α = 1 + 2 sin α + sin² α - cos² αBut 1 - cos² α = sin² α, so:Denominator = sin² α + 2 sin α + sin² α = 2 sin² α + 2 sin α = 2 sin α (sin α + 1)Yes, that's correct.So, denominator = 2 sin α (1 + sin α)Numerator = 2R (1 + sin α - cos α)So, c = [2R (1 + sin α - cos α)] / [2 sin α (1 + sin α)] = [R (1 + sin α - cos α)] / [sin α (1 + sin α)]Simplify numerator:1 + sin α - cos α = (1 - cos α) + sin αHmm, not sure if that helps. Alternatively, factor:1 + sin α - cos α = 1 - cos α + sin αI know that 1 - cos α = 2 sin²(α/2), and sin α = 2 sin(α/2) cos(α/2). So:1 - cos α + sin α = 2 sin²(α/2) + 2 sin(α/2) cos(α/2) = 2 sin(α/2) [sin(α/2) + cos(α/2)]Similarly, denominator:sin α (1 + sin α) = 2 sin(α/2) cos(α/2) * (1 + 2 sin(α/2) cos(α/2))Wait, this might complicate things more. Maybe it's better to leave it as is.So, c = [R (1 + sin α - cos α)] / [sin α (1 + sin α)]Alternatively, let's see if we can write this as:c = R [ (1 + sin α - cos α) / (sin α (1 + sin α)) ]Let me split the fraction:= R [ (1 + sin α) / (sin α (1 + sin α)) - cos α / (sin α (1 + sin α)) ]Simplify:= R [ 1 / sin α - cos α / (sin α (1 + sin α)) ]= R [ csc α - (cos α) / (sin α (1 + sin α)) ]Hmm, not sure if that helps. Maybe another approach.Wait, going back to the earlier expression: c = 2R / (1 + sin α + cos α). Maybe that's the simplest form.Alternatively, we can write 1 + sin α + cos α as √2 sin(α + 45°) + 1, but I don't think that necessarily simplifies it.Alternatively, let's rationalize the denominator by multiplying numerator and denominator by (1 + sin α - cos α), but we already did that and it didn't lead to much simplification.Alternatively, perhaps express in terms of tan(α/2). Let me try that.Let me recall that sin α = 2 tan(α/2) / (1 + tan²(α/2)), and cos α = (1 - tan²(α/2)) / (1 + tan²(α/2)). Let me denote t = tan(α/2). Then:sin α = 2t / (1 + t²)cos α = (1 - t²) / (1 + t²)So, 1 + sin α + cos α = 1 + 2t/(1 + t²) + (1 - t²)/(1 + t²) = [ (1 + t²) + 2t + (1 - t²) ] / (1 + t²) = (2 + 2t) / (1 + t²) = 2(1 + t) / (1 + t²)So, c = 2R / [2(1 + t)/(1 + t²)] = 2R * (1 + t²)/(2(1 + t)) ) = R (1 + t²)/(1 + t)Simplify (1 + t²)/(1 + t):(1 + t²)/(1 + t) = (1 + t²)/(1 + t) = [ (1 + t)(1 - t) + 2t ] / (1 + t) = (1 - t + 2t) = 1 + t, but that's not correct because (1 + t²) = (1 + t)(1 - t) + 2t. Wait, actually, (1 + t²) = (1 + t)^2 - 2t. Hmm, not sure.Wait, let me perform polynomial division:Divide 1 + t² by 1 + t.1 + t² divided by 1 + t:1 + t goes into t² + 1 how many times? Well, t² / t = t. Multiply (1 + t) by t: t + t². Subtract from t² + 1: (t² + 1) - (t² + t) = -t + 1.Now, the remainder is -t + 1, which is degree 1, same as divisor. So, we can write:(1 + t²)/(1 + t) = t + (-t + 1)/(1 + t)But that doesn't seem helpful. Alternatively, factor:1 + t² = (t + i)(t - i), but that's complex numbers, which isn't helpful here.Alternatively, perhaps leave it as is.So, c = R (1 + t²)/(1 + t). Since t = tan(α/2), we can write:c = R (1 + tan²(α/2)) / (1 + tan(α/2))But 1 + tan²(α/2) = sec²(α/2), so:c = R sec²(α/2) / (1 + tan(α/2)) = R / [cos²(α/2) (1 + tan(α/2)) ]Hmm, maybe express 1 + tan(α/2) in terms of sine and cosine:1 + tan(α/2) = 1 + sin(α/2)/cos(α/2) = [cos(α/2) + sin(α/2)] / cos(α/2)So, c = R / [cos²(α/2) * [cos(α/2) + sin(α/2)] / cos(α/2) ] = R / [cos(α/2) (cos(α/2) + sin(α/2)) ]Simplify:c = R / [cos(α/2) (cos(α/2) + sin(α/2)) ]Alternatively, factor out cos(α/2):c = R / [cos²(α/2) + cos(α/2) sin(α/2) ]But I don't see an immediate simplification. Maybe another approach.Wait, going back to the original expression: c = 2R / (1 + sin α + cos α). Maybe that's the simplest form. Alternatively, factor numerator and denominator.Alternatively, let's consider that 1 + sin α + cos α = 2 cos(α/2 - 45°). Wait, let me check:We know that sin α + cos α = √2 sin(α + 45°). So, 1 + sin α + cos α = 1 + √2 sin(α + 45°). Hmm, not sure if that helps.Alternatively, maybe express in terms of tangent of α/2:We have 1 + sin α + cos α = 1 + 2 sin(α/2) cos(α/2) + (cos²(α/2) - sin²(α/2)) = [cos²(α/2) + 2 sin(α/2) cos(α/2) + sin²(α/2)] = (cos(α/2) + sin(α/2))²Ah! That's a nice identity. So, 1 + sin α + cos α = (sin(α/2) + cos(α/2))²Therefore, c = 2R / (sin(α/2) + cos(α/2))²So, c = 2R / (sin(α/2) + cos(α/2))²That seems like a good simplification. So, maybe that's the answer.Alternatively, we can write sin(α/2) + cos(α/2) as √2 sin(α/2 + 45°), but again, not sure if necessary.So, in conclusion, the length of the hypotenuse c is 2R divided by (sin(α/2) + cos(α/2)) squared.Alternatively, since (sin(α/2) + cos(α/2))² = 1 + sin α, but wait, no, because (sin x + cos x)^2 = 1 + 2 sin x cos x = 1 + sin(2x). So, in this case, x = α/2, so (sin(α/2) + cos(α/2))² = 1 + sin α.Wait, that's correct! Because:(sin(α/2) + cos(α/2))² = sin²(α/2) + 2 sin(α/2) cos(α/2) + cos²(α/2) = 1 + sin αBecause sin² + cos² = 1, and 2 sin cos = sin(2x), which here is sin α.So, (sin(α/2) + cos(α/2))² = 1 + sin αTherefore, c = 2R / (1 + sin α)Wait, that's a much simpler expression! So, c = 2R / (1 + sin α)Wait, let me verify that:We had c = 2R / (1 + sin α + cos α)But we also found that 1 + sin α + cos α = (sin(α/2) + cos(α/2))² = 1 + sin αWait, no, that's not correct. Wait, 1 + sin α + cos α = 1 + sin α + cos α, which we showed is equal to (sin(α/2) + cos(α/2))², which is 1 + sin α. Wait, that can't be because 1 + sin α + cos α is not equal to 1 + sin α unless cos α = 0, which it's not.Wait, I think I made a mistake in the identity. Let me re-examine.We have:(sin(α/2) + cos(α/2))² = sin²(α/2) + 2 sin(α/2) cos(α/2) + cos²(α/2) = 1 + sin αYes, that's correct. So, (sin(α/2) + cos(α/2))² = 1 + sin αBut earlier, we had 1 + sin α + cos α = (sin(α/2) + cos(α/2))² + cos α - cos α = ?Wait, no, that's not helpful. Wait, actually, 1 + sin α + cos α is equal to (sin(α/2) + cos(α/2))² + cos α - sin α? No, that's not correct.Wait, perhaps I confused the identity. Let me double-check.We have:1 + sin α = (sin(α/2) + cos(α/2))²Yes, that's correct.But 1 + sin α + cos α is different. Let's compute:1 + sin α + cos α = 1 + sin α + cos αWe can write this as 1 + sin α + cos α = (sin(α/2) + cos(α/2))² + cos α - sin αWait, no, that doesn't help.Alternatively, let's compute (sin(α/2) + cos(α/2))² = 1 + sin α, so 1 + sin α + cos α = (sin(α/2) + cos(α/2))² + cos αBut that doesn't seem helpful.Alternatively, perhaps express 1 + sin α + cos α in terms of sin and cos of α/2.Let me try:1 + sin α + cos α = 1 + 2 sin(α/2) cos(α/2) + (cos²(α/2) - sin²(α/2))= [cos²(α/2) + sin²(α/2)] + 2 sin(α/2) cos(α/2) + (cos²(α/2) - sin²(α/2))= 1 + 2 sin(α/2) cos(α/2) + cos²(α/2) - sin²(α/2)= 1 + sin α + cos αWait, that just brings us back. Hmm.Alternatively, maybe factor 1 + sin α + cos α as (1 + cos α) + sin α.We know that 1 + cos α = 2 cos²(α/2), and sin α = 2 sin(α/2) cos(α/2). So:1 + sin α + cos α = 2 cos²(α/2) + 2 sin(α/2) cos(α/2) = 2 cos(α/2) [cos(α/2) + sin(α/2)]So, 1 + sin α + cos α = 2 cos(α/2) (cos(α/2) + sin(α/2))Therefore, c = 2R / [2 cos(α/2) (cos(α/2) + sin(α/2))] = R / [cos(α/2) (cos(α/2) + sin(α/2)) ]Hmm, that's another expression, but not necessarily simpler.Alternatively, let's take the earlier result: c = 2R / (1 + sin α + cos α). Maybe that's the simplest form.Alternatively, we can write 1 + sin α + cos α as √2 sin(α + 45°) + 1, but I don't think that helps.Alternatively, let's consider that 1 + sin α + cos α = 2 cos(α/2 - 45°). Wait, let me check:We know that sin α + cos α = √2 sin(α + 45°), so 1 + sin α + cos α = 1 + √2 sin(α + 45°). Hmm, not sure.Alternatively, maybe use the identity:1 + sin α + cos α = 2 cos(α/2 - 45°)Wait, let me compute 2 cos(α/2 - 45°):= 2 [cos(α/2) cos(45°) + sin(α/2) sin(45°)] = 2 [ (cos(α/2) + sin(α/2)) / √2 ] = √2 (cos(α/2) + sin(α/2))So, 2 cos(α/2 - 45°) = √2 (cos(α/2) + sin(α/2))But we have 1 + sin α + cos α = (sin(α/2) + cos(α/2))² = 1 + sin αWait, no, earlier we saw that (sin(α/2) + cos(α/2))² = 1 + sin α, but 1 + sin α + cos α is different.Wait, perhaps I'm overcomplicating this. Let me just stick with c = 2R / (1 + sin α + cos α). That seems to be the simplest form.Alternatively, if I want to express it in terms of tangent, let's see:From c = 2R / (1 + sin α + cos α), let's divide numerator and denominator by cos α:c = 2R / [1 + sin α + cos α] = 2R / [cos α (sec α + tan α + 1)]But that might not help.Alternatively, let's express in terms of tan(α/2):We have:sin α = 2t / (1 + t²)cos α = (1 - t²) / (1 + t²)So, 1 + sin α + cos α = 1 + 2t/(1 + t²) + (1 - t²)/(1 + t²) = [ (1 + t²) + 2t + (1 - t²) ] / (1 + t²) = (2 + 2t) / (1 + t²) = 2(1 + t)/(1 + t²)So, c = 2R / [2(1 + t)/(1 + t²)] = R (1 + t²)/(1 + t)So, c = R (1 + t²)/(1 + t), where t = tan(α/2)Alternatively, factor numerator:1 + t² = (t + i)(t - i), but that's complex.Alternatively, write as:c = R (1 + t²)/(1 + t) = R [ (1 + t²) / (1 + t) ] = R [ (1 + t²) / (1 + t) ]But 1 + t² = (1 + t)^2 - 2t, so:c = R [ ( (1 + t)^2 - 2t ) / (1 + t) ] = R [ (1 + t) - 2t / (1 + t) ] = R [1 + t - 2t / (1 + t) ]Hmm, not helpful.Alternatively, perform polynomial division:Divide 1 + t² by 1 + t:1 + t² = (1 + t)(t - 1) + 2So, (1 + t²)/(1 + t) = t - 1 + 2/(1 + t)So, c = R (t - 1 + 2/(1 + t)) = R [ (t - 1) + 2/(1 + t) ]But t = tan(α/2), so:c = R [ tan(α/2) - 1 + 2/(1 + tan(α/2)) ]Hmm, not sure if that's helpful.Alternatively, let me consider that 1 + t² = sec²(α/2), so:c = R sec²(α/2) / (1 + tan(α/2)) = R / [cos²(α/2) (1 + tan(α/2)) ]But 1 + tan(α/2) = (sin(α/2) + cos(α/2))/cos(α/2), so:c = R / [cos²(α/2) * (sin(α/2) + cos(α/2))/cos(α/2) ] = R / [cos(α/2) (sin(α/2) + cos(α/2)) ]Which is the same as earlier.So, in conclusion, after trying various approaches, the simplest expression I can get is c = 2R / (1 + sin α + cos α). Alternatively, c = R (1 + tan²(α/2)) / (1 + tan(α/2)), but that might not be simpler.Alternatively, since 1 + sin α + cos α = 2 cos(α/2 - 45°), as we saw earlier, then c = 2R / [2 cos(α/2 - 45°)] = R / cos(α/2 - 45°)But that might be another way to write it.So, c = R sec(α/2 - 45°)But I'm not sure if that's more useful.Alternatively, since α is an acute angle, α/2 - 45° could be negative or positive depending on α.Wait, if α is acute, say α < 90°, then α/2 < 45°, so α/2 - 45° is negative, so sec(α/2 - 45°) = 1 / cos(45° - α/2) = sec(45° - α/2)So, c = R sec(45° - α/2)Which is another way to write it.So, c = R / cos(45° - α/2)That's a nice expression.Alternatively, using the identity cos(A - B) = cos A cos B + sin A sin B, so:cos(45° - α/2) = cos 45° cos(α/2) + sin 45° sin(α/2) = (√2/2)(cos(α/2) + sin(α/2))So, c = R / [ (√2/2)(cos(α/2) + sin(α/2)) ] = (2R)/√2 / (cos(α/2) + sin(α/2)) = √2 R / (cos(α/2) + sin(α/2))But we already had c = 2R / (1 + sin α + cos α), and 1 + sin α + cos α = (sin(α/2) + cos(α/2))², so c = 2R / (sin(α/2) + cos(α/2))²But √2 R / (cos(α/2) + sin(α/2)) is another expression.Wait, let me check:From c = R / cos(45° - α/2) = R / [ (√2/2)(cos(α/2) + sin(α/2)) ] = (2R)/√2 / (cos(α/2) + sin(α/2)) = √2 R / (cos(α/2) + sin(α/2))Yes, that's correct.So, c = √2 R / (sin(α/2) + cos(α/2))Alternatively, since sin(α/2) + cos(α/2) = √2 sin(α/2 + 45°), we can write:c = √2 R / [√2 sin(α/2 + 45°)] = R / sin(α/2 + 45°)So, c = R / sin(α/2 + 45°)That's another way to express it.So, in summary, the length of the hypotenuse c can be expressed in multiple ways:1. c = 2R / (1 + sin α + cos α)2. c = R (1 + tan²(α/2)) / (1 + tan(α/2))3. c = √2 R / (sin(α/2) + cos(α/2))4. c = R / sin(α/2 + 45°)All of these are equivalent expressions. Depending on the context, one might be preferable over the others.But perhaps the simplest form is c = 2R / (1 + sin α + cos α). Alternatively, using the identity 1 + sin α + cos α = 2 cos(α/2 - 45°), so c = R / cos(α/2 - 45°), which is also simple.Alternatively, if I want to write it in terms of tangent, perhaps c = R / sin(α/2 + 45°), which is also concise.But I think the most straightforward expression is c = 2R / (1 + sin α + cos α). So, I'll go with that.**Final Answer**The length of the hypotenuse is boxed{dfrac{2R}{1 + sin alpha + cos alpha}}.