Area of Triangle: How to Calculate

  • Thread starter Thread starter huntingrdr
  • Start date Start date
  • Tags Tags
    Area Triangle
AI Thread Summary
The area of a triangle can be calculated using the formula A = 1/2 a*b*sin(theta), where a and b are the lengths of two sides and theta is the included angle. To derive this formula, one can start with the standard area formula A = 1/2 b*h and relate the height to the sine of the angle. By dropping a perpendicular height from the vertex opposite side a, the height can be expressed as h = b*sin(theta). It is important to remember that the triangle in question is not necessarily a right triangle, which affects how the height is determined. Understanding these relationships allows for the correct application of the area formula for any triangle.
huntingrdr
Messages
24
Reaction score
0

Homework Statement


Show that the area of a triangle with sides of lengths a and b with included angle theta is...

A = 1/2 a*b*sin(theta)

I am not sure what this question is really asking. Any help?


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
i believe strongly that you can figure it out.
 
If the area of a triangle is 1/2 b*h, and sin theta = opp/hyp, then that means a=h and b=b. Correct? Now how do I prove that the area is 1/2 a*b*sin(theta). I drew a right triangle and labeled the sides a and b, and the angle theta.
 
It isn't necessarily a right triangle. Try making side "a" the base and dropping a height "h" down from the opposite corner.
 
GunnaSix said:
It isn't necessarily a right triangle. Try making side "a" the base and dropping a height "h" down from the opposite corner.
Furthermore, since the OP's problem states only that it is a triangle, he/she should not assume that it is a right triangle.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Replies
9
Views
3K
Replies
5
Views
2K
Replies
16
Views
2K
Replies
3
Views
3K
Replies
9
Views
2K
Replies
5
Views
3K
Back
Top