Question

24. 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° to 60° as he walks
towards the building. Find the distance he walked towards the building.​

Answer 1

★ Figure refer to attachment

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° to 60° as he walks towards the building.

TO FIND

Find the distance he walked towards the building.

SOLUTION

• Height of the boy = 1.5 m
• Height of the building = 30 m
• BG = CD
• DE = GF
• BF = CE

→ AB = AC - BC = 30 - 1.5 = 28.5m

In ∆ABF

→ tan 30° = AB/BF

→ 1/√3 = 28.5/BF

→ BF = 28.5√3 m

Now, In ∆ABG

→ tan 60° = AB/BG

→ √3 = 28.5/BG

→ BG = 28.5/√3

So, distance he walked towards a building

→ DE = CE - CD

→ DE = 28.5√3 - 28.5/√3

→ DE = 28.5*3 - 28.5/√3

→ DE = 85.5 - 28.5/√3

→ DE = 57/√3

→ DE = 19 × √3 × √3/√3

→ DE = 19√3m

Hence, distance travelled towards building is 19√3 m

Answer 2

✧ Given:-

• Height of boy(PQ) :- 1.5m
• Distance b/w boy and building(AB) :- 30m
• The angle of elevation increases from 30° to 60° as he walks towards the building.

✧ To find:-

• Distance when he walked towards the building.

✧ Solution:-

Angle of elevation from the initial point (Q) to top of building:- 30°

• Hence, <APC = 30°

Change in angle of elevation from P to R is 60°

• Hence, <ASC = 60°

★ PQ // CB

• So, PQ = CB = 1.5m

• Now, AC = AB - CB

• AC = 30 - 1.5

• AC = 28.5m

Also, BC // QB

• So, PS = QR and SC = RB

• Since, ACP = 90° becz tower is vertical.

In right angled ∆ APC :-

• $\sf{ \tan{P} = \dfrac{AC}{PC}}$

• $\sf{ \tan{30} = \dfrac{28.5}{PC}}$

• $\sf{ \dfrac{1}{ \sqrt{3}} = \dfrac{28.5}{PC}}$

• $\sf{ PC = 28.5 \sqrt{3}}$

In right angled ∆ ASC:-

• $\sf{ \tan{S} = \dfrac{AC}{SC}}$

• $\sf{ \tan{60} = \dfrac{28.5}{SC}}$

• $\sf{ SC = \dfrac{28.5}{ \sqrt{3}}}$

Now, PC = PS + SC

Putting value:-

• $\sf{ 28.5 \sqrt{3} = PS + \dfrac{28.5}{3}}$

• $\sf{ 28.5 \sqrt{3} - \dfrac{28.5}{3} = PS}$

• $\sf{ PS = \dfrac{28.5 \sqrt{3} × \sqrt{3} - 28.5}{ \sqrt{3}}}$

• $\sf{ PS = \dfrac{28.5(3 - 1)}{ \sqrt{3}}}$

• $\sf{ PS = \dfrac{28.52}{ \sqrt{3}}}$

Multiplying √3 in both numerator and denominator:-

• $\sf{ PS = \dfrac{28.5 × 2}{ \sqrt{3}} × \dfrac{ \sqrt{3}}{ \sqrt{3}}}$

• $\sf{ PS = \dfrac{28.5 × \sqrt{3} × 2}{3}}$

• $\sf{ PS = \dfrac{57 × \sqrt{3}}{2}}$

• $\sf{ PS = 19 \sqrt{3}}$

• Since, QR = PS

• QR = 19√3 metre

$\bold{\underline{\sf{\red{\dag \; Hence, \; he \; walked \; 19 \sqrt{3} \; metre \; towards \; the \; building.}}}}$

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