Math Question

There you go - just as well everyone wasn't in a narky mood.    

N=16 degrees 

Please present you question better - I cannot be bothered with all that scrolling. 

since your question has not been answered i assume I am notthe only one that feels this way.

Triangle LMN has vertexes at L(-1, -6), M(1, -6), and N(1, 1). Find the measure of angle N to the nearest degree. Answer D)

I.

$$\\ \overline{LM}= \sqrt{ {[ (-1)-(1) ]}^2 + {[ (-6)-(-6) ]}^2 } = \sqrt{ {( -2) }^2 + {(0) }^2 } =2 \\\\ \overline{MN}= \sqrt{ {[ (1)-(1) ]}^2 + {[ (-6)-(1) ]}^2 } = \sqrt{ {( 0 )}^2 + { (-7) }^2 } =7 \\\\ \tan{(N)} = \dfrac{\overline{LM}}{\overline{MN}}=\frac{2}{7}\\\\ N=\tan^{-1}{(\frac{2}{7})} = 15.9453959009\ensurement{^{\circ}}\approx 16\ensurement{^{\circ}}$$

The answer is D) $$16\ensurement{^{\circ}}$$

Triangle LMN has vertexes at L(-1, -6), M(1, -6), and N(1, 1). Find the measure of angle N to the nearest degree.

Let's translate N so that it lies at (0,0)...then L= (-2, -7) and M = (0, -7).....this will make the calculation of the distances easier.......

So NL = √(4 + 49) = √(53)

And NM = √49 = 7

And ML = √4 = 2

And LMN forms a right triangle, with angle NML = 90

So.....using the Law of Sines, we have

NL / sin 90 = ML / sin(LNM)

√(53) = 2/sin(LNM)

sin(LNM) = 2/√(53)

sin-1[2/√(53] = LNM ≈15.945° ≈ 16°