Let's call the entire angle between the 10 and 5 sides to be x
Applying to cosine rule:
[tex]9 {}^{2} = 10 {}^{2} + 5 {}^{2} - 2(10)(5)cos(x) \\ 81 = 100 + 25 - 100cos(x) \\ 100cos(x) = 125 - 81 \\ cos(x) = \frac{44}{100} = 0.44 \\ x = arccos(0.44)\approx63.9[/tex]
We can conclude that the angle inside the small triangle is half of x, which i will denote by y
[tex]y = \frac{x}{2} = \frac{63.9}{2} = 31.95[/tex]
Similarly to find the angle which i will denote by z at the right lower corner of the small or large triangle, apply the cosine law again:
[tex]10{}^{2} = 5 {}^{2} + 9 {}^{2} - 2(9)(5)cos(z) \\ 100 = 25 + 81 - 90cos(z) \\ 90cos(z) = 106 - 100\\ cos(z) = \frac{6}{90} = 0.06666 \\ z = arccos(0.06666)\approx86.2[/tex]
The last angle in our small triangle which i will denote by w, will be 180 minus z and y
[tex]w = 180 - (y + z) \\ w = 180 - (31.95 + 86.2) \\ w = 61.85[/tex]
Let t be our missing side,
applying the law of sines:
[tex] \frac{5}{sin(w)} = \frac{t}{sin(y)} \\ t = \frac{5sin(y)}{sin(w)} = \frac{5sin(31.95)}{sin(86.2)} [/tex]
[tex]t \approx2.652[/tex]
