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  B=30  x3=50 m  tanA=y:x1 tanB=y:x2  x22=x32+x12  y=x1 tanA=x2 tanB=35.355 m  x1 tanA=x2 tanB x12 tan2 A=x22 tan2 B  x12 tan2 A=(x32+x12) tan2 B  t1=(tanA)2=(tan45 )2=12=1 t2=(tanB)2=(tan30 )2=0.577352=0.33333  x12 t1=x32 t2+x12 t2 x12 (t1t2)=x32 t2  x1=x3 t2t1t2=50 0.333310.333325 2 m35.3553 m  y=x1 t1=35.3553 125 235.3553   Correctness test:  x2=y/t2=35.3553/0.333325 6 m61.2372 m  x22=x32+x12=502+35.3553225 6 m61.2372 mA=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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