Determinants Matrices Question 85
Question: If A, B, and C are the angles of a triangle and $ \begin{vmatrix} 1 & 1 & 1 \\ 1+\sin A & 1+\sin B & 1+\sin C \\ \sin A+{{\sin }^{2}}A & \sin B+{{\sin }^{2}}B & \sin C+{{\sin }^{2}}C \\ \end{vmatrix}=0, $ then the triangle must be
Options:
A) Isosceles
B) Equilateral
C) Right-angled
D) None of these
Show Answer
Answer:
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 = \begin{vmatrix} 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{vmatrix} $ 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) $
$ \begin{vmatrix} 1 & 0 & 0 \\ 1+\sin A & 1 & 1 \\ \sin A+{{\sin }^{2}}A & \sin B+\sin A & \sin C+\sin A \\ \end{vmatrix} $
$ =(sinB-sinA)(sinC-sinA)(sinC-sinB) $ 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.
BETA
