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?

Result

c1 =  119.942 m
c2 =  56.728 m

Solution:

a=60 m A=30  b=102 m  a2=b2+c22 b c cos(A)  k=2 b cosA=2 b cos30 =2 102 cos30 =2 b 0.866025=176.66918  a2=b2+c2kc  602=1022+c2176.669182372 c c2+176.669c6804=0 c2176.669c+6804=0  p=1;q=176.669;r=6804 D=q24pr=176.6692416804=3995.99999999 D>0  c1,2=q±D2p=176.67±39962 c1,2=88.33459119±31.6069612585 c1=119.941552445=119.942 m c2=56.7276299275   Factored form of the equation:  (c119.941552445)(c56.7276299275)=0a=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 \ b \cdot \ 0.866025=176.66918 \ \\ \ \\ a^2=b^2+c^2-k*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}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 176.67 \pm \sqrt{ 3996 } }{ 2 } \ \\ c_{1,2}=88.33459119 \pm 31.6069612585 \ \\ c_{1}=119.941552445=119.942 \ \text{m} \ \\ c_{2}=56.7276299275 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -119.941552445) (c -56.7276299275)=0

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