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Question: If the lines \( x{\sin ^2}A + y\sin A + 1 = 0 \\\ x{\sin ^2}B + y\sin B + 1 = 0 \\\ x{\si...

If the lines
xsin⁡2A+ysin⁡A+1=0 xsin⁡2B+ysin⁡B+1=0 xsin⁡2C+ysin⁡C+1=0 x{\sin ^2}A + y\sin A + 1 = 0 \\\ x{\sin ^2}B + y\sin B + 1 = 0 \\\ x{\sin ^2}C + y\sin C + 1 = 0 \\\
Are concurrent where A, B, C are angles of triangle then ΔABC\Delta ABC must be-
A.Equilateral triangle
B.Isosceles Triangle
C.Right angle Triangle
D.No such triangle exists

Explanation

Solution

The condition of concurrency which states that if three linesa1x1+b1y1+c1=0{a_1}{x_1} + {b_1}{y_1} + {c_1} = 0 , a2x2+b2y2+c2=0{a_2}{x_2} + {b_2}{y_2} + {c_2} = 0 and a3x3+b3y3+c3=0{a_3}{x_3} + {b_3}{y_3} + {c_3} = 0 are concurrent then
\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}} \\\ {{a_2}}&{{b_2}}&{{c_2}} \\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0 to solve the lines. If two angles are equal then it is an isosceles triangle, if three angles are equal then it is an equilateral triangle. If one angle equals 90∘{90^ \circ } then it is a right angle triangle.If none of the above conditions exists then there is no such triangle.

Complete step-by-step answer:
Given that these lines are concurrent-
xsin⁡2A+ysin⁡A+1=0 xsin⁡2B+ysin⁡B+1=0 xsin⁡2C+ysin⁡C+1=0 x{\sin ^2}A + y\sin A + 1 = 0 \\\ x{\sin ^2}B + y\sin B + 1 = 0 \\\ x{\sin ^2}C + y\sin C + 1 = 0 \\\
We know that that Three linesa1x1+b1y1+c1=0{a_1}{x_1} + {b_1}{y_1} + {c_1} = 0 , a2x2+b2y2+c2=0{a_2}{x_2} + {b_2}{y_2} + {c_2} = 0 and a3x3+b3y3+c3=0{a_3}{x_3} + {b_3}{y_3} + {c_3} = 0 are concurrent only if there determinant is equal to zero which is given as-
\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}} \\\ {{a_2}}&{{b_2}}&{{c_2}} \\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0
On comparing the given lines and the lines of the condition of concurrency we get,
a1=sin⁡2A{a_1} = {\sin ^2}A ,b1=sin⁡A{b_1} = \sin A, c1=c2=c3=1{c_1} = {c_2} = {c_3} = 1, a2=sin⁡2B{a_2} = {\sin ^2}B ,b2=sin⁡B{b_2} = \sin B ,a3=sin⁡2C{a_3} = {\sin ^2}C and b3=sin⁡C{b_3} = \sin C
On putting these values in the condition we get,
\Rightarrow \left| {\begin{array}{*{20}{c}} {{{\sin }^2}A}&{\sin A}&1 \\\ {{{\sin }^2}B}&{\sin B}&1 \\\ {{{\sin }^2}C}&{\sin C}&1 \end{array}} \right| = 0
On solving the determinant we get,
\Rightarrow {\sin ^2}A\left| {\begin{array}{*{20}{c}} {\sin B}&1 \\\ {\sin C}&1 \end{array}} \right| - \sin A\left| {\begin{array}{*{20}{c}} {{{\sin }^2}B}&1 \\\ {{{\sin }^2}C}&1 \end{array}} \right| + 1\left| {\begin{array}{*{20}{c}} {{{\sin }^2}B}&{\sin B} \\\ {{{\sin }^2}C}&{\sin C} \end{array}} \right| = 0
On solving further-
⇒sin⁡2A(sin⁡B−sin⁡C)−sin⁡A(sin⁡2B−sin⁡2C)+1(sin⁡2Bsin⁡C−sin⁡2Csin⁡B)=0\Rightarrow {\sin ^2}A\left( {\sin B - \sin C} \right) - \sin A\left( {{{\sin }^2}B - {{\sin }^2}C} \right) + 1\left( {{{\sin }^2}B\sin C - {{\sin }^2}C\sin B} \right) = 0
We know that,a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
So using the identity and taking the common terms out of the bracket in the third term we get,
⇒sin⁡2A(sin⁡B−sin⁡C)−sin⁡A(sin⁡B−sin⁡C)(sin⁡B+sin⁡C)+sin⁡Bsin⁡C(sin⁡B−sin⁡C)=0\Rightarrow {\sin ^2}A\left( {\sin B - \sin C} \right) - \sin A\left( {\sin B - \sin C} \right)\left( {\sin B + \sin C} \right) + \sin B\sin C\left( {\sin B - \sin C} \right) = 0
On taking(sin⁡B−sin⁡C)\left( {\sin B - \sin C} \right) common we get,
⇒(sin⁡B−sin⁡C)[sin⁡2A−sin⁡A(sin⁡B+sin⁡C)+sin⁡Bsin⁡C]=0\Rightarrow \left( {\sin B - \sin C} \right)\left[ {{{\sin }^2}A - \sin A\left( {\sin B + \sin C} \right) + \sin B\sin C} \right] = 0
Now on solving the terms inside the bracket,
⇒(sin⁡B−sin⁡C)[sin⁡2A−sin⁡Asin⁡B−sin⁡Asin⁡C+sin⁡Bsin⁡C]=0\Rightarrow \left( {\sin B - \sin C} \right)\left[ {{{\sin }^2}A - \sin A\sin B - \sin A\sin C + \sin B\sin C} \right] = 0
On taking common we get,
⇒(sin⁡B−sin⁡C)[sin⁡A(sin⁡A−sin⁡B)−sin⁡C(sin⁡A−sin⁡B)]=0\Rightarrow \left( {\sin B - \sin C} \right)\left[ {\sin A\left( {\sin A - \sin B} \right) - \sin C\left( {\sin A - \sin B} \right)} \right] = 0
On taking(sin⁡A−sin⁡B)\left( {\sin A - \sin B} \right) common we get,
⇒(sin⁡B−sin⁡C)(sin⁡A−sin⁡B)[sin⁡A−sin⁡C]=0\Rightarrow \left( {\sin B - \sin C} \right)\left( {\sin A - \sin B} \right)\left[ {\sin A - \sin C} \right] = 0
On equating the multiplication terms to zero we get,
⇒(sin⁡B−sin⁡C)=0 or (sin⁡A−sin⁡B)=0 or [sin⁡A−sin⁡C]=0\Rightarrow \left( {\sin B - \sin C} \right) = 0{\text{ or }}\left( {\sin A - \sin B} \right) = 0{\text{ or }}\left[ {\sin A - \sin C} \right] = 0
So we get,
⇒sin⁡B=sin⁡C or sin⁡A=sin⁡B or sin⁡A=sin⁡C\Rightarrow \sin B = \sin C{\text{ or }}\sin A = \sin B{\text{ or }}\sin A = \sin C
This means that any of the two angles of the triangles are equal
So the triangle is an isosceles triangle.

Hence, the correct answer is B.

Note: The condition of concurrency of three lines means that they pass through the same point. If instead of angles a quadratic equation was given with integer values then we could have solved the question by solving the first two lines and finding the value of x and y. Then we would have checked if these values satisfied the third line. If they did then the lines would be concurrent.