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Chapter 9: Problem 29

Compute each angle of the given triangle. Where necessary, use a calculator and round to one decimal place. $$a=b=2 / \sqrt{3}, c=2$$

Short Answer

Expert verified
Gamma is 70.5 degrees, Alpha and Beta are 54.8 degrees each.

Step by step solution

01

Set up the Trigonometry Rule

In this problem, we will use the cosine rule to find the angles of the triangle. The cosine rule states that for a triangle with sides a, b, and c, and the angle opposite side c being \( \gamma \), the formula is given by \( c^2 = a^2 + b^2 - 2ab \cos(\gamma) \).
02

Compute Angle Gamma (\( \gamma \))

Substituting the given values into the cosine rule: \[ c^2 = a^2 + b^2 - 2ab \cos(\gamma) \] \[ 2^2 = \left( \frac{2}{\sqrt{3}} \right)^2 + \left( \frac{2}{\sqrt{3}} \right)^2 - 2 \left( \frac{2}{\sqrt{3}} \right) \left( \frac{2}{\sqrt{3}} \right) \cos(\gamma) \] Solving: \[ 4 = \frac{4}{3} + \frac{4}{3} - \frac{8}{3} \cos(\gamma) \] \[ 4 = \frac{8}{3} - \frac{8}{3} \cos(\gamma) \]\[ \frac{8}{3} \cos(\gamma) = \frac{8}{3} - 4 \]\[ \cos(\gamma) = \frac{4}{3} - 1 \]\[ \cos(\gamma) = \frac{1}{3} \]Using a calculator to find the angle for \( \cos(\gamma) = \frac{1}{3} \), we find \( \gamma \approx 70.5^\circ \).
03

Compute Angles Alpha (\( \alpha \)) and Beta (\( \beta \))

Since the triangle is isosceles with sides \( a = b \), angles \( \alpha \) and \( \beta \) will be equal. To find one of those angles, use: \[ \alpha = \beta = \frac{180^\circ - \gamma}{2} \]Substituting: \[ \alpha = \beta = \frac{180^\circ - 70.5^\circ}{2} \]\[ \alpha = \beta = \frac{109.5^\circ}{2} \]\[ \alpha = \beta \approx 54.75^\circ \]
04

Rounding Off

Round the calculated angles to one decimal place as requested. Therefore, \( \gamma = 70.5^\circ \) and \( \alpha = \beta = 54.8^\circ \).

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Key Concepts

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

The Cosine Rule
The cosine rule is a vital tool in trigonometry used for calculating the angles of a triangle when the side lengths are known. It's especially helpful when dealing with non-right triangles, like the isosceles triangle in this exercise. The formula states:\[ c^2 = a^2 + b^2 - 2ab \cos(\gamma) \]Here, \( c \) is the side opposite the angle \( \gamma \), which you want to calculate. For triangles where you need to find an angle and you know all sides, the cosine rule is your best friend. In this particular problem, the sides \( a \) and \( b \) are equal. This helps simplify calculations since you can directly substitute these values into the equation to get \( \cos(\gamma) \). Once you have the value of \( \cos(\gamma) \), use a calculator to find \( \gamma \). Remember to convert back from cosines to angles, using either degrees or radians based on your calculator settings.
Understanding Isosceles Triangles
An isosceles triangle is one of the simplest types of triangles to recognize because it has two sides of equal length. In our example, both \( a \) and \( b \) are equal with the value \( 2/\sqrt{3} \). This equality also means that the two angles opposite these sides are equal.In any isosceles triangle:
  • Two sides are equal.
  • Two angles are equal (those opposite the equal sides).
Because of these properties, calculations become easier. If you know one angle, the other two are straightforward to find. For example, once you calculate \( \gamma \), use symmetry to know that \( \alpha \) and \( \beta \) are equal. In this exercise, determining even one angle gives you the complete angle picture of the triangle.
Calculating Angles in Triangles
Knowing how to calculate angles in a triangle is a fundamental trigonometric skill. Firstly, remember that the sum of all angles in any triangle is always \( 180^\circ \). This is a crucial fact because once you know two angles, you can easily find the third. In the context of this problem:
  • We initially calculated \( \gamma \) using the cosine rule.
  • Since it鈥檚 an isosceles triangle, \( \alpha = \beta \).
You use these relationships to find \( \alpha \) and \( \beta \) by subtracting \( \gamma \) from \( 180^\circ \) and dividing by 2:\[ \alpha = \beta = \frac{180^\circ - \gamma}{2} \]This simple step is essential for distributing the remaining degrees from \( 180^\circ \) evenly among the two equal angles, ensuring the internal logic of the triangle holds true.

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Most popular questions from this chapter

Find a unit vector having the same direction as the given vector. $$\langle-3,3\rangle$$

We study the dot product of two vectors. Given two vectors \(\mathbf{A}=\left\langle x_{1}, y_{1}\right\rangle\) and \(\mathbf{B}=\left\langle x_{2}, y_{2}\right\rangle,\) we define the dot product \(\mathbf{A} \cdot \mathbf{B}\) as follows: $$\mathbf{A} \cdot \mathbf{B}=x_{1} x_{2}+y_{1} y_{2}$$ For example, if \(\mathbf{A}=\langle 3,4\rangle\) and \(\mathbf{B}=\langle-2,5\rangle,\) then \(\mathbf{A} \cdot \mathbf{B}=(3)(-2)+(4)(5)=14 .\) Notice that the dot product of two vectors is a real number. For this reason, the dot product is also known as the scalar product. For Exercises \(61-63\) the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are defined as follows: $$\mathbf{u}=\langle-4,5\rangle \quad \mathbf{v}=\langle 3,4\rangle \quad \mathbf{w}=\langle 2,-5\rangle$$ (a) Compute \(\mathbf{v}+\mathbf{w}\) (b) Compute \(\mathbf{u} \cdot(\mathbf{v}+\mathbf{w})\) (c) Compute \(\mathbf{u} \cdot \mathbf{v}+\mathbf{u} \cdot \mathbf{w}\) (d) Show that for any three vectors \(\mathbf{A}, \mathbf{B},\) and \(\mathbf{C}\) we have \(\mathbf{A} \cdot(\mathbf{B}+\mathbf{C})=\mathbf{A} \cdot \mathbf{B}+\mathbf{A} \cdot \mathbf{C}\).

Compute the distance between the given points. (The coordinates are polar coordinates.) $$\left(2, \frac{2 \pi}{3}\right) \text { and }\left(4, \frac{\pi}{6}\right)$$

Show that the rectangular form of the equation $$r=\frac{a b}{(1-a \cos \theta)} \quad(a<1)$$is \(\left(1-a^{2}\right) x^{2}+y^{2}-2 a^{2} b x-a^{2} b^{2}=0\)

We study the dot product of two vectors. Given two vectors \(\mathbf{A}=\left\langle x_{1}, y_{1}\right\rangle\) and \(\mathbf{B}=\left\langle x_{2}, y_{2}\right\rangle,\) we define the dot product \(\mathbf{A} \cdot \mathbf{B}\) as follows: $$\mathbf{A} \cdot \mathbf{B}=x_{1} x_{2}+y_{1} y_{2}$$ For example, if \(\mathbf{A}=\langle 3,4\rangle\) and \(\mathbf{B}=\langle-2,5\rangle,\) then \(\mathbf{A} \cdot \mathbf{B}=(3)(-2)+(4)(5)=14 .\) Notice that the dot product of two vectors is a real number. For this reason, the dot product is also known as the scalar product. For Exercises \(61-63\) the vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are defined as follows: $$\mathbf{u}=\langle-4,5\rangle \quad \mathbf{v}=\langle 3,4\rangle \quad \mathbf{w}=\langle 2,-5\rangle$$ (a) Compute \(\mathbf{u} \cdot \mathbf{v}\) and \(\mathbf{v} \cdot \mathbf{u}\) (b) Compute \(\mathbf{v} \cdot \mathbf{w}\) and \(\mathbf{w} \cdot \mathbf{v}\) (c) Show that for any two vectors \(\mathbf{A}\) and \(\mathbf{B}\), we have \(\mathbf{A} \cdot \mathbf{B}=\mathbf{B} \cdot \mathbf{A} .\) That is, show that the dot product is commutative. Hint: Let \(\mathbf{A}=\left\langle x_{1}, y_{1}\right\rangle,\) and let \(\mathbf{B}=\left\langle x_{2}, y_{2}\right\rangle\)
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