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Could someone please explain to me how I would find this out?

I have a triangle and I need to find the length of the line that would split it down the middle so that the areas were even.

A = 105 degrees

B = 42 degrees

C = 33 degrees

AB = 3.2

AC = 3.9

BC = 5.7

The line needs to be draw across the triangle splitting angle A (the 105 degrees angle).

The total area of the triangle is $6.075km^2$ , so the two halves have to be equal in area.

Can anyone help me out here? Thank you!

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  • $\begingroup$ A truncation error occurs in your data set. Assuming that all the angles, plus AB = 3.2 and BC = 5.7 are all correct, then AC = .. by cosine law .. = 3.9(522). There are other combinations too. Should the correct combination be used, it is possible to get an answer matching that from the book. $\endgroup$ – Mick Aug 4 '14 at 4:04
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The line has to be drawn just through the midpoint of $BC$.

A median splits a triangle in two triangles having the same base and the same height, hence equal areas.

To find the length of a median you can use the formula: $$ m_a^2 = \frac{2b^2+2c^2-a^2}{4},$$ giving in your case: $$ m_a = \frac{1}{2}\sqrt{2\cdot 3.9^2+2\cdot 3.2^2-5.7^2}=2.145\ldots$$

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  • $\begingroup$ Thank you - I have done so using the Apollonius Theorem, and am coming up with the answer 2.15km. My book has the answer of 1.19km. I am completely stuck with how they got there!? $\endgroup$ – Dani Aug 4 '14 at 3:24
  • $\begingroup$ @Dani: I got the same answer by using the Stewart's theorem (en.wikipedia.org/wiki/Stewart%27s_theorem) hence I think we can assume the book is wrong. $\endgroup$ – Jack D'Aurizio Aug 4 '14 at 3:32
  • $\begingroup$ I don't believe any line segment of length 1.19(km) can even reach from any vertex of the triangle to its opposite edge, so it's hard to understand where that number could come from. $\endgroup$ – Steven Stadnicki Aug 4 '14 at 3:34
  • $\begingroup$ Ok... so it is likely that they have supplied me with the wrong answer? I got the same answer as you did. Thanks for your help $\endgroup$ – Dani Aug 4 '14 at 3:36
  • $\begingroup$ @StevenStadnicki is clearly right, the shortest height has length $\frac{2\cdot6.075}{5.7}=2.13\ldots$. $\endgroup$ – Jack D'Aurizio Aug 4 '14 at 3:36

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