Question

A 1.5 m tall boy is standing at same distance from a 30m tall building. The angle of allivation from his eyes to the top of the building increaseas from 30 to 60as he walks towards the building. Find the distance he walked

Answer 1

Given

A 1.5 m tall boy is standing at same distance from a 30m tall building. The angle of elevation from his eyes to the top of the building increaseas from 30°to 60°as he walks towards the building

To find

The distance the person walked

Explanation

From the figure,

Let AB be the person of height 1.5 m

CD=building of height 30m

Let the distance travelled by the person AA'=x

$CE=CD-ED$

since ED=AB

$=>CE=CD-AB$

$CE=30-1.5=28.5m$

Now

$\sf from \:\bigtriangleup AEC,$

It is a right angled triangle

$\angle EAC=30\degree$

$=>tan30\degree=\dfrac{CE}{AE}$

$=>\dfrac{1}{\sqrt{3}}=\dfrac{28.5}{AE}$

$=>\dfrac{1}{1.73}=\dfrac{28.5}{AE}$

$=>AE=1.73\times28.5$

$\underline{AE=49.3m}$

$\sf from \: \bigtriangleup A'EC,$

It is a right angled triangle

$\angle EA'C=60\degree$

$=>tan60\degree=\dfrac{CE}{A'E}$

$=>\sqrt{3}=\dfrac{28.5}{A'E}$

$=>1.73=\dfrac{28.5}{A'E}$

$=>A'E=\dfrac{28.5}{1.73}$

$\underline{A'E=16.4m}$

$\sf Distance \:travelled\: by\: the\: person \\ AA'=x=AE-A'E$

$x=49.3-16.4=32.9 m$

Distance travelled by the person=32.9m

Answer 2

ANSWER:-

Given:

A 1.5m tall boy is standing at some distance from a 30m tall building. The angle of elevation from his eyes to the top of the building increases from 30° from 60°, as he walks towards the building.

To find:

The distance he walked towards the building.

Solution:

Let AC be the observer of height 1.5m & BE be the building of height 30m.

The angle of elevation observes from the eyes increases from 30° to 60° to the top of the building.

⚫angle ECD= 30°

⚫angle EFD= 60°

⚫Let CF= R m & FD= M m

In right ∆EDF,

⚫DE= BE - BD

=) DE= 30 - 1.5

=) DE= 28.5m

$tan60 \degree = \frac{DE}{DF} \\ \\ = > \sqrt{3} = \frac{28.5}{M} \\ \\ = > \sqrt{3} M = 28.5 \\ \\ = > M = \frac{28.5 }{ \sqrt{3} } \\ [Rationalise] \\ = > \frac{28.5 \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } \\ \\ = > \frac{28.5 \sqrt{3} }{3} ...............(1)$

In right ∆EDC,

[tan30° = 1/√3]

$tan30 \degree = \frac{DE}{DC} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{28.5}{DF + CF} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{28.5}{M+ R} \\ [Cross \: multiplication] \\ = > R + M = 28.5 \sqrt{3} \\ \\ = > M = 28.5 \sqrt{3} - R................(2)$

Now,

Comparing equation (1) & (2), we get;

$= > \frac{28.5\sqrt{3}}{3} = 28.5 \sqrt{3} - R \\ \\ = > R = 28.5 \sqrt{3} - \frac{28.5 \sqrt{3} }{ 3 } \\ \\ = > R = \frac{3(28.5 \sqrt{3} ) - 28.5 \sqrt{3} }{3} \\ \\ = > R = \frac{85.5 \sqrt{3} - 28.5 \sqrt{3} }{3} \\ \\ = > R = \frac{57 \sqrt{3} }{3} \\ \\ = > R = 19 \sqrt{3} m$

Hence,

The distance he walked towards the building is (CF)=19√3m.

=) 19(1.732) [√3= 1.732]

=) 32.908m

Hope it helps ☺️