# 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.

Result

h =  11.567 m

#### Solution:

$c=30 \ \text{m} \ \\ A=18 \ ^\circ \ \\ B=10 \ ^\circ \ \\ t_{0}=\tan( A ^\circ \rightarrow\ \text{rad})=\tan( A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=\tan( 18 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=0.32492 \ \\ t_{1}=\tan( B ^\circ \rightarrow\ \text{rad})=\tan( B ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=\tan( 10 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=0.17633 \ \\ h/x=\tan A=t_{0} \ \\ h/(x+c)=\tan B=t_{1} \ \\ \ \\ x=h/t_{0} \ \\ h/(h/t_{0}+c)=t_{1} \ \\ h=t_{1} \cdot \ (h/t_{0}+c) \ \\ h=t_{1}/t_{0} \cdot \ h+c \cdot \ t_{1} \ \\ h(1 -t_{1}/t_{0})=c \cdot \ t_{1} \ \\ \ \\ h=\dfrac{ c \cdot \ t_{1} }{ 1 -t_{1}/t_{0} }=\dfrac{ 30 \cdot \ 0.1763 }{ 1 -0.1763/0.3249 } \doteq 11.5669 \doteq 11.567 \ \text{m}$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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