A normal human eye can see an object making an angle of ${1.8}^{o}$ at the eye. What is the approximate height of object which can be seen by an eye placed at a distance of $1 m$ from the eye?

1 m

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\frac{π}{2} m

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1.8 m

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π c m

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Solution

The correct option is B $\frac{π}{2} m$
$tan θ = \frac{O}{A}$

$tan {1.8}^{\circ} = \frac{h}{1}$

$h = 1 \times tan {1.8}^{\circ}$

$∴ h = tan {1.8}^{\circ}$

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A normal human eye can see an object making an angle of 1.8o at the eye. What is the approximate height of object which can be seen by an eye placed at a distance of 1 m from the eye?

The correct option is B π2 mtanθ=OAtan1.8∘=h1h=1×tan1.8∘∴h=tan1.8∘

A normal human eye can see an object making an angle of ${1.8}^{o}$ at the eye. What is the approximate height of object which can be seen by an eye placed at a distance of $1 m$ from the eye?

The correct option is B $\frac{π}{2} m$
$tan θ = \frac{O}{A}$

$tan {1.8}^{\circ} = \frac{h}{1}$

$h = 1 \times tan {1.8}^{\circ}$

$∴ h = tan {1.8}^{\circ}$