💡Okay, so I have this problem about a point P in the plane of triangle ABC. The condition given is |PB - PC| - |PB + PC - 2PA| = 0. I need to figure out what shape triangle ABC cannot be based on this condition. The options are obtuse, right, isosceles, or equilateral.First, let me try to understand the vector equation. It involves vectors from point P to the vertices of the triangle. The equation is |PB - PC| - |PB + PC - 2PA| = 0, which implies that |PB - PC| equals |PB + PC - 2PA|.Maybe I can rewrite these vectors in terms of position vectors relative to point A. Let me denote the position vectors of points A, B, C, and P as vectors a, b, c, and p respectively. Then, PB is b - p, PC is c - p, and PA is a - p.So, substituting these into the equation:| (b - p) - (c - p) | - | (b - p) + (c - p) - 2(a - p) | = 0Simplify the first term: (b - p) - (c - p) = b - cSimplify the second term: (b - p) + (c - p) - 2(a - p) = b + c - 2p - 2a + 2p = b + c - 2aSo the equation becomes |b - c| - |b + c - 2a| = 0, which means |b - c| = |b + c - 2a|Hmm, interesting. So the magnitude of vector (b - c) is equal to the magnitude of vector (b + c - 2a). Let me think about what these vectors represent.Vector (b - c) is the vector from point C to point B, so its magnitude is just the length of side BC.Vector (b + c - 2a) can be rewritten as (b - a) + (c - a) = AB + AC. So the magnitude of AB + AC is equal to the magnitude of BC.So |AB + AC| = |BC|Wait, that seems like a condition on the triangle. Let me square both sides to eliminate the absolute value:|AB + AC|² = |BC|²Expanding the left side: (AB + AC) · (AB + AC) = |AB|² + 2 AB · AC + |AC|²The right side is |BC|²So, |AB|² + 2 AB · AC + |AC|² = |BC|²But in triangle ABC, by the law of cosines, |BC|² = |AB|² + |AC|² - 2 |AB||AC| cos(angle at A)So, substituting that into the equation:|AB|² + 2 AB · AC + |AC|² = |AB|² + |AC|² - 2 |AB||AC| cos(angle at A)Simplify both sides:Left side: |AB|² + |AC|² + 2 AB · ACRight side: |AB|² + |AC|² - 2 |AB||AC| cos(angle at A)Subtract |AB|² + |AC|² from both sides:2 AB · AC = -2 |AB||AC| cos(angle at A)Divide both sides by 2:AB · AC = - |AB||AC| cos(angle at A)But wait, the dot product AB · AC is equal to |AB||AC| cos(angle at A). So we have:|AB||AC| cos(angle at A) = - |AB||AC| cos(angle at A)Which simplifies to:|AB||AC| cos(angle at A) + |AB||AC| cos(angle at A) = 0So, 2 |AB||AC| cos(angle at A) = 0Since |AB| and |AC| are lengths of sides, they can't be zero. Therefore, cos(angle at A) must be zero.Which means angle at A is 90 degrees. So triangle ABC must be a right triangle with a right angle at A.Wait, but the question is asking which shape triangle ABC cannot be. The options are obtuse, right, isosceles, or equilateral.But according to my deduction, triangle ABC must be a right triangle. So does that mean it cannot be something else? Or is it that it must be a right triangle, so the other options are possible?Wait, no. The condition given must hold for some point P in the plane. So if the condition implies that ABC must be a right triangle, then ABC cannot be something else? Or is it that ABC can be any triangle, but under this condition, it must be a right triangle.Wait, maybe I need to think differently. The condition is given for point P. So perhaps for any triangle ABC, there exists a point P satisfying the condition, except when ABC is a certain type.But from my earlier deduction, the condition seems to enforce that ABC is a right triangle. So if ABC is not a right triangle, then there is no such point P? Or maybe the other way around.Wait, let me re-examine the steps.We started with |PB - PC| = |PB + PC - 2PA|Then, after substitution, we got |b - c| = |b + c - 2a|Which led us to |AB + AC| = |BC|Then, squaring both sides, we ended up with angle at A being 90 degrees.Therefore, triangle ABC must be a right triangle at A.So, if the condition is satisfied, then ABC must be a right triangle. Therefore, if ABC is not a right triangle, then such a point P does not exist.But the question is phrased as: "Then the shape of triangle ABC cannot be:"So, given that point P exists in the plane of triangle ABC satisfying the condition, then ABC cannot be which shape.Wait, that is, if ABC is such that no point P satisfies the condition, then ABC cannot be that shape.But from my deduction, if ABC is a right triangle, then such a point P exists. So ABC can be a right triangle.But if ABC is not a right triangle, then such a point P does not exist.Therefore, the shape of triangle ABC cannot be something other than a right triangle.But the options are obtuse, right, isosceles, equilateral.Wait, but a right triangle is a specific case. An isosceles triangle can be right or not. Equilateral triangles are all acute. Obtuse triangles have one angle greater than 90 degrees.So, if ABC must be a right triangle, then it cannot be an obtuse triangle, or an equilateral triangle, or an isosceles triangle that is not right.Wait, but the options are separate: A is obtuse, B is right, C is isosceles, D is equilateral.So, if ABC must be a right triangle, then it cannot be an obtuse triangle, an isosceles triangle, or an equilateral triangle? But that doesn't make sense because a right triangle can also be isosceles if the two legs are equal.Wait, perhaps I need to think again.If ABC must be a right triangle, then it can be a right triangle, but it cannot be an obtuse triangle, or an equilateral triangle, or an isosceles triangle that is not right.But the options are separate, so perhaps the answer is that ABC cannot be an obtuse triangle, because if it were obtuse, then angle A would be greater than 90 degrees, which contradicts our earlier conclusion that angle A must be 90 degrees.Wait, but angle A is specifically 90 degrees. So if ABC is an obtuse triangle, then one of its angles is greater than 90 degrees, but not necessarily angle A. So maybe angle A is still 90 degrees, but another angle is obtuse.Wait, but in a triangle, only one angle can be obtuse. So if angle A is 90 degrees, then the other two angles must be acute, so the triangle cannot be obtuse.Therefore, if ABC is a right triangle, it cannot be obtuse. So the shape of ABC cannot be an obtuse triangle.But wait, the options are A: Obtuse, B: Right, C: Isosceles, D: Equilateral.So, based on the condition, ABC must be a right triangle, so it cannot be an obtuse triangle, an isosceles triangle, or an equilateral triangle? But that seems conflicting because a right triangle can be isosceles if the two legs are equal.Wait, perhaps the key is that ABC must be a right triangle, so it cannot be something else. But the options are about what it cannot be, not about what it must be.So, if ABC must be a right triangle, then it cannot be an obtuse triangle, because if it were obtuse, it wouldn't satisfy the condition. Similarly, it cannot be an equilateral triangle because equilateral triangles are all acute and not right. It also cannot be an isosceles triangle unless it's a right isosceles triangle.But the options are separate, so perhaps the answer is that ABC cannot be an obtuse triangle, because if it were, the condition would not hold.Wait, but in my earlier deduction, the condition enforces that ABC must be a right triangle, so if ABC is not a right triangle, then such a point P does not exist. Therefore, if ABC is an obtuse triangle, then such a point P does not exist, meaning ABC cannot be an obtuse triangle.Similarly, if ABC is equilateral, then such a point P does not exist because ABC is not a right triangle. Same with isosceles unless it's a right isosceles triangle.But the options are separate, so the question is asking which shape ABC cannot be, given that such a point P exists.Therefore, if ABC must be a right triangle, then it cannot be an obtuse triangle, an equilateral triangle, or an isosceles triangle that is not right.But since the options are separate, and the question is asking which shape ABC cannot be, the answer would be that ABC cannot be an obtuse triangle, because if it were, the condition would not hold.Wait, but I'm getting confused. Let me try to think differently.Suppose ABC is an obtuse triangle. Then, can there exist a point P such that |PB - PC| = |PB + PC - 2PA|?From my earlier deduction, for such a P to exist, ABC must be a right triangle. Therefore, if ABC is obtuse, such a P does not exist. Therefore, ABC cannot be obtuse.Similarly, if ABC is equilateral, such a P does not exist because ABC is not a right triangle. Same with isosceles unless it's a right isosceles triangle.But the options are A: Obtuse, B: Right, C: Isosceles, D: Equilateral.So, the question is asking which shape ABC cannot be. Since ABC must be a right triangle, it cannot be an obtuse triangle, an equilateral triangle, or an isosceles triangle that is not right.But the options are separate, so the answer would be that ABC cannot be an obtuse triangle, because if it were, the condition would not hold.Wait, but the problem says "the shape of triangle ABC cannot be". So, given that such a point P exists, ABC must be a right triangle, so it cannot be something else. Therefore, the answer is that ABC cannot be an obtuse triangle.But I'm not entirely sure. Maybe I should check with specific examples.Suppose ABC is a right triangle. Then, does there exist a point P such that |PB - PC| = |PB + PC - 2PA|?From my earlier steps, yes, because angle A is 90 degrees, so the condition holds.Now, suppose ABC is an obtuse triangle. Then, angle A is greater than 90 degrees. Would there exist a point P satisfying the condition?From my deduction, no, because the condition requires angle A to be 90 degrees. Therefore, such a P does not exist, so ABC cannot be obtuse.Similarly, if ABC is equilateral, angle A is 60 degrees, so the condition would not hold, meaning such a P does not exist, so ABC cannot be equilateral.But the options are separate, and the question is asking which shape ABC cannot be. So, since ABC must be a right triangle, it cannot be an obtuse triangle, an equilateral triangle, or an isosceles triangle that is not right.But the options are A: Obtuse, B: Right, C: Isosceles, D: Equilateral.So, the answer would be that ABC cannot be an obtuse triangle, because if it were, the condition would not hold.Wait, but I'm still a bit confused because an isosceles triangle can be right or not. If it's a right isosceles triangle, then it's allowed. But if it's an isosceles triangle that's not right, then it's not allowed.Similarly, an equilateral triangle is not allowed because it's not a right triangle.But the question is asking which shape ABC cannot be, given that such a point P exists. So, if ABC is a right triangle, it's allowed. If it's not, it's not allowed.Therefore, the shapes that ABC cannot be are obtuse, isosceles (unless it's right), and equilateral.But since the options are separate, and the question is asking which shape ABC cannot be, the answer would be that ABC cannot be an obtuse triangle.Wait, but I'm not sure. Maybe I should think about the implications again.If ABC is a right triangle, then such a P exists. If ABC is not a right triangle, then such a P does not exist. Therefore, ABC cannot be any shape other than a right triangle.But the options are A: Obtuse, B: Right, C: Isosceles, D: Equilateral.So, the question is asking which shape ABC cannot be. Since ABC must be a right triangle, it cannot be an obtuse triangle, an isosceles triangle that's not right, or an equilateral triangle.But the options are separate, so the answer would be that ABC cannot be an obtuse triangle.Wait, but I'm still not entirely confident. Maybe I should look for another approach.Alternatively, perhaps the condition |PB - PC| = |PB + PC - 2PA| can be interpreted geometrically.Let me consider point P such that |PB - PC| = |PB + PC - 2PA|.This can be rewritten as |PB - PC| = |PB + PC - 2PA|.Let me denote vector PA as vector a, PB as vector b, and PC as vector c.Then, the equation becomes |b - c| = |b + c - 2a|.But I'm not sure if this helps. Maybe I can think in terms of coordinates.Let me place point A at the origin for simplicity. Then, vector PA is just the position vector of P, say (x, y). Then, vectors PB and PC would be (x - b_x, y - b_y) and (x - c_x, y - c_y) respectively.But this might get too complicated. Maybe there's a better way.Wait, earlier I concluded that angle A must be 90 degrees. So, ABC must be a right triangle at A. Therefore, ABC cannot be an obtuse triangle because that would require angle A to be greater than 90 degrees, which contradicts the condition.Therefore, the answer is A: Obtuse triangle.But I'm still a bit unsure because I might have made a mistake in my earlier steps. Let me double-check.Starting from |PB - PC| = |PB + PC - 2PA|Expressed in vectors, this is |b - c| = |b + c - 2a|Then, |b - c|² = |b + c - 2a|²Expanding both sides:(b - c)·(b - c) = (b + c - 2a)·(b + c - 2a)Which gives:|b|² - 2b·c + |c|² = |b|² + 2b·c + |c|² - 4a·b - 4a·c + 4|a|²Simplify:Left side: |b|² + |c|² - 2b·cRight side: |b|² + |c|² + 2b·c - 4a·(b + c) + 4|a|²Subtract left side from right side:0 = 4b·c - 4a·(b + c) + 4|a|²Divide both sides by 4:0 = b·c - a·(b + c) + |a|²But since a is the position vector of A, which we placed at the origin, a = 0. Therefore, this simplifies to:0 = b·cSo, b·c = 0, meaning vectors AB and AC are perpendicular.Therefore, angle at A is 90 degrees, so ABC is a right triangle.Therefore, if ABC is not a right triangle, such a point P does not exist. Therefore, ABC cannot be an obtuse triangle, an isosceles triangle that's not right, or an equilateral triangle.But the options are separate, so the answer is that ABC cannot be an obtuse triangle.Wait, but if ABC is an isosceles triangle with angle A = 90 degrees, then it's a right isosceles triangle, which is allowed. So, ABC can be isosceles if it's also right.Similarly, ABC can be a right triangle, which is allowed.Therefore, the shape that ABC cannot be is an obtuse triangle.So, the answer is A: Obtuse triangle.

💡Okay, so I have this system of equations:1. ( x + y + z = a )2. ( x^2 + y^2 + z^2 = b^2 )3. ( xy = z^2 )And I need to find the condition on ( a ) and ( b ) such that ( x, y, z ) are positive and distinct. Hmm, let me try to figure this out step by step.First, I notice that the third equation relates ( xy ) to ( z^2 ). Maybe I can use this to express one variable in terms of the others. Let me solve for ( z ) from the third equation:( z = sqrt{xy} )But since ( z ) has to be positive, I can just write ( z = sqrt{xy} ).Now, let's substitute this expression for ( z ) into the first equation:( x + y + sqrt{xy} = a )Hmm, that looks a bit complicated. Maybe I can square both sides to eliminate the square root? Let's try that:( (x + y + sqrt{xy})^2 = a^2 )Expanding the left side:( x^2 + y^2 + xy + 2xsqrt{xy} + 2ysqrt{xy} + 2xy = a^2 )Wait, that seems messy. Maybe there's a better way. Let me think about the second equation:( x^2 + y^2 + z^2 = b^2 )Since ( z^2 = xy ), I can substitute that in:( x^2 + y^2 + xy = b^2 )Okay, so now I have two equations:1. ( x + y + sqrt{xy} = a )2. ( x^2 + y^2 + xy = b^2 )Maybe I can let ( s = x + y ) and ( p = xy ). That might simplify things.So, ( s + sqrt{p} = a ) and ( s^2 - p + p = b^2 ). Wait, let's see:From ( s = x + y ), we know that ( x^2 + y^2 = s^2 - 2p ). So substituting into the second equation:( s^2 - 2p + p = b^2 )Which simplifies to:( s^2 - p = b^2 )So now I have:1. ( s + sqrt{p} = a )2. ( s^2 - p = b^2 )Let me solve the first equation for ( s ):( s = a - sqrt{p} )Now substitute this into the second equation:( (a - sqrt{p})^2 - p = b^2 )Expanding ( (a - sqrt{p})^2 ):( a^2 - 2asqrt{p} + p - p = b^2 )Simplify:( a^2 - 2asqrt{p} = b^2 )Now, solve for ( sqrt{p} ):( -2asqrt{p} = b^2 - a^2 )Divide both sides by ( -2a ):( sqrt{p} = frac{a^2 - b^2}{2a} )Since ( sqrt{p} ) must be positive (because ( p = xy ) and ( x, y ) are positive), the right side must also be positive:( frac{a^2 - b^2}{2a} > 0 )Assuming ( a > 0 ) (since ( x, y, z ) are positive and their sum is ( a )), this implies:( a^2 - b^2 > 0 )So,( a^2 > b^2 )Which means:( |a| > |b| )But since ( a ) and ( b ) are given numbers, and ( x, y, z ) are positive, ( a ) must be positive. So,( a > b )Wait, but earlier I had ( sqrt{p} = frac{a^2 - b^2}{2a} ). Let me square both sides to find ( p ):( p = left( frac{a^2 - b^2}{2a} right)^2 )Simplify:( p = frac{(a^2 - b^2)^2}{4a^2} )Okay, so ( p = xy = frac{(a^2 - b^2)^2}{4a^2} ). Now, going back to ( s = a - sqrt{p} ):( s = a - frac{a^2 - b^2}{2a} )Simplify:( s = a - frac{a^2 - b^2}{2a} = frac{2a^2 - (a^2 - b^2)}{2a} = frac{a^2 + b^2}{2a} )So, ( s = frac{a^2 + b^2}{2a} ). Since ( s = x + y ), we have:( x + y = frac{a^2 + b^2}{2a} )And ( xy = frac{(a^2 - b^2)^2}{4a^2} )Now, ( x ) and ( y ) are roots of the quadratic equation:( t^2 - st + p = 0 )Which is:( t^2 - frac{a^2 + b^2}{2a} t + frac{(a^2 - b^2)^2}{4a^2} = 0 )Let me compute the discriminant to ensure that ( x ) and ( y ) are real and distinct:Discriminant ( D = left( frac{a^2 + b^2}{2a} right)^2 - 4 cdot 1 cdot frac{(a^2 - b^2)^2}{4a^2} )Simplify:( D = frac{(a^2 + b^2)^2}{4a^2} - frac{(a^2 - b^2)^2}{a^2} )Factor out ( frac{1}{4a^2} ):( D = frac{(a^2 + b^2)^2 - 4(a^2 - b^2)^2}{4a^2} )Expand the numerator:( (a^2 + b^2)^2 - 4(a^2 - b^2)^2 = (a^4 + 2a^2b^2 + b^4) - 4(a^4 - 2a^2b^2 + b^4) )Simplify:( = a^4 + 2a^2b^2 + b^4 - 4a^4 + 8a^2b^2 - 4b^4 )( = -3a^4 + 10a^2b^2 - 3b^4 )So,( D = frac{-3a^4 + 10a^2b^2 - 3b^4}{4a^2} )For ( x ) and ( y ) to be real and distinct, ( D > 0 ):( -3a^4 + 10a^2b^2 - 3b^4 > 0 )Let me factor this expression:Let me set ( u = a^2 ) and ( v = b^2 ), then the inequality becomes:( -3u^2 + 10uv - 3v^2 > 0 )Multiply both sides by -1 (which reverses the inequality):( 3u^2 - 10uv + 3v^2 < 0 )Factor the quadratic in ( u ):( 3u^2 - 10uv + 3v^2 = (3u - v)(u - 3v) )So,( (3u - v)(u - 3v) < 0 )This inequality holds when one factor is positive and the other is negative.Case 1: ( 3u - v > 0 ) and ( u - 3v < 0 )Which implies:( 3u > v ) and ( u < 3v )Substituting back ( u = a^2 ) and ( v = b^2 ):( 3a^2 > b^2 ) and ( a^2 < 3b^2 )Which simplifies to:( frac{a^2}{3} < b^2 < 3a^2 )Case 2: ( 3u - v < 0 ) and ( u - 3v > 0 )Which implies:( 3u < v ) and ( u > 3v )But this would mean ( u > 3v ) and ( 3u < v ), which is impossible because if ( u > 3v ), then ( 3u > 9v ), contradicting ( 3u < v ).Therefore, only Case 1 is valid, so:( frac{a^2}{3} < b^2 < 3a^2 )But earlier, we had ( a > b ) from ( sqrt{p} > 0 ). So combining these:From ( a > b ), we have ( b^2 < a^2 ).From the discriminant condition, ( b^2 > frac{a^2}{3} ).So overall, the condition is:( frac{a^2}{3} < b^2 < a^2 )Which can be written as:( frac{a}{sqrt{3}} < b < a )But since ( a ) and ( b ) are positive, we can write:( frac{a}{sqrt{3}} < b < a )Therefore, the numbers ( a ) and ( b ) must satisfy ( frac{a}{sqrt{3}} < b < a ) for the solutions ( x, y, z ) to be positive and distinct.Wait, let me double-check. If ( b^2 ) is between ( frac{a^2}{3} ) and ( a^2 ), then ( b ) is between ( frac{a}{sqrt{3}} ) and ( a ). That makes sense because if ( b ) is too small, the sum of squares might not allow for distinct positive solutions, and if ( b ) is too large, it might not satisfy the product condition.Also, we need to ensure that ( z = sqrt{xy} ) is distinct from ( x ) and ( y ). Since ( x ) and ( y ) are roots of the quadratic and are distinct (because ( D > 0 )), and ( z ) is different from both because ( z = sqrt{xy} ), which would only equal ( x ) or ( y ) if ( x = y ), but since ( x ) and ( y ) are distinct, ( z ) is also distinct.So, putting it all together, the condition is ( frac{a}{sqrt{3}} < b < a ).

💡First, identify the chips in each bag:- Bag X contains: 1, 4, 7.- Bag Y contains: 3, 5, 8.Next, calculate all possible sums by adding each chip from Bag X with each chip from Bag Y:- 1 + 3 = 4- 1 + 5 = 6- 1 + 8 = 9- 4 + 3 = 7- 4 + 5 = 9- 4 + 8 = 12- 7 + 3 = 10- 7 + 5 = 12- 7 + 8 = 15List all distinct sums and eliminate duplicates:- Distinct sums: 4, 6, 7, 9, 10, 12, 15.Count the number of unique sums:- There are 7 unique sums.Therefore, the number of different possible values for the sum is 7.

💡First, I identify the total number of sums attempted and the number of sums solved correctly.Next, I calculate the number of sums that were answered incorrectly by subtracting the correct sums from the total sums.Then, I determine the ratio of incorrect sums to correct sums by dividing the number of incorrect sums by the number of correct sums.Finally, I simplify the ratio by dividing both numbers by their greatest common divisor to express the ratio in its simplest form.

💡Okay, so I have this problem where there are two vectors, vector a and vector b. Vector a is given as (sinθ, -2) and vector b is (1, cosθ). They're saying these vectors are perpendicular to each other, and θ is between 0 and π/2. I need to find the value of sinθ + cosθ, and the options are given as A, B, C, D.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. Let me write that down.The dot product of a and b is (sinθ)(1) + (-2)(cosθ). So that's sinθ - 2cosθ. Since they're perpendicular, this should equal zero. So, sinθ - 2cosθ = 0.Hmm, okay. So, sinθ = 2cosθ. Maybe I can divide both sides by cosθ to get tanθ. Let's try that.Dividing both sides by cosθ gives tanθ = 2. So, tanθ is 2. That means θ is an angle whose tangent is 2. Since θ is between 0 and π/2, it's in the first quadrant, so all trigonometric functions will be positive there.Now, I need to find sinθ + cosθ. I think I can use the identity that (sinθ + cosθ)^2 = sin²θ + 2sinθcosθ + cos²θ. I know that sin²θ + cos²θ = 1, so that simplifies to 1 + 2sinθcosθ. Therefore, sinθ + cosθ is the square root of (1 + 2sinθcosθ).But I need to find sinθcosθ. I know that tanθ = 2, so maybe I can express sinθ and cosθ in terms of tanθ. Let me recall that if tanθ = 2, then we can think of a right triangle where the opposite side is 2 and the adjacent side is 1, so the hypotenuse would be sqrt(1² + 2²) = sqrt(5).Therefore, sinθ = opposite/hypotenuse = 2/sqrt(5) and cosθ = adjacent/hypotenuse = 1/sqrt(5). So, sinθcosθ would be (2/sqrt(5))(1/sqrt(5)) = 2/5.Wait, is that right? Let me check. If tanθ = 2, then yes, the sides are 1 and 2, hypotenuse sqrt(5). So, sinθ is 2/sqrt(5), cosθ is 1/sqrt(5). Multiplying them gives 2/5. Okay, that seems correct.So, going back to (sinθ + cosθ)^2 = 1 + 2*(2/5) = 1 + 4/5 = 9/5. Therefore, sinθ + cosθ is sqrt(9/5) which is 3/sqrt(5). Rationalizing the denominator, that's 3sqrt(5)/5.Looking at the options, D is 3sqrt(5)/5, so that should be the answer.Wait, let me just make sure I didn't make any mistakes. So, starting from the dot product, sinθ - 2cosθ = 0, which gives tanθ = 2. Then, constructing a right triangle with opposite 2, adjacent 1, hypotenuse sqrt(5). So, sinθ = 2/sqrt(5), cosθ = 1/sqrt(5). Then, sinθ + cosθ is (2 + 1)/sqrt(5) = 3/sqrt(5) = 3sqrt(5)/5. Yep, that seems correct.Alternatively, I could have used the identity for sinθ + cosθ directly. Since I know tanθ = 2, I can write sinθ + cosθ as sqrt(1 + tan²θ) * something? Wait, maybe not. Alternatively, I can express sinθ + cosθ in terms of tanθ.Let me think. If tanθ = 2, then sinθ = 2/sqrt(5) and cosθ = 1/sqrt(5), so adding them gives 3/sqrt(5). So, same result.Alternatively, maybe I can square sinθ + cosθ and then take the square root. So, (sinθ + cosθ)^2 = sin²θ + 2sinθcosθ + cos²θ = 1 + 2sinθcosθ. Since sinθcosθ is 2/5, then 1 + 2*(2/5) = 1 + 4/5 = 9/5. So, sqrt(9/5) is 3/sqrt(5) = 3sqrt(5)/5.Yeah, that's consistent. So, I think I did it right.Wait, just to be thorough, let me check if the dot product is indeed zero with these values. So, vector a is (sinθ, -2) = (2/sqrt(5), -2). Vector b is (1, cosθ) = (1, 1/sqrt(5)). The dot product is (2/sqrt(5))*1 + (-2)*(1/sqrt(5)) = 2/sqrt(5) - 2/sqrt(5) = 0. Perfect, that checks out.So, I think I'm confident that the answer is D: 3sqrt(5)/5.**Final Answer**boxed{D}

💡First, I need to determine the number of male workers in the workshop. Since there are a total of 30 workers and 10 of them are female, the number of male workers is 30 minus 10, which equals 20.Next, I need to apply stratified sampling to select 6 workers. Stratified sampling means that the proportion of each subgroup (male and female) in the sample should reflect their proportion in the entire population.To find out how many male workers should be selected, I calculate the proportion of male workers in the workshop. There are 20 male workers out of 30 total workers, which is a proportion of 20/30 or 2/3.Applying this proportion to the sample size of 6 workers, I multiply 6 by 2/3, which equals 4. Therefore, 4 male workers should be selected in the stratified sample.