Area - Trigonometry

Mar 2010
21
0
Hello Forum.

I'm having trouble with this problem. Hope this is the right section, and that I translate it right the right way, so you can understand...

We have a right triangle. Smallest leg - AB. Biggest leg - BC. Hypotenuse - CA. We know that the hypotenuse has 10 centimeters, and that x is the angle BCA. How do I prove that the area of this triangle is given by A(x) = 25 sin (2x).

Any help would be appreciated. Thank you.
 
Dec 2009
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1111
Hello Forum.

I'm having trouble with this problem. Hope this is the right section, and that I translate it right the right way, so you can understand...

We have a right triangle. Smallest leg - AB. Biggest leg - BC. Hypotenuse - CA. We know that the hypotenuse has 10 centimeters, and that x is the angle BCA. How do I prove that the area of this triangle is given by A(x) = 25 sin (2x).

Any help would be appreciated. Thank you.
Dear Alvy,

First of all in a right angled traingle the hypotinuse is the biggest leg. Therefore BC cannot be the longest leg. However we can still find the area of the traingle.

Area of the traingle= \(\displaystyle \frac{AB\times{BC}}{2}\)

Now, AB= CAsinx and BC= CAcosx

Area of the tringle= \(\displaystyle \frac{CA~sinx\times{CA~cosx}}{2}=\frac{10sinx\times{10cosx}}{2}\)

Hope you can continue from here.
 
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Prove It

MHF Helper
Aug 2008
12,883
4,999
Hello Forum.

I'm having trouble with this problem. Hope this is the right section, and that I translate it right the right way, so you can understand...

We have a right triangle. Smallest leg - AB. Biggest leg - BC. Hypotenuse - CA. We know that the hypotenuse has 10 centimeters, and that x is the angle BCA. How do I prove that the area of this triangle is given by A(x) = 25 sin (2x).

Any help would be appreciated. Thank you.
Since it's a right angle triangle

\(\displaystyle A(x) = \frac{b\,h}{2}\)

\(\displaystyle = \frac{(AB)(BC)}{2}\).



Recall that \(\displaystyle \sin{2x} = 2\sin{x}\cos{x}\).

Also recall that \(\displaystyle \sin{x} = \frac{\textrm{Opposite}}{\textrm{Hypotenuse}}\) and \(\displaystyle \cos{x} = \frac{\textrm{Adjacent}}{\textrm{Hypotenuse}}\).


So \(\displaystyle \sin{x} = \frac{AB}{10}\) and \(\displaystyle \cos{x} = \frac{BC}{10}\).


Therefore \(\displaystyle \sin{2x} = 2\sin{x}\cos{x}\)

\(\displaystyle = 2\left(\frac{AB}{10}\right)\left(\frac{BC}{10}\right)\)

\(\displaystyle = \frac{(AB)(BC)}{50}\).


Therefore \(\displaystyle 25\sin{2x} = \frac{(AB)(BC)}{2}\)

\(\displaystyle 25\sin{2x} = A(x)\).
 
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Mar 2010
21
0
Once again, thank you very much.