Question

If the lines
$ 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 $\Delta ABC$ must be-
A.Equilateral triangle
B.Isosceles Triangle
C.Right angle Triangle
D.No such triangle exists

Answer 1

Hint: The condition of concurrency which states that if three lines${a_1}{x_1} + {b_1}{y_1} + {c_1} = 0$ , ${a_2}{x_2} + {b_2}{y_2} + {c_2} = 0$ and ${a_3}{x_3} + {b_3}{y_3} + {c_3} = 0$ 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 ${90^ \circ }$ 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-
$ 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 lines${a_1}{x_1} + {b_1}{y_1} + {c_1} = 0$ , ${a_2}{x_2} + {b_2}{y_2} + {c_2} = 0$ and ${a_3}{x_3} + {b_3}{y_3} + {c_3} = 0$ 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,
${a_1} = {\sin ^2}A$ ,${b_1} = \sin A$, ${c_1} = {c_2} = {c_3} = 1$, ${a_2} = {\sin ^2}B$ ,${b_2} = \sin B$ ,${a_3} = {\sin ^2}C$ and ${b_3} = \sin C$
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-
$ \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,${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
So using the identity and taking the common terms out of the bracket in the third term we get,
$ \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$\left( {\sin B - \sin C} \right)$ common we get,
$ \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,
$ \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,
$ \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$\left( {\sin A - \sin B} \right)$ common we get,
$ \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,
$ \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,
$ \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.