Viewing angle

The observer sees a straight fence 60 m long at a viewing angle of 30°. It is 102 m away from one end of the enclosure.
How far is the observer from the other end of the enclosure?

Correct result:

c₁ =  119.9416 m
c₂ =  56.7276 m

Solution:

$$\begin{gathered} a=60\text{ m } \\ A=30{}^{\circ } \\ b=102\text{ m } \\ {a}^2=b^2+c^2-2\cdot b\cdot c\cdot \cos (A) \\ k=2\cdot b\cdot \cos A^{\circ }=2\cdot b\cdot \cos 30^{\circ }=2\cdot 102\cdot \cos 30^{\circ }=2\cdot 102\cdot 0.866025=176.66918 \\ {a}^2=b^2+c^2-k\ast c \\ 60^2=102^2+c^2-176.669182372\cdot c \\ -c^2+176.669c-6804=0 \\ {c}^2-176.669c+6804=0 \\ p=1;q=-176.669;r=6804 \\ D=q^2-4pr=176.669^2-4\cdot 1\cdot 6804=3995.99999999 \\ D>0 \\ {c}_{1,2}={\displaystyle \frac{-q\pm \sqrt{D}}{2p}}={\displaystyle \frac{176.67\pm \sqrt{3996}}{2}} \\ {c}_{1,2}=88.33459119\pm 31.6069612585 \\ {c}_1=119.941552445=119.9416\text{ m } \\ {c}_2=56.7276299275 \\ \text{ Factored form of the equation: } \\ (c-119.941552445)(c-56.7276299275)=0 \end{gathered}$$

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