The tower

The observer sees the tower's base 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 answer:

y₂ =  37.9832 m

Step-by-step explanation:

$$\begin{gathered} h=96\text{m} \\ A=30°10^{\prime }=30°+\frac{10}{60}°=30.1667°\doteq 30.1667 \\ B=20°50^{\prime }=20°+\frac{50}{60}°=20.8333°\doteq 20.8333 \\ h=y_1+y_2 \\ \tan A=y_1/x \\ \tan B=y_2/x \\ \tan A/\tan B=y_1/y_2 \\ \tan A/\tan B=(h-y_2)/y_2 \\ t=\tan A/\tan B=\tan 30.166666666667°/\tan 20.833333333333°=0.581235/0.38053=1.52744 \\ t=(h-y_2)/y_2 \\ {y}_2\cdot t=h-y_2 \\ {y}_2(t+1)=h \\ {y}_2=h/(t+1)=96/(1.5274+1)\doteq 37.9832\text{m} \\ \text{ Verifying Solution: } \\ {y}_1=h-y_2=96-37.9832\doteq 58.0168\text{m} \\ x=y_1/\tan A=y_1/\tan 30.166666666667°=58.0168/\tan 30.166666666667°=58.0168/0.581235=99.816\text{m} \end{gathered}$$

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