Q.1 The angle of depression of two ships from an aeroplane flying at the height of
7500 𝑚 are 30$°$ and 45$°$ . If both the ships are in the same line that one ship is exactly behind the other, find the side such distance between the ships.

 

(OR)

 

Q.2 The angle of elevation of the top 𝑄 of a vertical tower 𝑃𝑄 from a point 𝑋 on the ground is 60$°$. From a point 𝑦 40 𝑚 vertically above 𝑋, the angle of elevation of the top 𝑄 of the tower is 45$°$. Find the height of the tower 𝑃𝑄 and the distance 𝑃𝑋. (Use $\sqrt{3}$= 1. 73)

We are given the height at which the aeroplane is flying and the angle of depressions it makes with the two ships.

The height at which the aeroplane is flying, ℎ = 7500 𝑚

The angle of depression of the first ship, θ = 30$°$

And the angle of depression of the second ship, θ' = 45$°$

The figure depicting this condition is shown in the given figure,

Here, 𝑥 is the distance between the two ships. In ∆𝐴𝐵𝐶, ∠𝐵 is the right angle.
So, 𝑡𝑎𝑛 θ' =   𝐴𝐵/BC
⇒ 𝑡𝑎𝑛 45$°$ = 7500/𝑦
⇒ 1 = 7500 
⇒ 𝑦 = 7500 𝑚
Now, in ∆𝐴𝐵𝐷, ∠𝐵 is the right angle. So, 𝑡𝑎𝑛 θ =    𝐴𝐵/BD
⇒ 𝑡𝑎𝑛 30$°$ =7500/𝑥+𝑦

⇒    1/$\sqrt{3}$ =    7500/𝑥+7500
⇒ 𝑥 + 7500 = 7500 × $\sqrt{3}$
⇒ 𝑥 + 7500 = 7500 × 1. 73
⇒ 𝑥 + 7500 = 12975
⇒ 𝑥 = 12975 − 7500
⇒ 𝑥 = 5475 𝑚
Hence, the distance between the two ships is 5475 𝑚

(OR)

We are given the angle of elevation from the top of the tower at two points, and we have to find the height of the tower and the distance 𝑃𝑋.

The given situation is depicted in the figure:

The height of the tower is 𝑃𝑄.
∠𝑃𝑋𝑄 = 60$°$
∠𝑅𝑌𝑄 = 45$°$
The distance 𝑋𝑌 = 40 𝑚
As 𝑃𝑋𝑌𝑅 forms a rectangle, so 𝑋𝑌 = 𝑃𝑅
⇒ 𝑃𝑅 = 40 𝑚
Let 𝑅𝑄 be 𝑥.
Now, in
∆𝑄𝑅𝑌, 𝑡𝑎𝑛 45$°$ = 𝑄𝑅 / 𝑃𝑋
1 =  𝑥 /𝑃𝑋
⇒ 𝑥 = 𝑃𝑋
And in ∆𝑃𝑄𝑋, 𝑡𝑎𝑛 60$°$

 $\sqrt{3}$ =𝑃𝑄/PX  … eq(1)
 

$\sqrt{3}$ = 40+𝑥 / 𝑃𝑋

 $\sqrt{3}$ = 40+𝑥/x        (from eq(1))

⇒$\sqrt{3}$ 𝑥 = 40 + 𝑥
⇒ 1. 73𝑥 − 𝑥 = 40
⇒ 0. 73𝑥 = 40
  40 0.73
⇒ 𝑥 = 54. 79 𝑚
So, the height of the tower, 𝑃𝑄 = 40 + 𝑥
⇒ 𝑃𝑄 = 40 + 54. 79
⇒ 𝑃𝑄 = 94. 79 𝑚
And the distance 𝑃𝑋 = 𝑥
⇒ 𝑃𝑋 = 54. 79 𝑚
Hence, the height of the tower 𝑃𝑄 and the distance 𝑃𝑋 is 94. 79 𝑚 and 54. 79 𝑚, respectively.