Question

Q4. A group of students of class X visited Qutub Minar on an educational trip. The teacher and students had interest in history as well. The teacher narrated that Qutub Minar, a 73 m high tower of victory, was started to be build in 1198 by Qutub-ud-din Aibak. Based on the above information, answer the following questions: (3) (i) Calculate the angle of elevation if students are standing at a distance of 73 m away from Qutub Minar. (ii) If they want to click a picture of Qutub Minar at an angle of 60° then, at what distance they should stand? (iii) If they walk away from the point at which the angle of elevation was 60° in such a way that the angle changes to 30°, then find the distance covered by them. ​

Answer 1

1 = 45°

2= 25.24m

3rd I do not known but I hope u get ur answer

Answer 2

Answer:

(i) the angle of elevation if students are standing at a distance of 73 m away from Qutub Minar is 45°

(ii) the distance between the tower and the students is 42.14 m

(iii) then the distance covered by them, from an angle of 60° to 30° is 84.29 m

Step-by-step explanation:

given, height of Qutub Minar = 73 m

(i) distance between person and the tower = 73 m

to find: angle of elevation

tanθ = $\frac{73}{73}$ = 1

θ = 45°

thus, the angle of elevation if students are standing at a distance of 73 m away from Qutub Minar is 45°

(ii) given angle of elevation = 60°

to find: distance between the tower and the students

let the distance between the tower and the students be 'x'

tan 60 = $\frac{h}{x}$

√3 = $\frac{73}{x}$

x = $\frac{73}{\sqrt{3} }$  = 42.14

thus, the distance between the tower and the students is 42.14 m

(iii) let the distance between the tower and the students when the angle of elevation is 30° be (42.14+x)

tan 30 =  $\frac{height}{distance \ between \ the \ tower \ and \ the \ students}$

$\frac{1}{\sqrt{3} } = \frac{73}{42.14+x}$

42.14+x = 73√3

x = 73√3 - 42.14

x = 126.43-42.14 = 84.29

thus, then the distance covered by them, from an angle of 60° to 30° is 84.29 m