AP EAMCET · Maths · Properties of Triangles
In \(\triangle \mathrm{ABC}\), if \(\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=0\), then \(\cos \mathrm{A} \cos \mathrm{B}+\cos \mathrm{B} \cos \mathrm{C}+\cos \mathrm{C} \cos \mathrm{A}=\)
- A -1
- B \(\frac{3}{4}\)
- C \(\frac{9}{4}\)
- D 1
Answer & Solution
Correct Answer
(B) \(\frac{3}{4}\)
Step-by-step Solution
Detailed explanation
\( \left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right| = 3abc - (a^3+b^3+c^3) = 0 \) \( a^3+b^3+c^3 = 3abc \) Since \( a,b,c \) are sides of a triangle, \( a,b,c > 0 \). Thus, \( a=b=c \). \( \triangle \mathrm{ABC} \) is equilateral, so…
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