AP EAMCET · Maths · Properties of Triangles
In \(\triangle A B C\), if \(b \cos \theta=c-a\), (where \(\theta\) is an acute angle), then \((c-a) \tan \theta=\)
- A \(2 \sqrt{c a} \cos \frac{B}{2}\)
- B \(2 \sqrt{c a} \sin \frac{B}{2}\)
- C \(2 c a \cos \frac{B}{2}\)
- D \(2 c a \sin \frac{B}{2}\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{c a} \sin \frac{B}{2}\)
Step-by-step Solution
Detailed explanation
We have, \(b \cos \theta=c-a \Rightarrow \cos \theta=\frac{c-a}{b}\) \[ \begin{aligned} & \therefore \quad \sin \theta=\frac{\sqrt{b^2-(c-a)^2}}{b} \\ & \text { and } \tan \theta=\frac{\sqrt{b^2-(c-a)^2}}{(c-a)} \end{aligned} \] and…
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