Expert answers to "I have a non-right-angled isosceles triangle with two..."

High School
Geometry
Trigonometry
Non-right triangles and trigonometry

Fallbasiss4

Answered question

2022-01-21

I have a non-right-angled isosceles triangle with two longer sides, X, and a short base Y.
I know the length of the long sides, X.
I also know the acute, vertex angle opposite the base Y, lets

Answer & Explanation

euromillionsna

Beginner2022-01-22Added 16 answers

Step 1
Cut the iscoles triangle in half to get a two right triangles with opposite side

\frac{1}{2} y

and hypotenuse x.

\frac{1}{2} Y = \sin (\frac{1}{2} a) x

So apparently this is claiming

2 \sin (\frac{1}{2} a) = \tan a

which isnt

ataill0k

Beginner2022-01-23Added 18 answers

Step 1
Bisect angle a
$\sin \frac{a}{2} = \frac{y}{2 x}$
$\cos \frac{a}{2} = \frac{\sqrt{4 x^{2} - y^{2}}}{2 x}$
$\tan \frac{a}{2} = \frac{y}{\sqrt{4 x^{2} - y^{2}}}$
$\tan a = \frac{2 \tan \frac{a}{2}}{1 - \tan^{2} \frac{a}{2}}$
$\tan a = \frac{2 y}{\sqrt{4 x^{2} - y^{2}} (1 - \frac{y^{2}}{4 x^{2} - y^{2}})}$
$\tan a = \frac{y \sqrt{4 x^{2} - y^{2}}}{2 x^{2} - y^{2}}$
$\frac{y}{x}$ is small, $\frac{\sqrt{4 x^{2} - y^{2}}}{2 x^{2} - y^{2}} \approx \frac{1}{x}$
and $\tan a \approx \frac{y}{x}$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Trigonometry

Didn't find what you were looking for?