TS EAMCET · Maths · Properties of Triangles
In a \(\triangle A B C\), the mid-point of \(B C\) is \(D\). If \(A D\) is perpendicular to \(A C\), then \(\cos A \cos C=\)
- A \(\frac{1}{3} \frac{c^2+a^2}{a b}\)
- B \(\frac{2\left(c^2+a^2\right)}{a b}\)
- C \(\frac{2\left(c^2-a^2\right)}{3 a c}\)
- D \(\frac{3\left(a^2+b^2\right)}{2 b c}\)
Answer & Solution
Correct Answer
(C) \(\frac{2\left(c^2-a^2\right)}{3 a c}\)
Step-by-step Solution
Detailed explanation
\( \begin{aligned} & \text { In } \triangle A C D, \cos C=\frac{b}{\left(\frac{a}{2}\right)}=\frac{2 b}{a} \\ & \text { In } \triangle A B C, \cos C=\frac{a^2+b^2-c^2}{2 a b} \end{aligned} \) From Eqs. (i) and (ii), we get…
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