If in a triangle $$ABC,$$ \[\left| {\begin{array}{*{20}{c}}
1&{\sin A}&{{{\sin }^2}A}\\
1&{\sin B}&{{{\sin }^2}B}\\
1&{\sin C}&{{{\sin }^2}C}
\end{array}} \right| = 0\] then the triangle is
A.
equilateral or isosceles
B.
equilateral or right-angled
C.
right angled or isosceles
D.
None of these
Answer :
equilateral or isosceles
Solution :
\[\left| {\begin{array}{*{20}{c}}
1&{\sin A}&{{{\sin }^2}A}\\
1&{\sin B}&{{{\sin }^2}B}\\
1&{\sin C}&{{{\sin }^2}C}
\end{array}} \right| = 0\]
$$\eqalign{
& \Rightarrow \left( {\sin A - \sin B} \right)\left( {\sin B - \sin C} \right)\left( {\sin C - \sin A} \right) = 0 \cr
& \Rightarrow \sin A = \sin B{\text{ or }}\sin B = \sin C{\text{ or }}\sin C = \sin A \cr} $$
∴ atleast two of $$A, B, C$$ are equal.
Hence the triangle is isosceles or equilateral.
Releted MCQ Question on Algebra >> Matrices and Determinants
Releted Question 1
Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$ be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$ be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then
A.
$$C$$ is empty
B.
$$B$$ has as many elements as $$C$$
C.
$$A = B \cup C$$
D.
$$B$$ has twice as many elements as elements as $$C$$
Let $$a, b, c$$ be the real numbers. Then following system of equations in $$x, y$$ and $$z$$
$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$ has