Book error, or am i missing something? TRIG piece

  • Thread starter Thread starter rygza
  • Start date Start date
  • Tags Tags
    Book Error Trig
AI Thread Summary
The discussion revolves around the correct interpretation of a right triangle in relation to the sine function. The confusion stems from the notation used in the book, where "h" represents the vertical height (the opposite side) and "delta x" is identified as the hypotenuse. The formula sin(theta) = h/delta x is confirmed as correct, despite initial doubts about the notation. Clarification is provided that delta x corresponds to the line segment between points X0 and X1, which is indeed the hypotenuse. The conversation concludes with a participant expressing gratitude for the resolution of their misunderstanding.
rygza
Messages
37
Reaction score
0
In the attached picture there is a right triangle (the right angle is on the right bottom corner). The angle, theta, is on the left-most side. From what i remember, sin (theta) is equal to opposite/hypotenuse. If you can see in the picture, it has written: delta x = h/sin(theta)

so, sin(theta) = h/delta x

isn't this wrong?
 

Attachments

  • 2011-01-22_20.00.04.jpg
    2011-01-22_20.00.04.jpg
    11.9 KB · Views: 481
Physics news on Phys.org
The difference between X0 and X1 is delta x, which is the hypotenuse.

h is the height, which is the opposite part of sin(theta)=opposite/hypotenuse

So in this case, sin(theta) = h / delta x. Or to get the hypotenuse, delta x = h / sin(theta). The book is correct.

It is a poor choice of notation to use h in this case, as it is slightly confusing. However, h denotes the vertical height not the hypotenuse (which is delta x).
 
jarednjames said:
The difference between X0 and X1 is delta x, which is the hypotenuse.

h is the height, which is the opposite part of sin(theta)=opposite/hypotenuse

So in this case, sin(theta) = h / delta x. Or to get the hypotenuse, delta x = h / sin(theta). The book is correct.

It is a poor choice of notation to use h in this case, as it is slightly confusing. However, h denotes the vertical height not the hypotenuse (which is delta x).

but wouldn't delta x be the length of the bottom side? and then the hypotenuse would be = sqrt((delta x)^2 + (h)^2)
 
No, look where X0 and X1 are on the picture. The line between those points is delta x - which is the hypotenuse.
 
jarednjames said:
No, look where X0 and X1 are on the picture. The line between those points is delta x - which is the hypotenuse.

OOH yes now it makes sense :smile: I miss one little thing and it costs me 30minutes lol

THANK YOU!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top