Two groves

Two groves A, B are separated by a forest, both are visible from the hunting grove C, which is connected to both by direct roads. What will be the length of the projected road from A to B, if AC = 5004 m, BC = 2600 m and angle ABC = 53° 45 ’?

Result

c =  6080.931 m

Solution:

b=5004 m a=2600 m B=53+45/60=2154=53.75   c2=a2+c22accosβ  k=cosB=cos53.75 =0.59131  b2=a2+c22ack  50042=26002+c22 2600 c 0.591309648364 c2+3074.81c+18280016=0 c23074.81c18280016=0  p=1;q=3074.81;r=18280016 D=q24pr=3074.81241(18280016)=82574521.5907 D>0  c1,2=q±D2p=3074.81±82574521.592 c1,2=1537.40508575±4543.52620744 c1=6080.93129318 c2=3006.12112169   Factored form of the equation:  (c6080.93129318)(c+3006.12112169)=0  c=c1=6080.93136080.93136080.931 mb=5004 \ \text{m} \ \\ a=2600 \ \text{m} \ \\ B=53+45/60=\dfrac{ 215 }{ 4 }=53.75 \ ^\circ \ \\ \ \\ c^2=a^2 +c^2 −2ac \cos β \ \\ \ \\ k=\cos B ^\circ =\cos 53.75^\circ \ =0.59131 \ \\ \ \\ b^2=a^2 + c^2 - 2*a*c*k \ \\ \ \\ 5004^2=2600^2 + c^2 - 2 \cdot \ 2600 \cdot \ c \cdot \ 0.591309648364 \ \\ -c^2 +3074.81c +18280016=0 \ \\ c^2 -3074.81c -18280016=0 \ \\ \ \\ p=1; q=-3074.81; r=-18280016 \ \\ D=q^2 - 4pr=3074.81^2 - 4\cdot 1 \cdot (-18280016)=82574521.5907 \ \\ D>0 \ \\ \ \\ c_{1,2}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 3074.81 \pm \sqrt{ 82574521.59 } }{ 2 } \ \\ c_{1,2}=1537.40508575 \pm 4543.52620744 \ \\ c_{1}=6080.93129318 \ \\ c_{2}=-3006.12112169 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (c -6080.93129318) (c +3006.12112169)=0 \ \\ \ \\ c=c_{1}=6080.9313 \doteq 6080.9313 \doteq 6080.931 \ \text{m}

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