If the angle of elevation of the top of a tower of height 100mt from a point to its foot is tan−1(45) , then distance from the point to its foot is

θ=tan−1(45)⇒tanθ=45⇒ABBC=45AB=4K;BC=5K⇒4K=100⇒K=25;BC=5×25=125m

If the angle of elevation of the top of a tower of height 100mt from a point to its foot is tan−1(45) , then distance from the point to its foot is

θ=tan−1(45)⇒tanθ=45⇒ABBC=45AB=4K;BC=5K⇒4K=100⇒K=25;BC=5×25=125m

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If the angle of elevation of the top of a tower of height $100 m t$ from a point to its foot is $\tan^{- 1} (\frac{4}{5})$ , then distance from the point to its foot is

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Detailed Solution

$θ = \tan^{- 1} (\frac{4}{5})$

$\Rightarrow \tan θ = \frac{4}{5}$

$\Rightarrow \frac{A B}{B C} = \frac{4}{5}$

$A B = 4 K; B C = 5 K$

$\Rightarrow 4 K = 100$

$\Rightarrow K = 25; B C = 5 \times 25 = 125 m$

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If the angle of elevation of the top of a tower of height 100mt from a point to its foot is tan−1(45) , then distance from the point to its foot is

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If the angle of elevation of the top of a tower of height $100 m t$ from a point to its foot is $\tan^{- 1} (\frac{4}{5})$ , then distance from the point to its foot is