Visualizing the Area of a Triangle with Varying Angles

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Homework Statement



Show that for all [tex]\theta \epsilon (0, \pi)[/tex], the area of a triangle with side lengths a and b with included angle [tex]\theta is A = \frac{1}{2} a b sin \theta[/tex]. (Hint: You need to consider two cases)

Homework Equations


The Attempt at a Solution



I have just begun working on this problem.. not really sure where to start.

Does [tex]\theta \epsilon (0, \pi)[/tex] mean that the angle is > than 0 and < than pi?
Am I supposed to show that when the angle is less than or greater than the condition then the equation to find area is not valid?
 
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Yes, that's what (0,pi) means. The only cases where the area is not ab*sin(theta) is where sin(theta) might be negative. They aren't in (0,pi). What's the area in that case?
 
Can you give me a little more hint -_-;

What are the two cases that I need to consider?
 
zeion said:
Can you give me a little more hint -_-;

What are the two cases that I need to consider?

Use trig and A=bh/2. What's h in terms of a and the included angle? Draw a right triangle. And I'm really not sure what the 'two cases' they are talking about are.
 
h = b(sin theta)
or
h = b(sin 180 - theta)
 
So can I show this by drawing a picture?