formula 0.5 * b * c * sin(A) allows you to find the area of

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
theshadow
Posts: 102
Joined: Sun Feb 22, 2009 11:12 pm

formula 0.5 * b * c * sin(A) allows you to find the area of

Postby theshadow » Sat Jul 18, 2009 4:27 pm

The formula 0.5 * b * c * sin(A) allows you to find the area of an obtuse triangle, where b and c are two sides of the triangle and A is the included angle.

How can I create a proof and make a diagram to show that this formula works for A acute, A right, and A obtuse?

Thank you!

Honeysuckle588
Posts: 23
Joined: Wed Jul 15, 2009 4:07 pm
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Re: formula 0.5 * b * c * sin(A) allows you to find the area of

Postby Honeysuckle588 » Sat Jul 18, 2009 7:09 pm

I oriented my triangles with the side b along the horizontal (the initial side of angle A), and the side c as the terminal side. Then, drop a perpendicular line from the end of c down to the line that b sits on. It shouldn't take long for you to determine that the length of this vertical line is csin(A).

One of the fun things about proofs is knowing what you are allowed to assume and use. If you are allowed to use the well-known fact that the area of a triangle is 0.5*(base)*height) then you're essentially done.

If you can't use that as a known fact, then there is a little more work to do.

theshadow
Posts: 102
Joined: Sun Feb 22, 2009 11:12 pm

Re: formula 0.5 * b * c * sin(A) allows you to find the area of

Postby theshadow » Fri Jul 24, 2009 5:20 pm

Thank you. Using a set orientation was helpful.


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