The mast

The top of the pole we see at an angle of 45°. If we approach the pole by 10 m, we see the top of the pole at an angle of 60°. What is the height of the pole?

Result

h =  23.66 m

Solution:

tan⁡(60∘)=h/x tan⁡(45∘)=h/(x+10)  x=h/tan⁡(60∘)=h⋅ t1 t1=1/tan⁡(60∘→ rad)=1/tan⁡(60∘⋅ π180 )=1/tan⁡(60∘⋅ 3.1415926180 )=0.57735 t2=tan⁡(45∘→ rad)=tan⁡(45∘⋅ π180 )=tan⁡(45∘⋅ 3.1415926180 )=1  t2⋅ (h⋅ t1+10)=h t2⋅ h⋅ t1+10⋅ t2=h  h=10⋅ t21−t2⋅ t1=10⋅ 11−1⋅ 0.5774≐23.6603≐23.66 m\tan(60^\circ )=h/x \ \\ \tan(45^\circ )=h/(x+10) \ \\ \ \\ x=h / \tan(60^\circ )=h \cdot \ t_{1} \ \\ t_{1}=1/ \tan( 60 ^\circ \rightarrow\ \text{rad})=1/ \tan( 60 ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=1/ \tan( 60 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=0.57735 \ \\ t_{2}=\tan( 45 ^\circ \rightarrow\ \text{rad})=\tan( 45 ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=\tan( 45 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=1 \ \\ \ \\ t_{2} \cdot \ (h \cdot \ t_{1}+10)=h \ \\ t_{2} \cdot \ h \cdot \ t_{1}+10 \cdot \ t_{2}=h \ \\ \ \\ h=\dfrac{ 10 \cdot \ t_{2} }{ 1-t_{2} \cdot \ t_{1} }=\dfrac{ 10 \cdot \ 1 }{ 1-1 \cdot \ 0.5774 } \doteq 23.6603 \doteq 23.66 \ \text{m}

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