I need to find the length of the sides of a triangle. I have an angle and the area of the triangle.

I have the answer but I don't know how to figure it out so it doesn't help.

The area of the triangle is 18cm2 - the angle is 23 degrees.

Can someone explain how I can use the area and the angle to calculate the length of the sides?


  • $\begingroup$ Area and one angle are together not enough to determine the sides. Do we know more, such as triangle is right-angled? $\endgroup$ – André Nicolas Feb 8 '15 at 14:49
  • $\begingroup$ You have one angle and the area, so two pieces of information. It seems unlikely this is enough to determine the three sides. Perhaps you know something else? For example another angle? E.g. you were told the triangle is a right triangle... $\endgroup$ – hardmath Feb 8 '15 at 14:50

Not enough informations. Assume that $\widehat{ABC}=\widehat{A'B'C'}$ and $AB\cdot BC = A'B'\cdot B'C'$.

Then the triangles $ABC, A'B'C'$ have the same area and share an angle, but they are not necessarily congruent:

enter image description here


You are missing some critical piece of information.

To see that one angle and the area are not enough, pick all triangles that have an angle of $23$ degrees. Now shrink or dilate those triangles as necessary to get the area of $18$ cm². The results will not all have the same side lengths, as two triangles with one equal angle are not necessarily similar.


You obtain the 3 sides of a triangle by using this five equations (for a General triangle):

$a^2+b^2-2abcos(\alpha(a,b))=c^2$ (1)

$b^2+c^2-2bccos(\alpha(b,c))=a^2$ (2)

$a^2+c^2-2accos(\alpha(a,c))=b^2$ (3)

For the area: $A = \frac{1}{2}absin(\alpha(a,b))$ (4)

You also have the angle sum condition: $\alpha(a,b) + \alpha(b,c) +\alpha(a,c)= 180°$ (5)

From These you compute all other angles and the 3 sides of the triangle.

  • $\begingroup$ This doesn't work with the information given in the Question. $\endgroup$ – hardmath Feb 8 '15 at 14:59
  • $\begingroup$ You can write down as many equations as you like, but the given informations are not sufficient. See my answer for an obvious proof of this. $\endgroup$ – Jack D'Aurizio Feb 8 '15 at 15:06

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