Determinants Matrices Question 83
Question: If in a triangle ABC, $ \begin{vmatrix} 1 & \sin A & {{\sin }^{2}}A \\ 1 & \sin B & {{\sin }^{2}}B \\ 1 & \sin C & {{\sin }^{2}}C \\ \end{vmatrix}=0 $ then the triangle is
Options:
A) Equilateral or isosceles
B) Equilateral or right-angled
C) Right angled or isosceles
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ \begin{vmatrix} 1 & \sin A & {{\sin }^{2}}A \\ 1 & \sin B & {{\sin }^{2}}B \\ 1 & \sin C & {{\sin }^{2}}C \\ \end{vmatrix}=0 $
$ \Rightarrow (sinA-sinB)(sinB-sinC)(sinC-sinA)=0 $
$ \Rightarrow \sin A=\sin Bor\sin B=\sin Cor\sin C=\sin A $
$ \therefore $ at least two of A, B, C are equal. Hence the triangle is isosceles or equilateral.
BETA
