Question

A person of normal eye sight can read points at such a distance that the letters subtend an angle of 5' at the eye. find the height of the letters he can read at a distance of 2640m. ​

Answer 1

Given:

Letters subtend an angle of 5' (read as 5 minutes.)

Distance between eye and letters is 2640 m.

To find:

The height of letters = ?

Solution:

Please refer to the attached figure which depicts the given situation in the form of a right angle triangle $\triangle ABC$.

A is the location of eye.

The letters are at point B and point C subtends the angle of 5' on eye.

We have to find the side BC of the triangle (Height of the letters).

We can use trigonometric identities to solve for side BC.

First of all, let us convert given angle to degrees:

$\angle A = 5' = \dfrac{5}{60}^\circ =0.083^\circ$

Now, let us use tangent:

$tan\theta = \dfrac{Perpendicular}{Base}$

In the given triangle:

$tanA = \dfrac{BC}{AB}\\\Rightarrow BC = AB \times tanA\\\Rightarrow BC = 2640 \times tan (0.083)\\\Rightarrow BC = 2640 \times 0.001448\\\Rightarrow BC = \bold{3.824\ m}$

So, the height of letters is 3.824 m.

Answer 2

Step-by-step explanation:

Given:

Letters subtend an angle of 5' (read as 5 minutes.)

Distance between eye and letters is 2640 m.

To find:

The height of letters = ?

Solution:

Please refer to the attached figure which depicts the given situation in the form of a right angle triangle \triangle ABC△ABC .

A is the location of eye.

The letters are at point B and point C subtends the angle of 5' on eye.

We have to find the side BC of the triangle (Height of the letters).

We can use trigonometric identities to solve for side BC.

First of all, let us convert given angle to degrees:

\angle A = 5' = \dfrac{5}{60}^\circ =0.083^\circ∠A=5

′

=

60

5

∘

=0.083

∘

Now, let us use tangent:

tan\theta = \dfrac{Perpendicular}{Base}tanθ=

Base

Perpendicular

In the given triangle:

\begin{lgathered}tanA = \dfrac{BC}{AB}\\\Rightarrow BC = AB \times tanA\\\Rightarrow BC = 2640 \times tan (0.083)\\\Rightarrow BC = 2640 \times 0.001448\\\Rightarrow BC = \bold{3.824\ m}\end{lgathered}

tanA=

AB

BC

⇒BC=AB×tanA

⇒BC=2640×tan(0.083)

⇒BC=2640×0.001448

⇒BC=3.824 m

So, the height of letters is 3.824 m.

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