# Prove that $A_{T}= \frac{1}{2}a \cdot b \cdot \sin(\gamma)= \frac{1}{2}a \cdot c \cdot \sin(\beta)=\frac{1}{2}b \cdot c \cdot \sin(\alpha)$

Prove the following formula for the area $A_{T}$ of a triangle $ABC$:

$$A_{T}= \frac{1}{2}a \cdot b \cdot \sin(\gamma)= \frac{1}{2}a \cdot c \cdot \sin(\beta)=\frac{1}{2}b \cdot c \cdot \sin(\alpha)$$

Assume we have a rectangular triangle.

Then $\sin(\alpha)= \frac{\text{opposite}}{\text{hypotenuse}}=\frac{a}{c}$

$\sin(\beta)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{b}{c}$

$\sin(\gamma)=?$ I don't know :

But so far we would have:

$$A_{T}=\frac{1}{2}a \cdot b \cdot \sin(\gamma) = \frac{1}{2}a \cdot c \cdot \frac{b}{c}= \frac{1}{2}b \cdot c \cdot \frac{a}{c}$$

So

$$A_{T}= \frac{1}{2}a \cdot b \cdot \sin(\gamma) = \frac{1}{2}ab=\frac{1}{2}ab$$

I think I almost got it but how to do it correctly? :D

• start with the equation $$A_T=\frac{1}{2}gh_T$$ Nov 23, 2016 at 17:03

Hint:

note that:

$b\sin \gamma$ is the height with respect the side $a$

$a\sin \beta$ is the height with respect the side $c$

$c\sin \alpha$ is the height with respect the side $b$

From the figure:

If $a$ is the basis than $AD$ is the relative heigt and the trisangle $ADC$ is rectangle in $D$ and $AD= b \sin \gamma$, so the area of $ABC$ is $Area=\frac{1}{2}ab \sin \gamma$.

You can do the same using the oter sides as a basis.

• sry i don't understand : / It's not possible to solve it as I tried above? Nov 23, 2016 at 17:26
• I've added a figure. I don't relly understand your method, but it seems to me that your confusion come from the fact that you are thinking to $ABC$ as a rectangual triangle. But it's not. Nov 23, 2016 at 18:09