A) Isosceles
B) Equilateral
C) Right-angled
D) None of these
Correct Answer: A
Solution :
[a] Using \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\] and \[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\] in the given determinant, we have \[\Delta =\left| \begin{matrix} 1 & 0 & 0 \\ 1+\sin A & \sin B-\sin A & \sin C-\sin A \\ \sin A+{{\sin }^{2}}A & {{\sin }^{2}}B-{{\sin }^{2}}A & {{\sin }^{2}}C-{{\sin }^{2}}A \\ \end{matrix} \right|\] Now taking \[\sin B-\sin A\] common from \[{{C}_{2}}\] and sin C - sin A common from \[{{C}_{3}},\] we have \[\Delta =(sinB-sinA)(sinC-sinA)\] \[\left| \begin{matrix} 1 & 0 & 0 \\ 1+\sin A & 1 & 1 \\ \sin A+{{\sin }^{2}}A & \sin B+\sin A & \sin C+\sin A \\ \end{matrix} \right|\] \[=(sin\,\,B-sin\,\,A)(sin\,\,C-sin\,\,A)(sin\,\,C-sin\,\,B)\] As the determinant is zero, we must have sin B = sin A or sin A or sin C = sin A or sin C = sin B, that is B = A or C = A or C = B. In all three cases we will have an isosceles triangle.
You need to login to perform this action.
You will be redirected in
3 sec
