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Magazine Chapter 10: Problem 4
Find, to the nearest degree, the measurements of the interior angles of the triangle having vertices at \((1,0),(-3,2)\), and \((2,3)\)
Short Answer
Expert verified
The measures of the interior angles are approximately 42掳, 101掳, and 37掳.
Step by step solution
01
- Find the lengths of the triangle's sides
Use the distance formula to find the lengths of each side. The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]. Calculate the distances for each pair of vertices: \[ AB, BC, AC \].
02
- Calculate side AB
For vertices \( A(1,0) \) and \( B(-3,2) \), \[ AB = \sqrt{(-3 - 1)^2 + (2 - 0)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \].
03
- Calculate side BC
For vertices \( B(-3,2) \) and \( C(2,3) \), \[ BC = \sqrt{(2 + 3)^2 + (3 - 2)^2} = \sqrt{25 + 1} = \sqrt{26} \].
04
- Calculate side AC
For vertices \( A(1,0) \) and \( C(2,3) \), \[ AC = \sqrt{(2 - 1)^2 + (3 - 0)^2} = \sqrt{1 + 9} = \sqrt{10} \].
05
- Use the Law of Cosines to find the angles
The Law of Cosines states that for a triangle with sides \(a, b, c\), and corresponding angles \(\alpha, \beta, \gamma\): \[ \cos(\alpha) = \frac{b^2 + c^2 - a^2}{2bc} \]. Use this to find angles \(\alpha, \beta, \gamma\).
06
- Calculate Angle A
Using \( AC \) as \( a \), \( AB \) as \( b \), and \( BC \) as \( c \), angle \( A \) is \[ \cos(\alpha) = \frac{(2\sqrt{5})^2 + (\sqrt{26})^2 - (\sqrt{10})^2}{2 \cdot 2\sqrt{5} \cdot \sqrt{26}} = \frac{20 + 26 - 10}{2 \times 2\sqrt{5} \times \sqrt{26}} = \frac{36}{4\sqrt{130}} = \frac{9}{\sqrt{130}} \. \] Calculate \( \alpha \): \[ \alpha = \cos^{-1}\left(\frac{9}{\sqrt{130}}\right) \].
07
- Calculate Angle B
Using \( AB \) as \( c \), \( BC \) as \( a \), and \( AC \) as \( b \), angle \( B \) is \[ \cos(\beta) = \frac{10 + 26 - 20}{2 \cdot \sqrt{26} \cdot \sqrt{10}} = \frac{16}{2\sqrt{260}} = \frac{8}{\sqrt{260}}, \] Calculate \( \beta \): \[ \beta = \cos^{-1}\left(\frac{8}{\sqrt{260}}\right) \].
08
- Calculate Angle C
Using \( AC \) as \( c \), \( BC \) as \( b \), and \( AB \) as \( a \), angle \( C \) is \[ \cos(\gamma) = \frac{20 + 10 - 26}{2 \cdot 2\sqrt{5} \cdot \sqrt{10}} = \frac{4}{4\sqrt{50}} = \frac{1}{\sqrt{50}}, \] Calculate \( \gamma \): \[ \gamma = \cos^{-1}\left(\frac{1}{\sqrt{50}}\right) \].
09
- Find the nearest degree
Round each angle to the nearest degree using a calculator.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
distance formula
The **distance formula** is a cornerstone in **triangle geometry**. It helps in calculating the length of a side between two points in a coordinate plane. The formula is:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] With this formula, we determine each side of the triangle. For example:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] With this formula, we determine each side of the triangle. For example:
- For points \(A(1,0)\) and \(B(-3,2)\):
\[ AB = \sqrt{(-3 - 1)^2 + (2 - 0)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \] - For points \(B(-3,2)\) and \(C(2,3)\):
\[ BC = \sqrt{(2 + 3)^2 + (3 - 2)^2} = \sqrt{25 + 1} = \sqrt{26} \] - For points \(A(1,0)\) and \(C(2,3)\):
\[ AC = \sqrt{(2 - 1)^2 + (3 - 0)^2} = \sqrt{1 + 9} = \sqrt{10} \] These calculations give the triangle's side lengths, which are essential for finding the angles.
law of cosines
The **Law of Cosines** helps in finding the angles of a triangle when all sides are known. The formula is:
\[ \cos(\alpha) = \frac{b^2 + c^2 - a^2}{2bc} \] For angle calculations:
\[ \cos(\alpha) = \frac{b^2 + c^2 - a^2}{2bc} \] For angle calculations:
- Angle \(A\) Calculation:
Using sides \(AC\), \(AB\), and \(BC\):
\[ \cos(\alpha) = \frac{(2\sqrt{5})^2 + (\sqrt{26})^2 - (\sqrt{10})^2}{2 \cdot 2\sqrt{5} \cdot \sqrt{26}} = \frac{36}{4\sqrt{130}} = \frac{9}{\sqrt{130}} \] Then, solve for \(\alpha\) using inverse cosine. - Angle \(B\) Calculation:
Using sides \(AB\), \(BC\), and \(AC\):
\[ \cos(\beta) = \frac{10 + 26 - 20}{2 \cdot \sqrt{26} \cdot \sqrt{10}} = \frac{8}{\sqrt{260}} \] Then, solve for \(\beta\) using inverse cosine. - Angle \(C\) Calculation:
Using sides \(AC\), \(BC\), and \(AB\):
\[ \cos(\gamma) = \frac{20 + 10 - 26}{2 \cdot 2\sqrt{5} \cdot \sqrt{10}} = \frac{1}{\sqrt{50}} \] Then, solve for \(\gamma\) using inverse cosine.
triangle geometry
**Triangle geometry** involves understanding properties like side lengths and angles. In our exercise, we have:
- **Vertices**: \((1,0), (-3,2), (2,3)\)
- **Sides**: Are calculated using the distance formula as **\( AB, BC, and AC \)**
- The sum of interior angles of a triangle is always **180 degrees**
- Angles can be calculated using laws like the **Law of Cosines** or **Law of Sines**
interior angles calculation
The **interior angles** of a triangle are the angles **inside** it. To calculate these:
- Use side lengths obtained through the distance formula
- Apply the Law of Cosines to find each angle
- By calculating: \(\alpha, \beta, \gamma\), first find their cosines
- Convert these to angles using an inverse cosine function
