# Area - Trigonometry

#### Alvy

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.

#### Sudharaka

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.

Alvy

#### Prove It

MHF Helper
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 $$\displaystyle 25\sin{2x} = \frac{(AB)(BC)}{2}$$

$$\displaystyle 25\sin{2x} = A(x)$$.

Alvy

#### Alvy

Once again, thank you very much.