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Chapter 1: Problem 51

A chimpanzee sitting against his favorite tree gets up and walks \(51 \mathrm{~m}\) due east and \(39 \mathrm{~m}\) due south to reach a termite mound, where he eats lunch. (a) What is the shortest distance between the tree and the termite mound? (b) What angle does the shortest distance make with respect to due east?

Short Answer

Expert verified
The shortest distance is approximately 64.22 meters, and the angle with east is about 37.31掳.

Step by step solution

01

Understanding the Problem

To solve this exercise, we need to determine the shortest distance, which is the straight-line distance, between the two positions. The chimpanzee travels east and then south, creating a right triangle. We'll use the Pythagorean theorem to calculate the hypotenuse of this triangle. Additionally, we'll calculate the angle that this hypotenuse makes with the east direction.
02

Apply the Pythagorean Theorem

The chimpanzee's path follows two perpendicular directions, east and south, creating a right-angled triangle. The shortest distance is the hypotenuse. Using the Pythagorean theorem, we find the hypotenuse: \[h = \sqrt{(51)^2 + (39)^2}\] \[= \sqrt{2601 + 1521}\] \[= \sqrt{4122}\] \[\approx 64.22 \text{ meters}\].
03

Calculate the Angle with East

We use trigonometry to find the angle that the hypotenuse makes with the east direction. The tangent of this angle \(\theta\) is the opposite side over the adjacent side, which are the south and east displacements:\[\tan(\theta) = \frac{39}{51}\]To find \(\theta\), calculate:\[\theta = \tan^{-1}\left(\frac{39}{51}\right)\]Using a calculator, \(\theta \approx 37.31^\circ\).

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Key Concepts

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

Trigonometry
Trigonometry is a branch of mathematics that studies the relationships between the angles and sides of triangles. It's particularly useful for right triangles, where one of the angles is exactly 90 degrees. Understanding trigonometry helps in solving problems involving angles and distances, like in our exercise with the chimpanzee.
In the chimpanzee's journey, trigonometry assists in finding the angle and distance in a path that's not directly traveled. For instance, when the chimpanzee walks east and then south, a right triangle forms. By knowing the lengths of two sides of this triangle, trigonometry allows us to find unknown values such as the hypotenuse or angles using functions like sine, cosine, and tangent.
These functions relate to the sides of a right triangle in specific ways:
  • Sine (\( ext{sin}\)) - opposite side over the hypotenuse
  • Cosine (\( ext{cos}\)) - adjacent side over the hypotenuse
  • Tangent (\( ext{tan}\)) - opposite side over the adjacent side
Trigonometry, hence, is pivotal for calculating the needed angle and ensuring solutions are both precise and practical.
Right Triangle
A right triangle is a type of triangle that contains a 90-degree angle. This unique property makes it particularly useful for various calculations because it allows us to apply the Pythagorean Theorem.
In the context of the chimpanzee's path, the right triangle comes into play when the chimpanzee moves east and then south. These movements form the two legs of the triangle, with the shortest path back to the tree being the hypotenuse.
Right triangles have specific relationships:
  • The longest side, opposite the right angle, is called the hypotenuse.
  • The other two sides are called the legs. One leg is adjacent to the angle of interest, and the other is opposite.
  • The Pythagorean Theorem states that in a right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse: \[a^2 + b^2 = c^2\]
These relationships turn right triangles into a useful tool for solving spatial problems.
Angle Calculation
Calculating angles is vital when determining directions or understanding shapes and spaces. In the chimpanzee exercise, angle calculation involves determining the angle the shortest path (hypotenuse) makes with one of the legs (east direction).
For this, the tangent function is employed, which relates two sides of a right triangle鈥攖he opposite (south) and adjacent (east) sides. The formula used is:\[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\]
Once the tangent of the angle is known, finding the angle itself requires the inverse tangent or arctan function.
Using a calculator, this helps in converting the ratio into an angle in degrees. In our example:
  • Plug in the lengths of the opposite (\(39\) meters south) and adjacent side (\(51\) meters east).
  • Calculating \(\theta \approx 37.31^\circ\)
  • This angle is the measure between the shortest path and the east direction.
Understanding how to calculate angles helps in navigation and ensuring correct pathways in geometry.

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Most popular questions from this chapter

Interactive Solution \(1.49\) at presents the solution to a problem that is similar to this one. Vector \(\overrightarrow{\mathrm{A}}\) has a magnitude of 145 units and points \(35.0^{\circ}\) north of west. Vector \(\overrightarrow{\mathrm{B}}\) points \(65.0^{\circ}\) east of north. Vector \(\overrightarrow{\mathrm{C}}\) points \(15.0^{\circ}\) west of south. These three vectors add to give a resultant vector that is zero. Using components, find the magnitudes of (a) vector \(\overrightarrow{\mathrm{B}}\) and (b) vector \(\overrightarrow{\mathrm{C}}\).

The drawing shows sodium and chloride ions positioned at the corners of a cube that is part of the crystal structure of sodium chloride (common table salt). The edge of the cubs is \(0.281 \mathrm{nm}\left(1 \mathrm{nm}=1\right.\) nanometer \(\left.=10^{-9} \mathrm{~m}\right)\) in length. Find the distance (in nanometers) between the sodium ion located at one corner of the cube and the chloride ion located on the diagonal at the opposite corner.

To review the solution to a similar problem, consult Interactive Solution \(1.37\) at . The magnitude of the force vector \(\vec{F}\) is \(82.3\) newtons. The \(x\) component of this vector is directed along the \(+x\) axis and has a magnitude of \(74.6\) newtons. The \(y\) component points along the \(+y\) axis. (a) Find the direction of \(\overrightarrow{\mathbf{F}}\) relative to the \(+x\) axis. (b) Find the component of \(\overrightarrow{\mathbf{F}}\) along the \(+y\) axis.

The displacement vector \(\overrightarrow{\mathrm{A}}\) has scalar components of \(A_{x}=80.0 \mathrm{~m}\) and \(A_{y}=60.0 \mathrm{~m}\) The displacement vector \(\overrightarrow{\mathbf{B}}\) has a scalar component of \(B_{x}=60.0 \mathrm{~m}\) and a magnitude of \(B=75.0 \mathrm{~m} .\) The displacement vector \(\overrightarrow{\mathrm{C}}\) has a magnitude of \(C=100.0 \mathrm{~m}\) and is directed at an angle of \(36.9^{\circ}\) above the \(+x\) axis. Two of these vectors are equal. Determine which two, and support your choice with a calculation.

The components of vector \(\overrightarrow{\mathbf{A}}\) are \(A_{x}\) and \(A_{v}\) (both positive), and the angle that it makes with respect to the positive \(x\) axis is \(\theta\). (a) Does increasing the component \(A_{x}\) (while holding \(A_{y}\) constant) increase or decrease the angle \(\theta\) ? (b) Does increasing the component \(A_{y}\) (while holding \(A_{x}\) constant) increase or decrease the angle \(\theta\) ? Account for your answers. The components of displacement vector \(\overrightarrow{\mathrm{A}}\) are \(A_{x}=12 \mathrm{~m}\) and \(A_{y}=12 \mathrm{~m}\). Find \(\theta\). (b) The components of displacement vector \(\overrightarrow{\mathrm{A}}\) are \(A_{x}=17 \mathrm{~m}\) and \(A_{y}=12 \mathrm{~m}\). Find \(\theta\). (c) The components of displacement vector \(\overrightarrow{\mathbf{A}}\) are \(A_{x}=12 \mathrm{~m}\) and \(A_{y}=17 \mathrm{~m}\). Find \(\theta\). Be sure that your answers are consistent with your answers to the Concept
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