Power line pole

From point A, the power line pole is seen at an angle of 18 degrees. From point B to which we get when going from point A 30m away from the column at an angle of 10 degrees. Find the height of the power pole.


h =  11.567 m


c=30 m A=18  B=10  t0=tan(A rad)=tan(A π180 )=tan(18 3.1415926180 )=0.32492 t1=tan(B rad)=tan(B π180 )=tan(10 3.1415926180 )=0.17633 h/x=tanA=t0 h/(x+c)=tanB=t1  x=h/t0 h/(h/t0+c)=t1 h=t1 (h/t0+c) h=t1/t0 h+c t1 h(1t1/t0)=c t1  h=c t11t1/t0=30 0.176310.1763/0.324911.566911.567 mc=30 \ \text{m} \ \\ A=18 \ ^\circ \ \\ B=10 \ ^\circ \ \\ t_{0}=\tan( A ^\circ \rightarrow\ \text{rad})=\tan( A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=\tan( 18 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=0.32492 \ \\ t_{1}=\tan( B ^\circ \rightarrow\ \text{rad})=\tan( B ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=\tan( 10 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=0.17633 \ \\ h/x=\tan A=t_{0} \ \\ h/(x+c)=\tan B=t_{1} \ \\ \ \\ x=h/t_{0} \ \\ h/(h/t_{0}+c)=t_{1} \ \\ h=t_{1} \cdot \ (h/t_{0}+c) \ \\ h=t_{1}/t_{0} \cdot \ h+c \cdot \ t_{1} \ \\ h(1 -t_{1}/t_{0})=c \cdot \ t_{1} \ \\ \ \\ h=\dfrac{ c \cdot \ t_{1} }{ 1 -t_{1}/t_{0} }=\dfrac{ 30 \cdot \ 0.1763 }{ 1 -0.1763/0.3249 } \doteq 11.5669 \doteq 11.567 \ \text{m}

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