The angles of depression of the top and bottom of a 8 m tall building from the top of a tower are 30° and 45° respectively. Find the height of the tower and the distance between the tower and the building.

OR

As observed from the top of a lighthouse, 75 m high from the sea level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Let building be AB and tower be CD

Height of building, AB = 8 m

Let height of tower = CD

and, distance between tower and building = AC

Angle of depression to top of building, ∠QDB = 30°

Angle of depression to bottom of building, ∠QDA = 45°

We have to find

Height of Tower i.e. CD

and distance between tower and building i.e. AC

Construction: Draw BP parallel to AC & DQ

Lines DQ & BP are parallel and BD is the transversal

∠DBP = ∠QDB [Alternate angles]

∠DBP = 30°

Lines DQ & AC are parallel and AD is the transversal

∠DAC = ∠QDA [Alternate angles]

∠DBP = 45°

Now,

AC and BD are parallel lines

So, AC = BD

Similarly, AB and CP are also parallel lines

So, CP = AB = 8m

Also,

Since DC ⊥ AC

∠DPB = ∠DCA = 90°

In right Δ PBD, we have

⇒ BP = DP√3 …(i)

In right Δ CAD, we have

[∵ AC = BP]

⇒ BP = CD …(ii)

From (i) and (ii), we have

DP√3 = CD

⇒ DP√3 = DP + PC

⇒ DP√3 = DP + AB [∵ AB = PC]

⇒ DP√3 = DP + 8

⇒ DP√3 – DP = 8

⇒ DP(√3 – 1) = 8

Rationalizing

[∵ (a – b)(a + b) = (a² – b²)]

⇒ DP = 4(√3 + 1)

Now,

CD = DP + PC

= 4(√3 + 1) + 8

= 4√3 + 4 + 8

= 4√3 + 12

= 4(√3 + 3)

So, height of tower, CD = 4(√3 + 3)m

From eq.(i), we have

BP = DP√3

= 4(√3 + 1)√3

= 4√3(√3 + 1)

= 4 × 3 + 4√3

= 12 + 4√3

= 4(3 + √3)m

So, AC = BP = 4(3 + √3)m

∴ Distance between two buildings = AC = 4(3 + √3)m

OR

Given: Height of lighthouse, AD = 75 m

Angle of depression of first ship, ∠PAC = 45°

Angle of depression of second ship, ∠PAB = 30°

To find: distance between two ships i.e. BC

Lines PA & BD are parallel and AB is the transversal

∠ABD = ∠PAB [Alternate angles]

So, ∠ABD = 30°

Lines PA & BD are parallel and AC is the transversal

∠ACD = ∠PAC [Alternate angles]

So, ∠ACD = 45°

Since, lighthouse is perpendicular to ground

∠ADB = 90°

In right Δ ACD, we have

⇒ CD = 75m

In right Δ ABD, we have

⇒ BD = 75√3

⇒ BC + CD = 75√3

⇒ BC + 75 = 75√3

⇒ BC = 75(√3 – 1)

∴ Distance between two ships = 75(√3 – 1)m