The tower

The observer sees the base of the tower 96 meters high at a depth of 30 degrees and 10 minutes and the top of the tower at a depth of 20 degrees and 50 minutes. How high is the observer above the horizontal plane on which the tower stands?

Correct result:

y2 =  37.983 m

Solution:

h=96 m A=30+10/60=1816≐30.1667 ∘ B=20+50/60=1256≐20.8333 ∘  h=y1+y2 tan⁡A=y1/x tan⁡B=y2/x  tan⁡A/tan⁡B=y1/y2 tan⁡A/tan⁡B=(h−y2)/y2  t=tan⁡A∘/tan⁡B∘=tan⁡30.166666666667∘ /tan⁡30.166666666667∘ =0.581235/0.581235=1.52744  t=(h−y2)/y2 y2⋅ t=h−y2  y2(t+1)=h  y2=h/(t+1)=96/(1.5274+1)≐37.9832≐37.983 m   Correctness test:  y1=h−y2=96−37.9832≐58.0168 m x=y1/tan⁡A∘=y1/tan⁡30.166666666667∘ =58.016831708774/tan⁡30.166666666667∘ =58.016831708774/0.581235=99.816h=96 \ \text{m} \ \\ A=30+10/60=\dfrac{ 181 }{ 6 } \doteq 30.1667 \ ^\circ \ \\ B=20+50/60=\dfrac{ 125 }{ 6 } \doteq 20.8333 \ ^\circ \ \\ \ \\ h=y_{1}+y_{2} \ \\ \tan A=y_{1}/x \ \\ \tan B=y_{2}/x \ \\ \ \\ \tan A/\tan B=y_{1}/y_{2} \ \\ \tan A/\tan B=(h-y_{2})/y_{2} \ \\ \ \\ t=\tan A ^\circ /\tan B ^\circ =\tan 30.166666666667^\circ \ /\tan 30.166666666667^\circ \ =0.581235/0.581235=1.52744 \ \\ \ \\ t=(h-y_{2})/y_{2} \ \\ y_{2} \cdot \ t=h-y_{2} \ \\ \ \\ y_{2}(t+1)=h \ \\ \ \\ y_{2}=h / (t+1)=96 / (1.5274+1) \doteq 37.9832 \doteq 37.983 \ \text{m} \ \\ \ \\ \text{ Correctness test: } \ \\ y_{1}=h - y_{2}=96 - 37.9832 \doteq 58.0168 \ \text{m} \ \\ x=y_{1} / \tan A ^\circ =y_{1} / \tan 30.166666666667^\circ \ =58.016831708774 / \tan 30.166666666667^\circ \ =58.016831708774 / 0.581235=99.816

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