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If the lines xsin2A + ysinA + 1 = 0, xsin2B + ysinB + 1 = 0, xsin2C + ysinC + 1 = 0 are concurrent where A, B, C are angles of triangle then ΔABC must be
- a)Equilateral
- b)Isosceles
- c)Right angle
- d)No such triangle exist
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
If the lines xsin2A + ysinA + 1 = 0, xsin2B + ysinB + 1 = 0, xsin2C + ...
⇒ (sinA–sinB)(sinB–sinC)(sinC–sinC)=0
⇒ A = B or B = C or C = A
any two angles are equal
Δ is isosceles.
⇒ A = B or B = C or C = A
any two angles are equal
Δ is isosceles.
Most Upvoted Answer
If the lines xsin2A + ysinA + 1 = 0, xsin2B + ysinB + 1 = 0, xsin2C + ...
We can use the concept of concurrent lines in a triangle to solve this problem.
First, let's rewrite the given equations in a more standard form:
x*sin(2A) + y*sin(A) + 1 = 0 (equation 1)
x*sin(2B) + y*sin(B) + 1 = 0 (equation 2)
x*sin(2C) + y*sin(C) + 1 = 0 (equation 3)
Now, let's consider equation 1. The coefficient of x is sin(2A) and the coefficient of y is sin(A). Similarly, for equation 2, the coefficients of x and y are sin(2B) and sin(B), respectively. And for equation 3, the coefficients of x and y are sin(2C) and sin(C), respectively.
According to the concept of concurrent lines, if the lines represented by these equations are concurrent, then the ratio of the coefficients of x and y in each equation should be the same.
Therefore, we can set up the following equations:
sin(2A)/sin(A) = sin(2B)/sin(B) = sin(2C)/sin(C) (equation 4)
Now, let's simplify equation 4:
sin(2A)/sin(A) = 2*sin(A)*cos(A)/sin(A) = 2*cos(A)
sin(2B)/sin(B) = 2*sin(B)*cos(B)/sin(B) = 2*cos(B)
sin(2C)/sin(C) = 2*sin(C)*cos(C)/sin(C) = 2*cos(C)
Therefore, equation 4 becomes:
2*cos(A) = 2*cos(B) = 2*cos(C)
Dividing both sides by 2, we get:
cos(A) = cos(B) = cos(C)
Since A, B, and C are angles of a triangle, the sum of the angles is 180 degrees. Therefore, A + B + C = 180 degrees.
If cos(A) = cos(B) = cos(C), then A = B = C = 60 degrees.
Therefore, if the lines represented by the given equations are concurrent, then the angles of the triangle are all 60 degrees, which means the triangle is an equilateral triangle.
First, let's rewrite the given equations in a more standard form:
x*sin(2A) + y*sin(A) + 1 = 0 (equation 1)
x*sin(2B) + y*sin(B) + 1 = 0 (equation 2)
x*sin(2C) + y*sin(C) + 1 = 0 (equation 3)
Now, let's consider equation 1. The coefficient of x is sin(2A) and the coefficient of y is sin(A). Similarly, for equation 2, the coefficients of x and y are sin(2B) and sin(B), respectively. And for equation 3, the coefficients of x and y are sin(2C) and sin(C), respectively.
According to the concept of concurrent lines, if the lines represented by these equations are concurrent, then the ratio of the coefficients of x and y in each equation should be the same.
Therefore, we can set up the following equations:
sin(2A)/sin(A) = sin(2B)/sin(B) = sin(2C)/sin(C) (equation 4)
Now, let's simplify equation 4:
sin(2A)/sin(A) = 2*sin(A)*cos(A)/sin(A) = 2*cos(A)
sin(2B)/sin(B) = 2*sin(B)*cos(B)/sin(B) = 2*cos(B)
sin(2C)/sin(C) = 2*sin(C)*cos(C)/sin(C) = 2*cos(C)
Therefore, equation 4 becomes:
2*cos(A) = 2*cos(B) = 2*cos(C)
Dividing both sides by 2, we get:
cos(A) = cos(B) = cos(C)
Since A, B, and C are angles of a triangle, the sum of the angles is 180 degrees. Therefore, A + B + C = 180 degrees.
If cos(A) = cos(B) = cos(C), then A = B = C = 60 degrees.
Therefore, if the lines represented by the given equations are concurrent, then the angles of the triangle are all 60 degrees, which means the triangle is an equilateral triangle.
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If the lines xsin2A + ysinA + 1 = 0, xsin2B + ysinB + 1 = 0, xsin2C + ...
Both sides are equal so we are called that isosceles triangle
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If the lines xsin2A + ysinA + 1 = 0, xsin2B + ysinB + 1 = 0, xsin2C + ysinC + 1 = 0 are concurrent where A, B, C are angles of triangle then ΔABC must bea)Equilateralb)Isoscelesc)Right angled)No such triangle existCorrect answer is option 'B'. Can you explain this answer?
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