Stick shadow angle

The meter stick is located on the meridian plane and deviated from the horizontal plane to the north by an angle of magnitude 70°. Calculate the length of the shadow cast by a meter stick at true noon if the Sun culminates at an angle of 41°03'.

Final Answer:

s =  1.132 m

Step-by-step explanation:

$$\begin{gathered} a=1\text{m} \\ \alpha =70{}^{\circ } \\ \beta =41°3^{\prime }=41°+\frac{3}{60}°=41.05°=41.05 \\ \cos \alpha =x:a \\ \sin \alpha =y:a \\ x=a\cdot \cos \alpha =a\cdot \cos 70°=1\cdot \cos 70°=1\cdot 0.34202=0.34202\text{m} \\ y=a\cdot \sin \alpha =a\cdot \sin 70°=1\cdot \sin 70°=1\cdot 0.939693=0.93969\text{m} \\ \tan \beta =y:z \\ z=y/\tan \beta =y/\tan 41.05°=0.9397/\tan 41.05°=0.9397/0.87082=1.07909\text{m} \\ {s}^2=x^2+z^2 \\ s=\sqrt{x^2+z^2}=\sqrt{0.342^2+1.0791^2}=1.132\text{m} \end{gathered}$$

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