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Chapter 9: Problem 41

Round each answer to one decimal place. (a) Let \(m\) and \(n\) be positive numbers, with \(m>n .\) Furthermore, suppose that in triangle \(A B C\) the lengths \(a\) \(b,\) and \(c\) are given by $$ \begin{array}{c} a=2 m n+n^{2} \quad b=m^{2}-n^{2} \\ c=m^{2}+n^{2}+m n \end{array} $$ Show that \(\cos C=-1 / 2,\) and conclude that \(\angle C=120^{\circ}\) (b) Give an example of a triangle in which the lengths of the sides are whole numbers and one of the angles is \(120^{\circ} .\) (Specify the three sides; you needn't find the other angles.)

Short Answer

Expert verified
In triangle ABC, \( \angle C = 120^\circ \), possible side lengths are 5, 3, 7.

Step by step solution

01

Apply the Cosine Law

In triangle ABC, we apply the Cosine Law: \[ c^2 = a^2 + b^2 - 2ab \cos C \]Substitute the given expressions for \(a\), \(b\), and \(c\) into the equation:
02

Solve for Cosine C

Substitute:\[ (m^2+n^2+mn)^2 = (2mn+n^2)^2 + (m^2-n^2)^2 - 2(2mn+n^2)(m^2-n^2) \cos C \] When we expand and simplify both sides, the left-hand side becomes \(m^4 + 2m^3n + 3m^2n^2 + 2mn^3 + n^4\). The right-hand side simplifies to the same expression when \(\cos C = -\frac{1}{2}\). Thus, \(\cos C = -\frac{1}{2}\) is verified.
03

Conclude Angle C is 120 Degrees

Since \(\cos C = -\frac{1}{2}\), this corresponds to an angle of \(120^\circ\) because \(\cos 120^\circ = -\frac{1}{2}\). Thus, \(\angle C = 120^\circ\).
04

Example with Integer Side Lengths

We need to find integers \(m\) and \(n\) such that \(a\), \(b\), and \(c\) are integers. Consider \(m=2, n=1\). Then:- \(a = 2(2)(1) + 1^2 = 5\)- \(b = 4 - 1 = 3\)- \(c = 4 + 1 + 2 = 7\).The triangle with sides \(5, 3, 7\) has an angle of \(120^\circ\).

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Key Concepts

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

Triangle Geometry
When studying triangles, it is essential to understand some fundamental properties that help in analyzing and solving problems. Triangles are three-sided polygons characterized by their sides and angles. Some key properties include:
  • Sum of Interior Angles: In any triangle, the sum of the three interior angles is always 180 degrees.
  • Types of Triangles: Triangles can be classified based on their sides, such as equilateral (all sides equal), isosceles (two sides equal), and scalene (all sides different). They can also be classified based on their angles, like acute (all angles less than 90 degrees), right (one angle is 90 degrees), and obtuse (one angle is more than 90 degrees).
To explore the angles, lengths, and relationships within triangles, mathematical theorems and laws such as the Cosine Law are used, which aid in further understanding complex properties of triangles.
Angle Properties
In triangle problems, understanding angle properties is crucial. The Cosine Law, as used in the problem, relates the angles of a triangle to the lengths of its sides. The law is given by:\[ c^2 = a^2 + b^2 - 2ab \cos C \]This equation helps determine the angle when the lengths of all sides are known, and vice versa. For example, the exercise demonstrates that if the sides of a triangle are expressed through specific formulas, it can be shown mathematically that \( \cos C = -\frac{1}{2} \), leading to an angle of 120 degrees.
This is because one of the key angle properties is that the cosine of 120 degrees is indeed -1/2. Such relationships are useful in reconstructing a triangle or in verifying provided information about a triangle's geometry.
Solving Equations
Solving equations is an integral part of applying geometric principles to problems. For instance, using the expression from the Cosine Law:\[ (m^2+n^2+mn)^2 = (2mn+n^2)^2 + (m^2-n^2)^2 - 2(2mn+n^2)(m^2-n^2) \cos C \], transforming and simplifying both sides of the equation is essential to finding solutions, such as computing \( \cos C \).
In this exercise, simplification verified that \( \cos C = -\frac{1}{2} \). Proper comprehension of algebraic manipulation, as well as squared terms and factorization techniques, is key when tackling these equations.
Thus, understanding how to manipulate and solve for unknowns in equations bridges the gap between triangle sides and angles, enabling accurate determination of the entire triangle's geometry.

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Most popular questions from this chapter

Graph the polar curves. $$r=(\sin \theta) / \theta, \quad 0 < \theta \leq 4 \pi$$

Use the given information to find the cosine of each angle in \(\triangle \overline{A B C}\) \(a=17 \mathrm{cm}, b=8 \mathrm{cm}, c=15 \mathrm{cm}\) (For this particular tri- angle, you can check your answers, because there is an alternative method of solution that does not require the law of cosines.)

Let \(A\) and \(B\) be nonzero vectors. (a) If \(\mathbf{A} \cdot \mathbf{B}=0,\) explain why \(\mathbf{A}\) and \(\mathbf{B}\) are perpendicular. (b) If \(A\) and \(B\) are perpendicular, explain why \(A \cdot B=0\) (a) and (b) together show that two nonzero vectors are perpendicular if and only if their dot product is zero.

Compute each angle of the given triangle. Where necessary, use a calculator and round to one decimal place. $$a=7, b=8, c=13$$

Let \(\theta\) (where \(0 \leq \theta \leq \pi\) ) denote the angle between the two nonzero vectors \(\mathbf{A}\) and \(\mathbf{B}\). Then it can be shown that the cosine of \(\theta\) is given by the formula $$\cos \theta=\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}$$ (See Exercise 77 for the derivation of this result.) In Exercises \(65-70,\) sketch each pair of vectors as position vectors, then use this formula to find the cosine of the angle between the given pair of vectors. Also, in each case, use a calculator to compute the angle. Express the angle using degrees and using radians. Round the values to two decimal places. (a) \(\mathbf{A}=\langle-8,2\rangle\) and \(\mathbf{B}=\langle 1,-3\rangle\) (b) \(\mathbf{A}=\langle-8,2\rangle\) and \(\mathbf{B}=\langle-1,3\rangle\)
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