TS EAMCET · Maths · Properties of Triangles
If \(A(1,2,-3), B(2,3,-1), C(3,1,1)\) are the vertices of \(\triangle A B C\) then \(\left|\frac{\cos A}{\cos B}\right|=\)
- A \(\frac{3 \sqrt{3}}{4 \sqrt{2}}\)
- B \(\frac{3 \sqrt{3}}{\sqrt{7}}\)
- C \(\frac{4 \sqrt{2}}{3 \sqrt{3}}\)
- D \(\frac{\sqrt{7}}{3 \sqrt{3}}\)
Answer & Solution
Correct Answer
(B) \(\frac{3 \sqrt{3}}{\sqrt{7}}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } A B=\sqrt{1+1+4}=\sqrt{6}=c ; B C=\sqrt{1+4+4}=3=a \\ & A C=\sqrt{4+1+16}=\sqrt{21}=b \\ & \cos A=\frac{b^2+c^2-a^2}{2 b c}=\frac{21+6-9}{2 \sqrt{2 \times 3 \times 3 \times 7}}=\frac{18}{6 \sqrt{14}}=\frac{3}{\sqrt{14}} \\ & \cos…
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