What is the Correct Method for Finding Height Using Geometry?

[Coderhk]

<< Mentor Note -- 2 threads merged >>

1. Homework Statement
I need help with question 43 it is attached as an attachment

Homework Equations

#a^2+b^2=c^2#
Sin(a)/A=sin(b)/B

The Attempt at a Solution

Well if I draw a line from the left corner of the bown down to the water it forms a right triangle which I can use to find the diameter of the water and from that I divide by 2 but that doesn't seem to be any of the choices

 

[Mark44]

Please post a better picture. In the one you attached I can see the bowl, but the text is too small to read. A screen shot of just the problem and not the rest of the page would be better.

 

[Coderhk]

Homework Statement

I need help with question 43 it is attached as an attachment

Homework Equations

#a^2+b^2=c^2#
Sin(a)/A=sin(b)/B

The Attempt at a Solution

Well if I draw a line from the left corner of the bown down to the water it forms a right triangle which I can use to find the diameter of the water and from that I divide by 2 but that doesn't seem to be any of the choices

Edit: I'll type the problem out.

Water in a hemisphere bowl of With a diameter of 30 cm Begins to pour out when the bowl is tilted to an angle theta of 17 degrees. How deep is the water in the bowl? Round to nearest tenth of cm.

 

[Coderhk]

Please post a better picture. In the one you attached I can see the bowl, but the text is too small to read. A screen shot of just the problem and not the rest of the page would be better.

I'll type it out here...water in a hemiaphhem bowl with a diameter of 30 cm begins to pour out when the bowl is tilted to an angle theta of 17 degrees. How deep is the water in bowl?

 

[Coderhk]

I

I'll type it out here...water in a hemiaphhem bowl with a diameter of 30 cm begins to pour out when the bowl is tilted to an angle theta of 17 degrees. How deep is the water in bowl?

I meant hemisphere

 

[Mark44]

Create a right triangle whose left side is on the line from the lowest point on the bowl up to the midpoint of the tilted rim of the bowl. The base of this triangle is horizontal, running from the point on the bowl where the water is running out. Calculate the height of this triangle. From this height you can get the depth of the water. The value I get is one of the options, although I can just barely read them.

 

[berkeman]

I drew the bowl untilted and tilted about the center of the diameter. The line from the center of the top down to the left 17 degrees to the surface of the water forms a triangle with those two points and a point vertically below the center where the top of the water is. Call that vertical distance x and the depth of the water d. Can you say what x + d is equal to? And can you see how to solve for x?

 

[berkeman]

BTW, you can use the UPLOAD button in the lower right of the edit window to upload a PDF or JPEG image of the problem.

 

[Coderhk]

I

I drew the bowl untilted and tilted about the center of the diameter. The line from the center of the top down 17 degrees to the surface of the water forms a triangle with those two points and a point vertically below the center where the top of the water is. Call that vertical distance x and the depth of the water d. Can you say what x + d is equal to? And can you see how to solve for x?

Get it now thank you. I wasn't sure if I could make the assumption that the deepest part of the bowl is directly below the midpoint. But I see now. So the answer is 10.6

 

[berkeman]

So the answer is 10.6

Is that the correct answer? I get a different answer, but without seeing the figure, it's hard to know I did the drawing correctly.

 

[Mark44]

@Coderhk, please don't post the same question twice...
My answer, rounded, is 10.6 as well.

 

[Coderhk]

The possible choices are 4.4, 5.7, 10.4, 10.6,25.6

 

[Coderhk]

I attached the diagram

 

[berkeman]

My answer, rounded, is 10.6 as well.

I see what I did wrong now. I had a factor of 2 error. Now I get 10.614 too. Thanks Mark.