MHB Area and angles of iso triangle given find sides

[karush]

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

 

[MarkFL]

Consider the following diagram:

View attachment 5343

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

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

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

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

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]

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.

...

Hello, Admin. How do I post a thread? No bottom to click. I can only reply. (Worried)

 

[MarkFL]

...Hello, Admin. How do I post a thread? No bottom to click. I can only reply. (Worried)

If you browse to a forum, you will see, above and below the thread listing, large buttons labeled "+ Post New Thread" that will allow you to begin a new thread in that forum. :)