The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

Question 28
The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.
Let building be AB & tower be CD
Height of building = AB = 50 m
Let, Height of tower = PC
And, distance between tower and building = AC
Angle of depression to top of building = ∠ QDB = 30°
Angle of depression to bottom of building = ∠ QDA = 60°
We have to find
height of tower i.e. CD
And distance between tower &building i.e. AC
Draw BE parallel to AC & DQ
Lines DQ & BE are parallel,
And BD is the transversal
∠ DBE = ∠ QDB
∠ DBE = 30°
Now,
AC and BE are parallel lines.
So, AC = BE
Lines DQ & AC are parallel,
And AD is the transversal
∠ DAC = ∠ QDA
∠ DAC = 45°
Similarly, AB and CE are also parallel lines.
So, CD = AB
CD = 50 m
Also,
Since DC is perpendicular AC
∠ DEB = ∠ DCA = 90°
In right angle triangle DBE
tan B = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝐵)/(Side 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐵)
tan B = (" " 𝐷𝐸)/𝐵𝐸
tan 30° = 𝐷𝐸/𝐵𝐸
1/√3 = (" " 𝐷𝐸)/𝐵𝐸
BE = DE √3
In right angle triangle PAC
tan A = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐴)
tan 60° = (" " 𝐷𝐶)/𝐴𝐶
√3 = (" " 𝐷𝐶)/𝐴𝐶
AC = (" " 𝐷𝐶)/√3
"BE = " (" " 𝐷𝐶)/√3
From (1) & (2)
DE √3 = (" " 𝐷𝐶)/√3
DE√3 × √3 = DC
3DE = DC
3DE = DE + EC
3DE – DE = EC
2DE = EC
2DE = 50
DE = 50/2
DE = 25 m
∴ Height of tower = DC = DE + EC = 25 + 50 = 75 m
From (1)
BE = DE √3
Putting DE = 25
BE = 25√3 m
So, AC = BE = 25 √3 m
∴ Distance between two buildings = AC = 25√𝟑 m

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