The radius of the bowl R as a function of the angle theta is [tex]\mathrm{R}=\frac{3}{1-\sin \theta}[/tex]
Solution:
The figure is attached below
If we consider the centre of hemisphere be A
The radius be AC and AD
According to question,
A bowl in the shape of a hemispere is filled with water to a depth h=3 inches .i.e. BC = h = 3 inches
And radius of the bowl is R inches .i.e. R = AD =AC
Now , using trigonometric identities in triangle ABD we get
[tex]\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{A B}{A D}[/tex]
[tex]\begin{array}{l}{\sin \theta=\frac{A B}{R}} \\\\ {A B=R \sin \theta}\end{array}[/tex]
Since , AC = AB + BC
R = R Sinθ + 3
R - R Sinθ = 3
R (1 – Sinθ ) = 3
[tex]\mathrm{R}=\frac{3}{1-\sin \theta}[/tex]
Which is the required expression for the radius of the bowl R as a function of the angle theta
