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Ex 9.1, 6 - A 1.5 m tall boy is standing at some distance - Ex 9.1

  1. Chapter 9 Class 10 Some Applications of Trigonometry
  2. Serial order wise
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Ex 9.1 , 6 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. Boy is 1.5 m tall So, PQ = 1.5 m Building is 30m tall, So, AB = 30 m Given that, angle of elevation from initial point (Q) to top of building = 30° Hence, ∠APC = 30° Boy moves from point Q to point R. angle of elevation changes to 60°. Hence, ∠ASC = 60°, Here, PQ, & CB are parallel lines So, PQ = CB = 1.5 m Now, AC = AB – CB AC = 30 – 1.5 AC = 28.5 m Also, PC & QB are parallel So, PS = QR & SC = RB Since tower is vertical, ∠ ACP = 90° Now, PC = PS + SC Putting values 28.5 √3 = PS + 28.5/√3 28.5 √3 – 28.5/√3 = PS PS = 28.5 √3 – 28.5/√3 PS = (28.5√3 × √(3 ) −28.5)/√3 PS = (28.5 × 3 − 28.5)/√3 PS = (28.5 (3 − 1))/√3 PS = (28.5 × 2)/√3. Multiplying √3 in both numerator and denominator PS = (28.5 × 2)/√3 × √3/√3 PS = (28.5 × √3 ×2)/3 PS = (57 × √3 )/3 PS = 19√3 metre Since QR = PS QR = 19√3 metre Hence, he walked 19√3 metre towards the building.

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