Question

Find the length which at a distance of 5280 m will subtend an angle of 1’ at the eye.

Answer 1

Hint: Draw a diagram for the given information in question and use trigonometric ratio $''\tan \theta ''$ for finding the relation between given quantity and unknown.
As 1’ is a very small angle, use approximation.

Complete step-by-step answer:

Let the required length be ‘l’.

We can use the formula of $\tan \theta $ (for $\angle A$) to find the relation between BC and AC.

We know, $\tan \theta =\dfrac{Perpendicular}{Base}$

$\tan A=\dfrac{BC}{AC}$

Given, $\angle A=1'$

As angle $\angle A$ is very small, we have to use approximation.

We know for small angles: $\sin \theta \approx \tan \theta \approx \theta $

$\begin{align}

  & \Rightarrow \tan 1'=\dfrac{BC}{AC} \\

 & \Rightarrow 1'=\dfrac{BC}{AC} \\

 & \Rightarrow {{\left( \dfrac{1}{60} \right)}^{{}^\circ }}=\dfrac{BC}{AC} \\

\end{align}$

There are 60 minutes in 1 degree;

i.e. $60'=1{}^\circ $

$\Rightarrow 1'={{\left( \dfrac{1}{60} \right)}^{{}^\circ }}$

Now, we have to convert the degree into radian.

Formula: $\left( x\ \text{degree} \right)\times \dfrac{\pi }{180}=\left( \text{equivalent radian} \right)$

On changing ${{\left( \dfrac{1}{60} \right)}^{{}^\circ }}$into radian, equation will become;

$\Rightarrow {{\left( \dfrac{1}{60} \right)}^{{}^\circ }}\times \dfrac{\pi
}{180}=\dfrac{BC}{AC}$

$\Rightarrow \dfrac{1}{60}\times \dfrac{\pi }{180}=\dfrac{l}{5280}$

Using cross multiplication, equation will become;

$\Rightarrow 60\times 180\times l=5280\times \pi $

Dividing both sides by $\left( 60\times 180 \right)$, equation will become;

$\begin{align}

  & \Rightarrow l=\dfrac{5280\times \pi }{60\times 180} \\

 & \Rightarrow l=\dfrac{5280\times 3.14}{60\times 180}\left( \pi =3.14 \right) \\

 & \Rightarrow l=\dfrac{16579.2}{60\times 180} \\

 & \Rightarrow l=1.535 \\

\end{align}$

Hence the length which at a distance of 5280 m subtends an angle of 1’ at eye = 1.525m.

Note: Another method:

\[\begin{align}

  & \theta =\dfrac{l}{r} \\

 & 1'=\dfrac{l}{5280} \\

 & {{\left( \dfrac{1}{60} \right)}^{{}^\circ }}\times \dfrac{\pi }{180}=\dfrac{l}{5280} \\

 & \Rightarrow l=\dfrac{5280\times \pi }{60\times 180} \\

 & \Rightarrow l=1.535m \\

\end{align}\]

A student can make mistakes by forgetting to change the angle from degrees to radian. When we are applying the formula, we are taking other quantities in their S.I unit. So, we have to take $\theta $ also in its S.I unit. S.I unit of $\theta $ is radian.