Question

India gate (formerly known as the All India war memorial) is located near Karthavya path (formerly Rajpath) at New Delhi. It stands as a memorial to 74187 soldiers of the Indian Army, who gave their life in the first world war. This 42 m tall structure was designed by Sir Edwin Lutyens in the style of Roman triumphal arches. A student named Shreya, who is 1 m tall, visited India Gate as part of her study tour.

1. What is the angle of elevation from Shreya’s eye to the top of India Gate if she is standing at a distance of 41 m away from the India Gate?   (1)
2. If Shreya observes the angle of elevation from her eye to the top of India Gate to be 60°, then how far is she standing from the base of the India Gate?   (1)
3. 1. If the angle of elevation from Shreya’s eye changes from 45° to 30°, when she moves some distance back from the original position. Find the distance she moves back.   (2)
      OR
   2. If Shreya moves to a point which is at a distance of `41/sqrt(3)` m from the India Gate, then find the angle of elevation made by her eye to the top of the India Gate.  (2)

Answer 1

i.

Shreya is 1 m tall.

So, PQ = 1 m

India gate is 42 m tall.

So, AB = 42 m

Now, AC = AB − BC

Since PQ = BC

AC = AB − PQ

= 42 − 1

= 41 m

Also, Shreya is standing at a distance of 41 m away.

So, QB = PC = 41 m

In right angle triangle APC,

tan P = `"Side opposite to angle P"/"Side adjacent to angle P"`

= `(AC)/(PC)`

= `41/41`

= 1

Thus, ∠P = 45°

So, the required angle of elevation is 45°.

ii.

In right angle triangle AMC,

Tan M = `"Side opposite to angle M"/"Side adjacent to angle M"`

tan 60° = `(AC)/(MC)`

`sqrt3 = 41/(MC)`

MC = `41/sqrt3`

Multiplying `sqrt3` in both numerator and denominator.

MC = `41/sqrt3 xx sqrt3/sqrt3`

MC = `(41 sqrt3)/3` m

∴ Shreya is standing at a distance of `(41sqrt(3))/3` m.

 iii. (A)

She moves back, so she is going from left to right.

We need to find distance HF.

Also, by symmetry HF = GE

In right angle triangle AGC,

tan G = `"Side opposite to angle G"/"Side adjacent to angle G"`

tan 30° = `(AC)/(GC)`

`1/sqrt3 = 41/(GC)`

GC = `41sqrt3`

In right angle triangle AEC,

tan E = `"Side opposite to angle E"/"Side adjacent to angle E"`

tan 45° = `(AC)/(EC)`

1 = `41/(EC)`

EC = 41

Now, GC = GE + EC

Putting values:

`41sqrt3` = GE + 41

`41sqrt3 - 41` = GE

GE = `41(sqrt3 - 1)`

∴ The distance she moved back is `41(sqrt3 - 1)` m.

OR

(B)

Now, Shreya is standing at a distance of `41/sqrt3` m away.

So, YB = XC = `41/sqrt3` m

In right angle triangle AXC,

tan X = `"Side opposite to angle X"/"Side adjacent to angle X"`

= `(AC)/(XC)`

= `41/(41/sqrt3)`

= `41 xx sqrt3/41`

= `sqrt3`

Thus, ∠X = 60°

So, the required angle of elevation is 60°.