# 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:

$h=96 \ \text{m} \ \\ A=30+10/60=\dfrac{ 181 }{ 6 } \doteq 30.1667 \ ^\circ \ \\ B=20+50/60=\dfrac{ 125 }{ 6 } \doteq 20.8333 \ ^\circ \ \\ \ \\ h=y_{1}+y_{2} \ \\ \tan A=y_{1}/x \ \\ \tan B=y_{2}/x \ \\ \ \\ \tan A/\tan B=y_{1}/y_{2} \ \\ \tan A/\tan B=(h-y_{2})/y_{2} \ \\ \ \\ t=\tan A ^\circ /\tan B ^\circ =\tan 30.1666666667^\circ \ /\tan 30.1666666667^\circ \ =0.581235/0.581235=1.52744 \ \\ \ \\ t=(h-y_{2})/y_{2} \ \\ y_{2} \cdot \ t=h-y_{2} \ \\ \ \\ y_{2}(t+1)=h \ \\ \ \\ y_{2}=h / (t+1)=96 / (1.5274+1) \doteq 37.9832 \doteq 37.983 \ \text{m} \ \\ \ \\ \text{ Correctness test: } \ \\ y_{1}=h - y_{2}=96 - 37.9832 \doteq 58.0168 \ \text{m} \ \\ x=y_{1} / \tan A ^\circ =y_{1} / \tan 30.1666666667^\circ \ =58.0168317089 / \tan 30.1666666667^\circ \ =58.0168317089 / 0.581235=99.816$ 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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