# TrigonometryArea and angles of iso triangle given find sides

#### karush

##### Well-known member
Given an Isosceles triangle with the area of 100 with internal angles of
$$40^o, 70^o,70^o$$
$A=\frac{1}{2}bh$
so
$100=\frac{1}{2}\left(z\sin\left({70^O}\right)\right)\left(2z\cos\left({70^o}\right)\right)$
$z$ is length of one the equal sides

At least started here

Last edited:

#### MarkFL

Staff member
Consider the following diagram:

Now, using the Law of Sines, we may state:

$$\displaystyle A=B\frac{\sin(a)}{\sin(b)}$$

If we denote the area of the triangle with $A_T$, then we may also state:

$$\displaystyle A_T=\frac{1}{2}B^2\sin(a)$$

Now, just solve for $B$, and then you will know $A$ as well. Then you will have formulas that you can plug into the given data.

#### stud17

##### New member
Given an Isosceles triangle with the area of 100 with internal angles of
$$40^o, 70^o,70^o$$
$A=\frac{1}{2}bh$
so
$100=\frac{1}{2}\left(z\sin\left({70^O}\right)\right)\left(2z\cos\left({70^o}\right)\right)$
$z$ is length of one the equal sides

At least started here
I think what you have done is correct just remember that $\sin(2x) = 2\cos(x)\sin(x)$ to simplify.