Hi!
How can I count the surface area of an isosceles triangle + lengths of all sides + count the angles ?
I only know the height of the isosceles triangle and the base angle too.
Thanks for your help!
Let b denote the length of the base, l the length of the two equal legs, h the height and $\displaystyle \alpha$ the angle at the base.
You can split the isosceles triangle into two congruent right triangles. Then you have:
$\displaystyle \tan(\alpha) = \dfrac{h}{\frac12 b}$ Solve for b.
$\displaystyle \sin(\alpha) = \dfrac hl$ Solve for l
The area is
$\displaystyle a=\dfrac12 \cdot b \cdot h$
Plug in the values you now know.
$\displaystyle
tan \phi = \frac {h}{\frac{1}{2}b} \Rightarrow \frac {1}{2}b = \frac {h}{tan \phi} \Rightarrow b = \frac {2h}{tan \phi}
$
You said you knew the angle and the height.. If you know these 2 you can calculate the base b from the equation posted by earthboth. (I just turned it a bit incase you were confused by it)
I just don't get it
Please don't spoil my task, but give me an example if the height of the isosceles triangle was: 200m, 250m, 300m... (You can choose one)
and the base angle was: 15 degrees, 20, 25...(You can choose one)
How will I find out what is the surface area of it?
Lengths of all sides of it?
Count the angles?
Please give an example with numbers?
I'd really appreciate it!
1. Why didn't you post a question using numbers?
2. I've attached a sketch of an isosceles triangle. H denotes the height and b the base of the triangle.
3. Example: $\displaystyle |\alpha| = 42^\circ$ , $\displaystyle |h| = 12\ cm$ (I use these values so you have the opportunity to control the results by a (more or less) exact construction)
4. Results:
$\displaystyle \tan(42^\circ)=\dfrac{12\ cm}{\frac12 b}~\implies~ b = 2 \cdot \dfrac{12\ cm}{\tan(42^\circ)} \approx 21.61\ cm$
$\displaystyle \dfrac hl=\sin(42^\circ)~\implies~l=\dfrac{12\ cm}{\sin(42^\circ)}\approx 17.93\ cm$
area:
$\displaystyle a = \frac12 \cdot b \cdot h~\implies~a=\frac12 \cdot 21.61\ cm \cdot 12\ cm \approx 129.66\ cm^2$