Ex 9.1 Q6 - A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60°

Ex 9.1 Q6 - A 1.5 m tall boy is standing at some distance from a 30 m 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.

Ex 9.1 Q6 - A 1.5 m tall boy is standing at some distance from a 30 m 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.

Solution:-

Let,
     *PQ be the height of the boy
     *AB be the height of the building

Ex 9.1 Q6 - A 1.5 m tall boy is standing at some distance from a 30 m 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.

Given,
     *Height of the boy = PQ = CB = 1.5 m
     *Height of the building = AB = 30 m

   ∴ AC = AB - CB
   ∴ AC = 30 - 1.5
   ∴ AC = 28.5 m       ------- (i)

   *Angle of elevation from starting point Q to the top of the building = ∠APC = 30°

   *Angle of elevation from final point S to the top of the building = ∠ARC = 60°

➙ In ∆APC, ∠ACP = 90°,

   ∴ tanP =
Opposite side of ∠P / Adjacent side of ∠P
   ∴ tan30° =
AC / PC
   ∴ tan30° =
28.5 / PC
       { from (i) }
   ∴
1 / √3
=
28.5 / PC
       { ∵ tan30° = 1/√3 }
   ∴ PC = 28.5√3         ------- (ii)

➙ In ∆ARC, ∠ACR = 90°,

   ∴ tanR =
Opposite side of ∠R / Adjacent side of ∠R
   ∴ tan60° =
AC / RC
   ∴ tan60° =
28.5 / RC
        { from (i) }
   ∴ √3 =
28.5 / RC
        { ∵ tan60° = √3 }
   ∴ RC =
28.5 / √3

*Multiplying Numerator and Denominator by √3,

   ∴ RC =
28.5 / √3
×
√3 / √3
   ∴ RC =
28.5√3 / 3
   ∴ RC = 9.5√3       ------- (iii)

Now,

   *The distance walked by the boy towards the building
   = QS
   = PR
   = PC - RC
   = 28.5√3 - 9.5√3       { from (i), (ii) }
   = 19√3

Hence,
          The distance walked by the boy towards the building is 19√3 m.
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Try This..

✒ A kite is flying at a height of 60 m. The angle of elevation is 30°. Find the length of the string.

✒ The angle of elevation to the top of a building from two points on the ground is 30° and 60°. Find the distance between the two points.

✒ A man standing 40 m away from a tower sees the top of the tower at 45°. Find the height of the tower.

❌ Common Mistakes

▸ Forgetting to subtract the height of the observer (like the 1.5 m of the boy here).

▸ Using wrong trigonometric values (like confusing tan 30° and tan 60°).

Skipping diagrams:
▸ A rough sketch can save you from silly errors.

📝 Related Questions:-


Queries Solved:-

Class 10 Ex 9.1
Ex 9.1 Q6 Class 10
Class 10 Ex 9.1 Q6
Class 10 Chap 9 Ex 9.1 Q6
Class 10 Some Applications Of Trigonometry

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