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!

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- Feb 24th 2009, 07:14 AMWendyMTriangle question
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! - Feb 24th 2009, 07:25 AMearboth
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. - Feb 24th 2009, 07:44 AMWendyM
Thanks. But I only know the angle of the base so how can i get the solve for b by multiplying ½ * b if I don't know the length of the base just the base angle.

- Feb 24th 2009, 08:26 AMmetlx
$\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) - Feb 24th 2009, 08:29 AMReckoner
- Feb 25th 2009, 09:58 AMWendyM
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! - Feb 25th 2009, 10:29 AMearboth
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:

$\displaystyle a = \frac12 \cdot b \cdot h~\implies~a=\frac12 \cdot 21.61\ cm \cdot 12\ cm \approx 129.66\ cm^2$