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As the bowl is hemispherical in share, tilting it does not change the shape of the water, and so its depth remains the same.

When the bowl has been tilted 35o, the distance the lip of the bowl has been lowered can be found using the Sine ratio. This can be subtracted from the height the lip was above the bottom the the bowl (namely the radius of the bowl) to find how deep the water is.

The angle is 35o.

The hypotenuse is the radius of the bowl.

The opposite side is the unknown drop.

sine 35o = drop/20 cm

⇒ drop = 20cm x sine 35o

≈ 11.47 cm

height = radius - drop

≈ 20 cm - 11.47 cm

= 8.53 cm

Q: How to solve using trig. Water in a hemispherical bowl with a diameter of 40 cm begins to pour out when the bowl is tilted through an angle of 35 degrees How deep is the water in the bowl?

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It would have been much simpler if this browser had halfway decent graphic facilities but it does not so you may have do a few sketches as you go along.. The bowl is tipped by 35 deg so the diameter of the bowl makes a 35 deg angle with the horizontal. The surface of the water must be horizontal and so that makes an angle of 35 deg with the tipped diameter. Therefore [in cross section] the water occupies a segment of a circle such that the angles between the water surface and the radii at both ends on the segment are 35 deg. Therefore the segment subtends an angle of 110 deg at the centre. Also, the radii are 20 cm. In the triangle formed by the water surface and the sides of the segment, the distance from the centre of the bowl to the water surface is 20*sin(35) = 11.47 cm. Therefore, the depth of the water is 20 - 11.47 = 8.53 cm.

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