# How to construct triangle LMN , such that LN = 8 cm and ∠ LMN = 80 degrees and LM- MN= 3 cm?

Feb 13, 2016

Use $L M = 7.461 c m$ and $M N = 4.461 c m$

#### Explanation:

Calling LM as $x$ and MN as $y$
$x - y = 3$ => $y = x - 3$

As $\angle L M N$ is the angle between x and y
Applying the Law of Cosines:

${8}^{2} = {x}^{2} + {y}^{2} - 2 x \cdot y \cdot \cos {80}^{\circ}$
$64 = {x}^{2} + {\left(x - 3\right)}^{2} - 2 x \left(x - 3\right) \cos {80}^{\circ}$
${x}^{2} + {x}^{2} - 6 x + 9 - 2 {x}^{2} \cos {80}^{\circ} + 6 x \cos {80}^{\circ} - 64 = 0$
$\left(2 - 2 \cos {80}^{\circ}\right) {x}^{2} + \left(- 6 + 6 \cos {80}^{\circ}\right) x - 55 = 0$

$\Delta = 36 + 36 {\cos}^{2} {80}^{\circ} - 72 \cos {80}^{\circ} + 440 - 440 \cos {80}^{\circ}$
$\Delta = 476 + 36 {\cos}^{2} {80}^{\circ} - 512 \cos {80}^{\circ}$
$\Delta \cong 388.178$ => $\sqrt{\Delta} \cong 19.702$

x=(6(1-cos 80^@)+-19.702)/(4(1-cos 80^@)

$\to {x}_{1} = 7.461 c m$ => $y = 7.461 - 3$ => $y = 4.461 c m$
$\to {x}_{2} = - 4.461$ (not valid)