JEE Mains · Maths · STD 11 - Trigonometrical equations
Let a vertical tower \(AB\) have its end \(A\) on the level ground. Let \(C\) be the mid-point of \(AB\) and \(P\) be apoint on the ground such that \(AP=2AB\) . If \(\angle BPC = \beta \) then \(\tan \beta \) is equal to :
- A \(\frac{4}{9}\)
- B \(\frac{6}{7}\)
- C \(\frac{1}{4}\)
- D \(\frac{2}{9}\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{9}\)
Step-by-step Solution
Detailed explanation
Let \(\angle APC\, = \,\alpha \) \(\tan \,\alpha \, = \,\frac{{AC}}{{AP}}\, = \,\frac{1}{2}\frac{{AB}}{{AP}}\, = \,\frac{1}{4}\) (\(\because C\) is the mid point \(\therefore AC=\frac {1}{2}AB\)) \(\Rightarrow \tan \alpha =\frac {1}{4}\) As…
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