This is about circles and trig please help ASAP

Question

A bowl is in the shape of a hemisphere (half sphere) with radius 10 cm. The surface of the water in the container has a radius of 7 cm. How deep is the water? Give your answer to one decimal place.

Answer 1

The bowl is a hemisphere which means that the distance to the bottom of the bowl itself is 10 cm.

 

The surface of the water is a circle with a radius of 7 cm.  So, any point of the edge of the bowl is 7 cm from the center of the water and 10 cm from center of the larger sphere (which includes the hemisphere of the bowl.)

 

There is a right triangle formed by the distance the water is below the center of the sphere (leg₁), the radius of the surface of the water (leg₂), and the radius of the hemisphere (hypotenuse).

 

The ratio of the radius of the water to the radius of the bowl is 7/10  or 0.70, that is the opposite over the hypotenuse of the angle found at the center of the sphere, so sin θ  = 0.70.

 

The distance of the surface of the surface of the water is below the center of the sphere is (10cm)(cos θ).  

 

cos θ = √(1 - sin² θ) = 

√(1 - 0.49) = √0.51 = 0.71

10cm (0.7141) = 7.1 cm below the center (first decimal place)

 

10cm - 7.1cm = 2.9 cm is the depth of the water (first decimal place)