Three pillars

On a straight road, three pillars are 6 m high at the same distance of 10 m. At what angle of view does Vlado see each pillar if it is 30 m from the first and his eyes are 1.8 m high?

Correct answer:

A =  11.4032 °
B =  8.5707 °
C =  6.8633 °

Step-by-step explanation:

$$\begin{gathered} n=3 \\ h=6\text{m} \\ a=10\text{m} \\ {l}_1=30\text{m} \\ o=1.8\text{m} \\ \tan \alpha =o:l_1 \\ \alpha =\frac{180°}{\pi }\cdot \arctan (o/l_1)=\frac{180°}{\pi }\cdot \arctan (1.8/30)\doteq 3.4336{}^{\circ } \\ \tan \beta =(h-o):l_1 \\ \beta =\frac{180°}{\pi }\cdot \arctan ((h-o)/l_1)=\frac{180°}{\pi }\cdot \arctan ((6-1.8)/30)\doteq 7.9696{}^{\circ } \\ A=\alpha +\beta =3.4336+7.9696=11.4032^{\circ }=11°24^{\prime }12" \end{gathered}$$

$$\begin{gathered} \theta =\arctan (o/(l_1+a))+\arctan ((h-o)/(l_1+a))=\arctan (1.8/(30+10))+\arctan ((6-1.8)/(30+10))\doteq 0.1496\text{rad} \\ B=\theta \to \text{°}=\theta \cdot \frac{180}{\pi }\text{°}=0.1496\cdot \frac{180}{\pi }\text{°}=8.571\text{°}=8.5707^{\circ }=8°34^{\prime }14" \end{gathered}$$

$$\begin{gathered} \Psi =\arctan (o/(l_1+2\cdot a))+\arctan ((h-o)/(l_1+2\cdot a))=\arctan (1.8/(30+2\cdot 10))+\arctan ((6-1.8)/(30+2\cdot 10))\doteq 0.1198\text{rad} \\ C=\Psi \to \text{°}=\Psi \cdot \frac{180}{\pi }\text{°}=0.1198\cdot \frac{180}{\pi }\text{°}=6.863\text{°}=6.8633^{\circ }=6°51^{\prime }48" \end{gathered}$$

Try another example