# Given the base and angles of an isosceles triangle, how to find length of the two sides?

I can't seem to find a textbook solution to this. It is always assumed that the length of the sides is know.

Isolceles triangle

So the base $a$ is known. The bottom angles where $\alpha$ and the two sides $b$ touch, are known.

What is $h$?

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solution sketch:

Note that $h$ divides the triangle into two right triangles (h is the perpendicular bisector (altitude) from the base to the opposite vertex); the angles line $h$ forms with $a$ are right angles. This gives you two right triangles, and you only need one of them to compute the values you need.

Call the two known (marked) angles $\theta$ (they are equal).

If you know the length of the side $b$: $\sin\theta = \dfrac{h}{b} \implies h = b\sin\theta$

If $a$ is known, $\tan\theta = \dfrac{h}{a/2} \implies h = \dfrac{a}{2}\tan \theta$.

Now, using the pythorean theorem to relate your sides, we know that

$$h^2+\left(\frac{a}{2}\right)^2\,=\;b^2\tag{pythogorean theorem}$$

If the length of $b$ is unknown, using the pythagorean theorem, then knowing $a/2$ and $h$ will allow you to solve for $b$.

Knowing $\,b\,$ and $\,h\,$ will allow you to solve for $\,\dfrac{a}{2}\,$ by the pythagorean theorem Then double the value of $a/2$ to get $a$.

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Thank you, this seems to work. Likewise Inquisitive's answer. It's part of larger problem and i didn't get the results i expected, but that was a fault on my part. Your answer is correct. – Adrian Jan 7 '13 at 15:17
You're welcome. Glad to help! – amWhy Jan 7 '13 at 15:20

I have explained it graphically below.

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from one of the following equations $$h^2+\left(\frac{a}{2}\right)^2=b^2$$ $$\sin\alpha=\frac{h}{b}$$ $$\cos\alpha=\frac{h/2}{b}=\frac{h}{2b}$$ $$\tan\alpha=\frac{h}{a/2}=\frac{2h}{a}$$ if you know two elements you can find the missing one

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You can calculate h

1) using a and b

(a/2) * (a/2) + h * h = b * b

2) using sine law or cosine law http://www.transtutors.com/math-homework-help/laws-of-triangle/

h/ sinα = b / sin 90

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## protected by Alexander Gruber♦Apr 24 '14 at 2:03

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