AP EAMCET · Maths · Properties of Triangles
If \(P\) is a point on the equation \(\mathbf{A D}\) of the \(\triangle A B C\), and \(\angle A B P=\frac{2 B}{3}\), then \(A P\) is equal to
- A \(C \sin \frac{B}{3}\)
- B \(2 C \sin \frac{B}{3}\)
- C \(C \sin \frac{2 B}{3}\)
- D \(2 C \sin \frac{2 B}{3}\)
Answer & Solution
Correct Answer
(B) \(2 C \sin \frac{B}{3}\)
Step-by-step Solution
Detailed explanation
\(\because A D\) is altitude of \(\triangle A B C\). \(\therefore \angle A D B=90^{\circ} \Rightarrow \angle A P B=90^{\circ}+\frac{B}{3}\) In \(\triangle A P B\), Using sine rule…
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