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

A student Shreya of height 1 m visited India Gate as a part of her study tour. 

i. What is the angle of elevation from Shreya’s eye to the top of India Gate, if she is standing at a distance of 41m away from the India Gate?
ii. If Shreya observes the angle of elevation from her eye to the top of India Gate to be 60°, then how far is the she standing from the base of the India Gate?
iii. 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.
iv. If Shreya moves to a point which is at a distance of (41/√3) m from the from India Gate, then find the angle of elevation made by her eye to the top of India Gate.

Q38 – Sample Question Paper – Set 1 – Maths Standard – CBSE 2026

Ans: 

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STEP BY STEP SOLUTION

(i) Angle of elevation when Shreya is 41m away from the India Gate:

Let’s start by making a diagram for our better understanding of the question:

In this diagram, AB is India Gate of height 42 m, PQ is Shreya of height 1 m. Distance of Q from point B is 41 m.

We need to find angle of elevation of point A from point P.

Now from point P, we draw a horizontal line PC which parallel to QB. It cuts AB at point C.

Since PC is parallel to QB, ∴ PC = QB = 41 m

∴ CB = PQ = 1 m

∴ AC = AB – CB = 42 – 1 = 41 m

Next in Δ APC, we need to find ∠ P, let’s consider it as θ.

Here in Δ APC, we have value of height as well as base, so we can apply tan θ formula to get value of θ

∵ tan θ = AC / PC

∴ tan θ = 41 / 41 = 1

We know that tan 45⁰ = 1

∴ tan θ = tan 45⁰

∴ θ = 45⁰

Therefore, angle of elevation is 45⁰

(ii) Distance from the base of the India Gate at AOE of 60⁰:

Let’s make a diagram for this case:

Here, we already have Height AB = 42 m, Angle θ = 60⁰ (given) and now distance of Q from B is to be calculated.

In Δ APC, tan θ = AC / PC

We have AC = 41 m (calculated above)

From the diagram, we can see that PC = BQ = D.

∴ tan 60⁰ = 41 / D

∴ √3 = 41 / D

∴ D = 41 / √3

Therefore, Shreya’s distance from India Gate is 41 / √3 m.

(iii) Distance moved back when AOE changed from changes from 45⁰ to 30⁰:

Let’s start by making a diagram for our better understanding of the question:

Here, when Shreya was standing at point Q, angle of elevation at point P was θ and we calculated it to be 45⁰ in part (i).

She moved back by distance D to point Q’ and angle of elevation at point P’ changed to 300. We need to find the distance D shown by Q’Q.

Since Q’Q = P’ P = D, if we find value of P’P, we will get value of Q’Q

In Δ AP’C, tan 30⁰ = AC / P’C

Here, AC = 41 m (calculated in part (i) above)

P’C = P’P + PC = P’P + QB = D + 41

∴ tan 300 = 41 / (D + 41)

∴ 1/ √3 = 41 / (D + 41)

∴ D + 41 = 41 √3

∴ D = 41 √3 – 41

∴ D = 41 (√3 – 1) m

Therefore, the distance moved back by Shreya is 41 (√3 – 1) m.

(iv) Angle of elevation when Shreya is 41 / √3 m away from the India Gate:

Let’s start by making a diagram for our better understanding of the question:

In this diagram, distance of Q from point B is 41 / √3 m.

In Δ APC, tan θ = AC / PC

Here, AC = 41 m (calculated above in part (i))

Since PC is parallel to QB, ∴ PC = QB = 41 / √3 m

∴ tan θ = 41 / (41 / √3)

∴ tan θ = 41 √3 / 41

∴ tan θ = √3

We know that for tan 60⁰ = √3

∴ tan θ = tan 60⁰

∴ θ = 60⁰

Therefore, angle of elevation is 60⁰

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