3.7: Special Triangles
- Page ID
- 190950
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Definitions and Theorems
A right triangle in which one angle is \( 30^{ \circ } \) and another angle is \( 60^{ \circ } \) is called a \( 30^{ \circ } \)-\( 60^{ \circ } \)-\( 90^{ \circ } \) triangle.
In any right triangle in which the two acute angles are \( 30^{ \circ } \) and \( 60^{ \circ } \), the hypotenuse is always twice the length of the shortest side (the side opposite the \( 30^{ \circ } \) angle), and the remaining side (opposite the \( 60^{ \circ } \) angle) is always \( \sqrt{3} \) times the shortest side.
- Proof
-
Consider an equilateral triangle with side lengths \( 2a \), as shown in the figure below.
Since this triangle is equilateral, each angle is \( 60^{ \circ } \). Drawing an altitude divides the triangle into two smaller triangles. Since the altitude is perpendicular to the base, each of these smaller triangles is a right triangle. Moreover, since one angle in each of these smaller right triangles is \( 60^{ \circ } \), the remaining angle must be \( 30^{ \circ } \). Furthermore, the side opposite the \( 30^{ \circ } \) angle must have length \( a \) (in the original equilateral triangle, the side opposite \( 60^{ \circ } \) had length \( 2a \), so it should make sense that the side opposite \( 30^{ \circ } \) is half of this). We now consider one of these smaller right triangles (see the figure below).
The missing side length, which we will call \( x \) for now, can be found using the Pythagorean Theorem.\[\begin{array}{rrcl}
& x^2 + a^2 & = & (2a)^2 \\
\implies & x^2 + a^2 & = & 4a^2 \\
\implies & x^2 & = & 3a^2 \\
\implies & x & = & a \sqrt{3} \\
\end{array} \nonumber \]
Since an isosceles right triangle has two angles that are \( 45^{ \circ } \), it is commonly called a \( 45^{ \circ } \)-\( 45^{ \circ } \)-\( 90^{ \circ } \) triangle.
In any right triangle in which the two acute angles are \( 45^{ \circ } \), the hypotenuse is always \( \sqrt{2} \) times the side length.
- Proof
-
The sides opposite the \( 45^{ \circ } \) angles in a \( 45^{ \circ } \)-\( 45^{ \circ } \)-\( 90^{ \circ } \) triangle must have the same length. Let \( a \) be the lengths of these sides and \( x \) be the length of the hypotenuse, as shown in the figure below.
Since this is a right triangle, we can use the Pythagorean Theorem to find the length of the hypotenuse in terms of \( a \).\[\begin{array}{rrcl}
& a^2 + a^2 & = & x^2 \\
\implies & 2 a^2 & = & x^2 \\
\implies & a \sqrt{2} & = & x \\
\end{array} \nonumber \]