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 m A=18 ∘ B=10 ∘ t0=tan⁡A∘=tan⁡18∘ =0.32492 t1=tan⁡B∘=tan⁡10∘ =0.176327=0.17633 h/x=tan⁡A=t0 h/(x+c)=tan⁡B=t1  x=h/t0 h/(h/t0+c)=t1 h=t1⋅ (h/t0+c) h=t1/t0⋅ h+c⋅ t1 h(1−t1/t0)=c⋅ t1  h=c⋅ t11−t1/t0=30⋅ 0.17631−0.1763/0.3249≐11.5669≐11.567 mc=30 \ \text{m} \ \\ A=18 \ ^\circ \ \\ B=10 \ ^\circ \ \\ t_{0}=\tan A ^\circ =\tan 18^\circ \ =0.32492 \ \\ t_{1}=\tan B ^\circ =\tan 10^\circ \ =0.176327=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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