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The angle of elevation of an object from a point P on the level ground is $α$ . Moving d metres on the ground towards the object, the angle of elevation is found to be $β$ . Then the height (in metres) of the object is

$dtan (α)$
$dcot (β)$
$\frac{d}{\cot (α) + \cot (β)}$
$\frac{d}{\cot (α) - \cot (β)}$

322.
A vertical pole subtends an angle $\tan^{- 1} (\frac{1}{2})$ at a point P on the ground. If the angles substended by the upper half and the lower half of the pole at P are respectively $α and β$ , then $(\tan (α), \tan (β))$ is equal to

$(\frac{1}{4}, \frac{1}{5})$

$(\frac{1}{5}, \frac{2}{9})$

$(\frac{2}{9}, \frac{1}{4})$

$(\frac{1}{4}, \frac{2}{9})$

$(\frac{2}{9}, \frac{1}{4})$

Let AC be a pole and point P be the position on the ground

$Given, θ = \tan^{- 1} (\frac{1}{2}) \Rightarrow \tan (θ) = \frac{1}{2} Also, θ = α + β \Rightarrow \tan (θ) = \tan (α + β) \Rightarrow \frac{1}{2} = \frac{\tan (α) + \tan (β)}{1 - \tan (α) \tan (β)} (a) When (\tan (α), \tan (β)) = (\frac{1}{4}, \frac{1}{5}) ∴ RHS = \frac{\frac{1}{4} + \frac{1}{5}}{1 - \frac{1}{4} \times \frac{1}{5}} = \frac{\frac{9}{20}}{\frac{19}{20}} = \frac{9}{19} \neq \frac{1}{2}, NOT SURE (b) When (\tan (α), \tan (β)) = (\frac{1}{5}, \frac{2}{9})$

$RHS = \frac{\frac{1}{5} + \frac{2}{9}}{1 - \frac{1}{5} \times \frac{2}{9}} = \frac{\frac{19}{45}}{\frac{43}{45}} = \frac{9}{19} \neq \frac{1}{2}, Not sure (c) When (\tan (α), \tan (β)) = (\frac{2}{9}, \frac{1}{4}) RHS = \frac{\frac{2}{9} + \frac{1}{4}}{1 - \frac{2}{9} \times \frac{1}{4}} = \frac{\frac{17}{36}}{\frac{34}{36}} = \frac{1}{2}, true$