Answer:
XY = 18.6 mts
Angle of Depression = 61°
Step-by-step explanation:
Let us start with finding out the Length of the XY
Y is directly on top of the center of the rectangular pool ABCD . Hence It must be on the top of the intersection point of the two diagonals. The length of XY will be half of Diagonal D .
Let us find Diagonal D by applying Pythagoras theorem in Triangle ABC
[tex]D^2=AB^2 +BC^2[/tex]
[tex]D^2=30^2+22^2[/tex]
[tex]D^2= 900+484[/tex]
[tex]D^2=1384[/tex]
[tex]D=\sqrt{1384}[/tex]
[tex]D=37.20[/tex] Approx
Hence Diagonal is 37.20 mts
Hence [tex]XY = \frac{D}{2}[/tex]
[tex]XY=\frac{37.20}{2}[/tex]
[tex]XY=18.60[/tex] mts
Hence we have our first answer as 18.6 Mts
Part 2:
Please refer to the image attached with this answer.
The eye of the diver , the eye of the man standing on the pool , and the diagonal of the pool makes a right triangle A'O'Y'
Where
[tex]O'Y' = 1.8-1.5+10 = 10.3[/tex]
A'O' = half of the diagonal = 18.60 ( From first part of the problem)
Hence in ΔA'O'Y' , we have to determine ∠A'Y'O' or ∠x
Here applying trigonometric ratios
[tex]\tan x = \frac{opposite}{adjacent}[/tex]
Opposite = A'O' = 18.60
Adjacent = Y'O' = 10.3
Hence
[tex]\tan x = \frac{18.60}{10.30}[/tex]
[tex]\tan x = 1.80[/tex]
[tex]x=\tan^{-1}(1.8)[/tex]
[tex]x=60.94 [/tex]
x≈61°
Hence Angle of depression for the diver towards the man standing at the pool edge would be 61°
