A decorative pottery bowl with a diameter of 30 cm is used as a garden ornament . A rain shower fills it with water to a maximum depth of 7 cm. The bowl is slowly tipped to remove the water. What angle will the rim of the bowl make with the horizontal when the water begins to spill out?
PLEASE HELP
Well, well, well, looks like we have a slippery situation here! Let me put on my clown glasses and calculate the angle for you.
Since we're dealing with a decorative bowl, not a clown car, let's get serious for a moment. To find the angle at which the water begins to spill out, we need to consider the relationship between the diameter, the depth of the water, and the angle.
I see we have a diameter of 30 cm, which means a radius of 15 cm. We also have a maximum depth of 7 cm.
To find the angle, we can use a little bit of trigonometry. The tangent of an angle is equal to the opposite side divided by the adjacent side, in this case, the depth divided by the radius.
So, the tangent of the angle we're looking for equals 7 cm / 15 cm. If we plug that into our handy dandy calculator, we get approximately 0.4667.
Now it's time to find the angle itself! To do that, we take the arctangent of 0.4667. Drumroll, please... And the angle is roughly 25.665 degrees.
Voila! The rim of the bowl will make an angle of approximately 25.665 degrees with the horizontal when the water starts spilling out. Just make sure you don't slip and spill anything else while you're at it!
To find the angle that the rim of the bowl makes with the horizontal, we can use trigonometry.
First, let's visualize the situation. We have a circular bowl, and the water inside it forms a circular shape as well. When the water starts spilling out, it will form a circular arc on the rim of the bowl. We need to find the angle that this circular arc makes with the horizontal.
Let's consider a right triangle formed by the center of the bowl, the point where the water just begins to spill out, and the point where the water level creates a right angle with the base of the bowl.
The radius of the circular bowl is half of its diameter, so the radius is 30 cm / 2 = 15 cm.
The maximum depth of the water is 7 cm, which is the vertical side of the right triangle.
The horizontal side of the right triangle is the radius of the bowl, which is 15 cm.
Now we can use trigonometric functions to find the angle that the circular arc makes with the horizontal.
The tangent function can be used to find the angle:
tan(angle) = opposite / adjacent
tan(angle) = 7 cm / 15 cm
angle = arctan(7 cm / 15 cm)
Using a calculator to find the arctan (inverse tangent) of 7 cm / 15 cm, we get:
angle ≈ 25.25 degrees
Therefore, the rim of the bowl will make an angle of approximately 25.25 degrees with the horizontal when the water begins to spill out.
To find the angle the rim of the bowl makes with the horizontal when the water begins to spill out, we can use trigonometry and basic geometry principles. Here is how you can approach the problem:
Step 1: Find the radius of the bowl
The diameter of the bowl is given as 30 cm. Divide it by 2 to find the radius.
Radius = Diameter / 2 = 30 cm / 2 = 15 cm
Step 2: Find the height of the water in terms of the radius
The maximum depth of the water is given as 7 cm. To find the height in terms of the radius, we need to consider the water level as a fraction of the total height of the bowl (which is twice the radius).
Height of water = (water depth) / (r + r) = 7 cm / (15 cm + 15 cm) = 7 cm / 30 cm = 0.2333
Step 3: Determine the angle using trigonometry
Since we now have the height of the water in terms of the radius, we can use trigonometry to find the angle. The tangent of the angle (θ) is given by the equation: tan(θ) = (height of water) / (radius).
tan(θ) = 0.2333 / 15 cm
Now, use the inverse tangent (arctan) function to find the angle θ:
θ = arctan(tan(θ))
θ = arctan(0.2333 / 15 cm)
Use a calculator to find the arctan value, which should be approximately 0.924 radians (or 52.99 degrees).
Therefore, the rim of the bowl will make an angle of approximately 0.924 radians or 52.99 degrees with the horizontal when the water begins to spill out.
correct
Let's say the bowl is a hemisphere. Then, draw a side view. The axis of symmetry of the sphere, and the radius to the edge at the top of the bowl form a right triangle with legs 15cm long.
If the right angle is considered to be the vertex angle of an isosceles triangle, the base of the triangle is the surface of the water when it just about to start over the rim.
So, drop an altitude such that the distance from the base of the triangle to the circle of the bowl is 7cm.
Then if the bowl has been tipped an angle θ,
cosθ = 7/15.