Intersection 6653

Two straight paths cross, making an angle alpha = 53 degrees 30'. There are two pillars on one of them, one at the intersection, the other at a distance of 500m from it. How far does one have to go from the intersection along the other road to see both poles in beta view?

a) alpha = beta
b) beta= 15 degrees?

Correct answer:

x₁ =  594.8228 m
x₂ =  1797.4287 m

Step-by-step explanation:

$$\begin{gathered} A=53+30/60=\frac{107}{2}=53.5{}^{\circ } \\ b=500\text{m} \\ {B}_1=A=53.5=\frac{107}{2}=53.5{}^{\circ } \\ v=b\cdot \sin A=b\cdot \sin 53.5°=500\cdot \sin 53.5°=500\cdot 0.803857=401.92843\text{m} \\ {x}_0=\sqrt{b^2-v^2}=\sqrt{500^2-401.9284^2}\doteq 297.4114\text{m} \\ {y}_0=v/\tan (B_1°\to \text{rad})=v/\tan (B_1°\cdot \frac{\pi }{180})=401.92843030861/\tan (53.5°\cdot \frac{3.1415926}{180})=297.41139\text{m} \\ {x}_1=x_0+y_0=297.4114+297.4114=594.8228\text{m} \end{gathered}$$

$$\begin{gathered} B_2=15{}^{\circ } \\ {y}_1=v/\tan (B_2°\to \text{rad})=v/\tan (B_2°\cdot \frac{\pi }{180})=401.92843030861/\tan (15°\cdot \frac{3.1415926}{180})=1500.01732\text{m} \\ {x}_2=x_0+y_1=297.4114+1500.0173=1797.4287\text{m} \end{gathered}$$

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