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\(\bullet$$\bullet\) A 1.55 -m-tall fisherman stands at the edge of a lake, being watched by a suspicious trout who is 3.50 \(\mathrm{m}\) from the fisherman in the horizontal direction and 45.0 \(\mathrm{cm}\) below the surface of the water. At what angle from the vertical does the fish see the top of the fisherman's head?

Short Answer

Expert verified
The angle from the vertical is approximately 29.74°.

Step by step solution

01

Understanding the problem

The problem involves a fisherman and a fish in a lake. The fish is observing the fisherman who is on land. We need to determine at what angle from the vertical the fish sees the top of the fisherman's head. We have the height of the fisherman, the horizontal distance from the fisherman to the fish, and the depth of the fish below the water surface.
02

Visualize the situation

Imagine a right triangle where the top of the triangle is the head of the fisherman and the other endpoints are at the fish's location. The vertical leg of this triangle is the sum of the fisherman's height plus the depth of the fish (converted to meters). The horizontal leg of the triangle is the distance between the fisherman and the fish.
03

Calculate vertical distance

Calculate the vertical distance from the top of the fisherman's head to the fish's position. This is equal to the fish's depth plus the fisherman's height:\[\text{Vertical distance} = \text{Height of fisherman} + \text{Depth of fish}\]\[= 1.55 \text{ m} + 0.45 \text{ m} = 2.00 \text{ m}\]
04

Calculate the angle using trigonometry

Use the tangent function to find the angle. The tangent of the angle is the ratio of the horizontal distance to the vertical distance:\[\tan(\theta) = \frac{\text{Horizontal Distance}}{\text{Vertical Distance}}\]\[\theta = \tan^{-1}\left(\frac{3.50 \text{ m}}{2.00 \text{ m}}\right)\]Use a calculator to find \( \theta \approx 60.26^{\circ} \).
05

Convert to "angle from the vertical"

Since the angle \(\theta\) is measured from the horizontal, the angle from the vertical is:\[90^{\circ} - 60.26^{\circ} = 29.74^{\circ}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angle Calculation
When solving physics problems involving angles, it's essential to identify what reference point the angle is being measured from, such as vertical or horizontal lines. In problems like the one with the fisherman and the trout, you need to use the geometry of a right triangle to find the angle.
First, you calculate one angle using trigonometry, such as the tangent function, which gives you an angle from the horizontal. This can be confusing if the angle asked for is from another reference, like the vertical. To find the angle from the vertical, subtract the horizontal angle from 90 degrees, because the vertical and horizontal axes are perpendicular.
This calculation turns into:
  • Determine the angle with the horizontal using trigonometry.
  • Subtract from 90° to get the angle with the vertical.
The process highlights how crucial it is to understand angle orientation, not just their size.
Right Triangle
Understanding the setup as a right triangle is pivotal in physics problems involving visual angles. A right triangle is a triangle with one angle at exactly 90 degrees. In our problem, the hypotenuse connects the top of the fisherman's head to the fish.
The vertical side of the triangle combines the fisherman's full height and the fish's depth. The horizontal side represents the distance between them across the lake. Establishing these relations helps in visualizing and simplifying complex scenarios into manageable calculations.
Utilizing right triangles is common in solving such problems because they model how light travels or how something is visually perceived in the simplest manner. They form the backbone of not just physics problems but numerous geometrical and engineering calculations.
Tangent Function
In trigonometry, the tangent function helps us relate angles to the lengths of the opposite side and adjacent side of a right triangle. In this particular exercise, we employ the tangent function to find out at what angle the trout sees the fisherman.
The tangent \( \tan(\theta) \) is defined as the ratio of the length of the opposite side to that of the adjacent side: \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). Here, the 'opposite' is the horizontal distance of 3.50 meters, and the 'adjacent' is the vertical distance of 2.00 meters.
Calculating \( \theta = \tan^{-1}(\frac{3.50}{2.00}) \) we find \( \theta \approx 60.26^{\circ} \) from the horizontal. Mastering the tangent function is essential as it applies widely in physics, from simple calculations to complex wave dynamics.
Visualization of Physics Problems
To solve a physics problem effectively, one of the first steps must always be visualization. Having a clear mental or even a physical diagram helps break down a complex problem into simpler, more approachable parts. In this case, the visualization involves imagining a right triangle with specific known measures.
By visualizing where the fish is and where the fisherman is in relation to each other, you can better grasp how the angle calculation should proceed. Whether using sketches, diagrams, or real-world analogies, visualizing allows us to ground abstract numbers into a concrete image.
  • It helps identify known and unknown variables.
  • It clarifies the relationships between elements of the problem.
  • It guides the method of solution, step by step.
Never underestimate the insight a well-made diagram can provide in tackling physics challenges.

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Most popular questions from this chapter

\(\bullet$$\bullet\) A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of 35.4 \(\mathrm{cm}\) and an electric field amplitude of \(5.40 \times 10^{-2} \mathrm{V} / \mathrm{m}\) at a distance of 250 \(\mathrm{m}\) from the antenna. Calculate: (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

\(\bullet\) Light inside the eye. The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of 1.34 . Visible light ranges in wavelength from 400 \(\mathrm{nm}\) (violet) to \(700 \mathrm{nm}(\mathrm{red}),\) as measured in air. This light travels through the vitreous humor and strikes the rods and cones at the surface of the retina. What are the ranges of (a) the wavelength, (b) the frequency, and (c) the speed of the light just as it approaches the retina within the vitreous humor?

If a light beam strikes a 10 \(\mathrm{cm}\) thick slab of glass (which is immersed in air) at an angle of \(30^{\circ}\) from the normal to the surface, what will be its angle to the normal when it leaves the back side of the slab? Assume that the slab has parallel sides and an index of refraction of \(1.5 .\) $$ \begin{array}{lllllll}{\text { A. } 0^{\circ}} & {\text { B. } 30^{\circ}} & {\text { C. } 60^{\circ}} & {\text { D. } 58.3^{\circ}}\end{array} $$

\(\bullet\) Most people perceive light having a wavelength between 630 \(\mathrm{nm}\) and 700 \(\mathrm{nm}\) as red and light with a wavelength between 400 \(\mathrm{nm}\) and 440 \(\mathrm{nm}\) as violet. Calculate the approximate frequency ranges for (a) violet light and (b) red light.

\(\bullet\) Laboratory lasers. He-Ne lasers are often used in physics demonstrations. They produce light of wavelength 633 \(\mathrm{nm}\) and a power of 0.500 \(\mathrm{mW}\) spread over a cylindrical beam 1.00 \(\mathrm{mm}\) in diameter (although these quantities can vary). (a) What is the intensity of this laser beam? (b) What are the maximum values of the electric and magnetic fields? (c) What is the average energy density in the laser beam?

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