# Thread: area of triangle given angle bisector

1. ## Re: area of triangle given angle bisector Originally Posted by Vinod Hello,

Mathematica gives a list of possible solutions, among them some with negative values.

After we eliminate those extra solutions we get two triangles satisfying the conditions given in the problem,
both with area = 60

2. ## Re: area of triangle given angle bisector Originally Posted by Idea Mathematica gives a list of possible solutions, among them some with negative values.

After we eliminate those extra solutions we get two triangles satisfying the conditions given in the problem,
both with area = 60
Hello,
If we put all the four equations given by SlipEternal in the Wolfram computational intelligence It gave the answer for the four unknown variables as follows $z(v)=26.4979,w=3.50207,x=5.31243,y=22.2655$. So the area of triangle assuming $base=y+4=26.2655,height=w=3.50207,$ we get the area of triangle 45.99 using the formula of the area of triangle $[\frac12*base*height]$ which is wrong and different from the correct answer 60

3. ## Re: area of triangle given angle bisector

Interesting!

Wolfram gives three solutions. Two solutions are acceptable and give an area = 60

The third solution w=3.50207 gives an angle BAN = 48.8 and angle NAC = 33.6 degrees
so AN is not the bisector

4. ## Re: area of triangle given angle bisector

Can someone confirm that there are two possible triangles that fit the requirements, both having an area of 60 ?

(All dimensions approximate, 4dp)

(1) AB =13.7262, BC = 8.7423, CA = 16.2738, /_BAC = 32.4933 deg.

(2) AB = 4.8666, BC = 24.6577, CA = 25.1334, /_BAC = 78.8351 deg.

5. ## Re: area of triangle given angle bisector Originally Posted by Idea Interesting!

Wolfram gives three solutions. Two solutions are acceptable and give an area = 60

The third solution w=3.50207 gives an angle BAN = 48.8 and angle NAC = 33.6 degrees
so AN is not the bisector
Hello,

The WolframAlpha computational intelligence gave these four solutions 1)v≈26.4979, w≈3.50207, x≈-5.31643, y≈22.2655;2)v≈26.4979, w≈3.50207, x≈5.31643, y≈22.2655;
3)v≈34.5687, w≈-4.56873, x≈-6.07234, y≈-38.2655; 4)v≈34.5687, w≈-4.56873, x≈-6.07234, y≈-38.2655.

Now tell me which two solutions are acceptable?
Secondly, You are telling $\overline{AN}$ is not the angle bisector. But your answer confirms that $\overline{AN}$ is the angle bisector and its length is 14.2971.

6. ## Re: area of triangle given angle bisector Originally Posted by BobP Can someone confirm that there are two possible triangles that fit the requirements, both having an area of 60 ?

(All dimensions approximate, 4dp)

(1) AB =13.7262, BC = 8.7423, CA = 16.2738, /_BAC = 32.4933 deg.

(2) AB = 4.8666, BC = 24.6577, CA = 25.1334, /_BAC = 78.8351 deg.

Hello,
Both the triangles have the area of ${60^\circ}$.

7. ## Re: area of triangle given angle bisector Originally Posted by Idea Interesting!

Wolfram gives three solutions. Two solutions are acceptable and give an area = 60

The third solution w=3.50207 gives an angle BAN = 48.8 and angle NAC = 33.6 degrees
so AN is not the bisector
Hello,

There is yet another equation to add in inputs to get correct solution i-e $\frac{y}{4}=\frac{z}{w}$.So WolframAlpha computational intelligence is correct, we omitted one input to add.

8. ## Re: area of triangle given angle bisector Originally Posted by SlipEternal Let's use letters to represent the different lengths:

$$\overline{AB} = w, \overline{AN} = x, \overline{NC}=y, \overline{AC} = z$$

Now, we have the following information:

By the Pythagorean Theorem:
$$w^2+4^2=x^2$$

By the Pythagorean Theorem:
$$(4+y)^2+w^2 = z^2$$

By the double angle formula for sine:
$$\dfrac{4+y}{z} = 2\dfrac{4}{x}\cdot \dfrac{w}{x}$$

Given:
$$w+z=30$$

That is four equations in four variables. So, this system may have a solution. You would need to solve it.
Hello,

There is one more equation to add which is $\frac{y}{4}=\frac{z}{w}$.