# How do you find the exact value of the third side given triangle ABC, AB=2.35, BC=6.24, mangleB=115?

May 18, 2018

$A C = \frac{\sqrt{444601 - 73320 \sqrt{2} + 73320 \sqrt{6}}}{100}$

#### Explanation:

We have the following situation:

We know the sides $c$, $a$ and the angle between them.

Generally, when we know this, our best bet is to apply the Law of cosines, which states that, in any triangle with sides $a$ and $c$ and angle between the $B$, the following relation is true:

${b}^{2} = {a}^{2} + {c}^{2} - 2 a c \cos B$

This is not only true for the side $b$. In fact, we have, these are equally as true:

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$
${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos A$

Finally, let us apply this relation in our triangle:

${b}^{2} = {a}^{2} + {c}^{2} - 2 a c \cos B$

b^2 = 2.35^2 +6.24^2 - 2*2.35*6.24*cos115°

In order to find cos 115°, we can write it as cos(45°+60°) and apply the Sum formula for cosine:

$\cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

:. cos115° = cos(45°+60^@) = sqrt2/2 * 1/2 - sqrt2/2 * sqrt3/2 =(sqrt2-sqrt6)/4

To find the exact for of $b$, we should rewrite $a$ and $c$ as the following:

${b}^{2} = {\left(\frac{235}{100}\right)}^{2} + {\left(\frac{624}{100}\right)}^{2} - 2 \cdot \frac{235}{100} \cdot \frac{624}{100} \cdot \frac{\sqrt{2} - \sqrt{6}}{4}$

${b}^{2} = \frac{444601}{10000} - \frac{293280}{10000} \frac{\sqrt{2} - \sqrt{6}}{4}$

${b}^{2} = \frac{444601 - 73320 \sqrt{2} + 73320 \sqrt{6}}{10000}$

color(red)(=> b=sqrt(444601-73320sqrt2+73320sqrt6)/100