# A triangle has sides A, B, and C. Sides A and B are of lengths 12 and 5, respectively, and the angle between A and B is pi/12. What is the length of side C?

Jun 12, 2016

$169 - 30 \cdot \sqrt{2} \cdot \left(\sqrt{3} + 1\right)$

#### Explanation:

We use Cosine-rule for the given triangle & get

${C}^{2}$=A^2+B^2-2AB(cosangle betwn. A & B)
$= {12}^{2} + {5}^{2} - 2 \cdot 12 \cdot 5 \cdot \cos \left(\frac{\pi}{12}\right)$
=144+25-120{sqrt(1/2(1+cos(2*pi/12))}
$= 169 - 120 \left\{\sqrt{\frac{1}{2} \left(1 + \frac{\sqrt{3}}{2}\right)}\right\}$
$= 169 - 120 \left\{\sqrt{\frac{2 + \sqrt{3}}{4}}\right\}$
$= 169 - \frac{120}{2} \left\{\sqrt{\left(2 + \sqrt{3}\right)}\right\}$
$= 169 - 60 \left(\sqrt{2 + \sqrt{3}}\right)$
=169-60(sqrt((4+2sqrt3)/2)
$= 169 - \left(\frac{60}{\sqrt{2}}\right) \cdot \left\{\sqrt{3 + 1 + 2 \sqrt{3}}\right\}$
$= 169 - \frac{60 \cdot \sqrt{2}}{2} \cdot \left[\sqrt{{\left(\sqrt{3}\right)}^{2} + {\left(\sqrt{1}\right)}^{2} + 2 \cdot \sqrt{3} \cdot \sqrt{1}}\right]$
$= 169 - 30 \cdot \sqrt{2} \cdot \sqrt{{\left(\sqrt{3} + \sqrt{1}\right)}^{2}}$
$= 169 - 30 \cdot \sqrt{2} \cdot \left(\sqrt{3} + 1\right)$

Jun 12, 2016

$\textcolor{red}{\text{A different method; just to demonstrate that you can}}$

I have taken you to the point where all you have to do is 'plug' the values into a calculator.

#### Explanation:

Always a good start to draw a diagram or sketch.

$\textcolor{b l u e}{\text{Method Plan}}$

Objective: determine $\angle b$ then by using the $\sin \left(\angle b\right)$ determine C
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Step 1: $\to$ determine h; then cP using $\sin \left(\angle c\right)$

Step 2:$\to$ 12 - cP = bP

Step 3:$\to \angle b = \arctan \left(\frac{h}{b P}\right)$

Step 4: $\implies \sin \left(\angle b\right) = \frac{h}{C} \text{ "=>" } C = \frac{h}{\sin} \left(\angle b\right)$

To reduce accumulated rounding error it is better do the calculation only at the end.
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$\textcolor{b l u e}{\text{Answering the question}}$

Step 1: $\text{ } \textcolor{b r o w n}{h = B \sin \left(\angle c\right) \to 5 \sin \left(\frac{\pi}{12}\right)}$

$\text{ } \textcolor{b r o w n}{c P = B \cos \left(\angle c\right) \to 5 \cos \left(\frac{\pi}{12}\right)}$
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Step 2:$\text{ } \textcolor{b r o w n}{b P = 12 - 5 \cos \left(\frac{\pi}{12}\right)}$

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Step 3: $\text{ } \textcolor{b r o w n}{\angle b = \arctan \left(\frac{h}{b P}\right) \to \angle b = \arctan \left(\frac{5 \sin \left(\frac{\pi}{12}\right)}{12 - 5 \cos \left(\frac{\pi}{12}\right)}\right)}$

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Step 4: " "color(brown)(C=h/sin(/_b)->color(blue)(C=h/[sin[arctan((5 sin(pi/12))/(12-5cos(pi/12))) ]))

$\textcolor{g r e e n}{\text{You will see arctan on your calculator as } {\tan}^{- 1}}$
So if $\tan \left(\theta\right) = x \text{ }$ then $\text{ } \arctan \left(x\right) = \theta$