First slide

Applications of trigonometry

Question

AB is a vertical pole. The end A is on the level ground. ‘C’ is the middle point of AB. P is a point on the level ground. The portion CB subtends an angle $β$ at P. If AP=nAB, then $\tan β =$

Moderate

A

$\frac{n}{n^{2} + 1}$
B

$\frac{n}{2 n^{2} + 1}$
C

$\frac{n}{n^{2} - 1}$
D

$\frac{n}{2 n^{2} - 1}$

Solution

We have $A B = h \Rightarrow A C = B C = h / 2$
$A l s o A P = n . A B = n h$
$\begin{array}{l} ∴ \tan α = \frac{h / 2}{n h} = \frac{1}{2 n}, \tan (α + β) = \frac{h}{n h} = \frac{1}{n} \\ N o w \tan β = \tan (α + β - α) = \frac{\tan (α + β) - \tan α}{1 + \tan (α + β) \tan α} = \frac{\frac{1}{n} - \frac{1}{2 n}}{1 + \frac{1}{n} . \frac{1}{2 n}} = \frac{n}{2 n^{2} + 1} \end{array}$

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