Question

A boy of height 150 cm tall is standing at some distance
from a 30 m tall building. The angle of elevation
increases from 30° to 60° as he walks towards the
building. Find the distance he walked towards the
building. [Use 3 = 1731​

Answer 1

Initially the angle of elevation of the top of the building of 30 m tall was 30°. Let the boy be at a distance d from the tower. given that the height of the boy is 150 cm = 1.5 cm.

Then,

$\displaystyle\longrightarrow\sf{\tan30^{\circ}=\dfrac{30-1.5}{d}}$

$\displaystyle\longrightarrow\sf{\dfrac{1}{\sqrt3}=\dfrac{28.5}{d}}$

$\displaystyle\longrightarrow\sf{d=28.5\sqrt3\ m}$

Taking $\displaystyle\sf{\sqrt3=1.731,}$

$\displaystyle\longrightarrow\sf{d=49.3335\ m}$

Let the boy move x distance (in m) towards the building making the angle of elevation 60°. Then the distance between him and the building will be (49.3335 - x) m.

Then,

$\displaystyle\longrightarrow\sf{\tan60^{\circ}=\dfrac{30-1.5}{49.3335-x}}$

$\displaystyle\longrightarrow\sf{\sqrt3=\dfrac{28.5}{49.3335-x}}$

$\displaystyle\longrightarrow\sf{49.3335-x=\dfrac{28.5}{1.731}}$

$\displaystyle\longrightarrow\sf{x=49.3335-\dfrac{28.5}{1.731}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{x=32.869\ m}}}$

Answer 2

QUESTION :-

•A boy of height 150 cm tall is standing at some distance from a 30 m tall building. The angle of elevation increases from 30° to 60° as he walks towards thebuilding. Find the distance he walked towards the building. [Use √3 = 1.732)

SOLUTION :-

➠ boy is 150cm tall, so PQ = 150cm.

➠ building is 30m tall, so AB = 30m.

➠ Given that, angle elevation from initial point to top of building = 30° ,hence, <APC = 30°

➠ boy move from point Q to point R. angle of elevation change to 60°.

➠ here ,PQ & CB are parallel lines, so, PQ = CB=150cm.

NOW,

➠ AC = CB- AB

➠ AC = 150 - 30

➠ AC = 120cm.

➠ also PC & QB is parallel So,

➠ .°. PS =QR & SC =RB.

➠ since, tower is vertical, <ACP=90°

➠ in right angle triangle APC.

➠tanP = AC / PC

➠ tan30° = 120 / PC

➠ 1/√3 = 120 /PC

➠ .°. PC = 120√3

➠ in right angle triangle ASC.

➠ tanS = AC/SC

➠tan 60° = 120/sc

➠ √3 = 120/Sc

➠ .°. SC = 120/√3

NOW,

➠ PC = PS + SC

PUTTING VALUES,

➠ 120√3 = PS + 120/√3

➠ PS = 120√3 - 120/√3

➠ PS = 120√3 ×√3 -120 / √3

➠ PS = 120×3 -120/√3

➠ PS = 120 ( 3-1)/√3

➠ PS = 120 ×2/√3

multiply √3 is both numerator and denominator.

➠ PS = 120 × 2 /√3 × √3/√3

➠ PS = 120 × √3 ×2/3

➠ PS = 40 × 2 ×√3

➠ PS = 80√3

➠ PS = 80 × 1.732 ( °.° √3 = 1.732)

➠ .°. PS = 138.56cm .

since, QR = PS

➠ .°. QR = 138.56cm.

So, he walked 138.56cm metres towards the building.