A 4.00 m long pole stands vertically in a lake having a depth of 2.50 m. The Sun is 40.0° above the horizontal. Determine the length of the pole's shadow on the bottom of the lake. Take the index of refraction for water to be 1.33.

Respuesta :

Answer:3.541 m

Step-by-step explanation:

 From Diagram we can say that

[tex]tan\left ( 50\right )=\frac{b}{1.5}[/tex]

b=1.787 m

also we know

[tex]\frac{sin50}{sin\theta}=\mu [/tex]

[tex]sin\theta =0.5745[/tex]

[tex]\theta =35.064^{\circ}[/tex]

[tex]also\ tan\theta =\frac{a}{2.5}[/tex]

a=1.754

thus shadow of pole[tex]=a+b=1.754+1.787=3.541 m[/tex]

Ver imagen nuuk

The shadow of the pole is; 3.617 m

What is trigonometric ratio?

From the diagram, we can say with the aid of trigonometric ratios that;

tan 50 = b/1.5

Thus; b = 1.5 * tan 50

b = 1.787 m

From snell's law, we can say that;

sin 50/sin θ = 1.33

Thus;

sin θ = (sin 50)/1.33

sin θ = 0.575973

θ = 35.1678°

Similarly;
tan θ = a/2.5

a = 2.5 * tan 35.1678°

a = 1.83

Shadow of pole = a + b

Shadow of pole = 1.83 + 1.787 = 3.617 m

Read more about Trigonometric ratios at; https://brainly.com/question/13276558

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