Q) The angle of elevation of the top of a building from a point A, on the ground, is 30 deg On moving a distance of 24 m towards its base to the point B, the angle of elevation changes to 60 deg. Find the height of the building and distance of point A from the base of the building. (Take √3 = 1.73)

(Q 34A – 30/3/3 – CBSE 2026 Question Paper)

Ans:

Step 1: Let’s draw the diagram for better understanding of the question:

Here, BC is the building and angle of elevation (AOE) from A is 30 deg. Let’s consider building is of ‘H’ height. After moving 24 m distance, from point D, AOE is now 60 deg. We need to find height H and Length DB.

Step 2: In Δ ABC, tan 30 =

∴

∴ P + 24 = H √3 …………(i)

Step 2: Now, In Δ CBD, tan 60 =

∴ √3 =  

∴ H = P √3 …………. (ii)

Step 3: By solving equation (i) & equation (ii), we get:

∴ P + 24 = (P√3) √3

∴ P + 24 = 3 P

∴ 3 P – P = 24

∴ 2 P = 24

∴ P = 12

(i) Height of the building:

From equation (ii), we have H = P √3

∵ P = 12 m           (from step 3)

∴ H = 12 √3 = 12 x 1.73          (given √3 = 1.73)

= 20.76

Therefore, the height of the building is 20.76 m

(ii) Distance of point A from the base of the building:

From the diagram, distance of point A from the base of the building is given by AB

∵ AB = AD + BD

∴ AB = 24 + P

∵ P = 12             (from step 3)

∴ AB = 24 + 12 = 36

Therefore, the distance of point A from the base of the building is 36 m.

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