# An observer

An observer standing west of the tower sees its top at an altitude angle of 45 degrees. After moving 50 meters to the south, he sees its top at an altitude angle of 30 degrees. How tall is the tower?

Correct result:

y =  35.355 m

#### Solution:

$A=45 \ ^\circ \ \\ B=30 \ ^\circ \ \\ x_{3}=50 \ \text{m} \ \\ \ \\ \tan A=y : x_{1} \ \\ \tan B=y : x_{2} \ \\ \ \\ x_{2}^2=x_{3}^2+x_{1}^2 \ \\ \ \\ y=x_{1} \cdot \ \tan A=x_{2} \cdot \ \tan B=35.355 \ \text{m} \ \\ \ \\ x_{1} \cdot \ \tan A=x_{2} \cdot \ \tan B \ \\ x_{1}^2 \cdot \ \tan^2 \ A=x_{2}^2 \cdot \ \tan^2 \ B \ \\ \ \\ x_{1}^2 \cdot \ \tan^2 \ A=(x_{3}^2+x_{1}^2) \cdot \ \tan^2 \ B \ \\ \ \\ t_{1}=(\tan A ^\circ )^2=(\tan 45^\circ \ )^2=1^2=1 \ \\ t_{2}=(\tan B ^\circ )^2=(\tan 30^\circ \ )^2=0.57735^2=0.33333 \ \\ \ \\ x_{1}^2 \cdot \ t_{1}=x_{3}^2 \cdot \ t_{2} +x_{1}^2 \cdot \ t_{2} \ \\ x_{1}^2 \cdot \ (t_{1}-t_{2})=x_{3}^2 \cdot \ t_{2} \ \\ \ \\ x_{1}=x_{3} \cdot \ \sqrt{ \dfrac{ t_{2} }{ t_{1}-t_{2} } }=50 \cdot \ \sqrt{ \dfrac{ 0.3333 }{ 1-0.3333 } } \doteq 25 \ \sqrt{ 2 } \ \text{m} \doteq 35.3553 \ \text{m} \ \\ \ \\ y=x_{1} \cdot \ \sqrt{ t_{1} }=35.3553 \cdot \ \sqrt{ 1 } \doteq 25 \ \sqrt{ 2 } \doteq 35.3553 \ \\ \ \\ \text{ Correctness test: } \ \\ x_{2}=y / \sqrt{ t_{2} }=35.3553 / \sqrt{ 0.3333 } \doteq 25 \ \sqrt{ 6 } \ \text{m} \doteq 61.2372 \ \text{m} \ \\ \ \\ x_{22}=\sqrt{ x_{3}^2+x_{1}^2 }=\sqrt{ 50^2+35.3553^2 } \doteq 25 \ \sqrt{ 6 } \ \text{m} \doteq 61.2372 \ \text{m}$

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