JEE Mains · Maths · STD 11 - Trigonometrical equations
The angle of elevation of the top of a tower from a point \(A\) due north of it is \(\alpha\) and from a point \(B\) at a distance of \(9\) units due west of \(A\) is \(\cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)\). If the distance of the point \(B\) from the tower is \(15\) units, then \(\cot \alpha\) is equal to.
- A \(\frac{6}{5}\)
- B \(\frac{9}{5}\)
- C \(\frac{4}{3}\)
- D \(\frac{7}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{6}{5}\)
Step-by-step Solution
Detailed explanation
\(\text { given } OB =15\) \(\cos \beta=\frac{3}{\sqrt{13}}\) \(\tan \beta=\frac{h}{15}\) \(\frac{2}{3}=\frac{h}{15}\) \(10= h\) \(OA ^{2}+ AB ^{2}=225\) \(OA ^{2}+81=225\) \(OA =12\) \(\tan \alpha=\frac{10}{12}\) \(\cot \alpha=\frac{12}{10}=\frac{6}{5}\)
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