Difference between revisions of "30-60-90 triangle"

Revision as of 14:51, 4 January 2025

A 30-60-90 triangle is a right triangle with a 30 degree angle, a 60 degree angle,a 90 degree angle. It is special because it side lengths are always in the same ratio. The length of the hypotenuse is twice the length of the shorter leg and the length of the longer leg is $\sqrt{3}$ the length of the shorter leg.

Simply put, the ratio in order of the sides opposite to the angles for any value of $y$ is $y$ : $y$ $\sqrt{3}$ : $2y$ . It is a simple "pneumonic" ratio that will help you remember the ratio and each part corresponds to the part of the 30-60-90 triangle. (For example: the x $/sqrt{3}$ corresponds to the $60°$ angle)

Proofs

Consider a right triangle with angles $30^\circ$, $60^\circ$, and $90^\circ$. Let:

$a$: side opposite the $30^\circ$ angle
$b$: side opposite the $60^\circ$ angle
$c$: hypotenuse

Using trigonometric identities

1. Hypotenuse $AC$

$\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{AC}$

And:

$\sin(30^\circ) = \frac{1}{2} = \frac{AB}{AC}$

$\boxed{\frac{{AB}}{{AC}} = \frac{1}{2} \text{ or } \bf{2AB=AC}}$

2. Side $AB$

$\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BC}{AC}$

Since $\cos(30^\circ) = \frac{\sqrt{3}}{2}$:

$\frac{\sqrt{3}}{2} = \frac{BC}{AC} \implies BC = AC \cdot \frac{\sqrt{3}}{2} \implies \boxed{\bf{BC = AB \cdot \sqrt{3}}}$

In conclusion, the side lengths are:

\begin{array}{l} \text{Opposite } 30^\circ: AB \\ \text{Opposite } 60^\circ: AB\sqrt{3} \\ \text{Hypotenuse } : 2AB \end{array}

Therefore, the side ratios are $\boxed{1:\sqrt{3}:2}$

This article is a stub. Help us out by expanding it.

@@ Line 7: / Line 7: @@
-==Proof of the 30-60-90 Triangle Side Ratios Using Trigonometry==
+==Proofs==
 Consider a right triangle with angles \(30^\circ\), \(60^\circ\), and \(90^\circ\). Let:
-- \(a\): side opposite the \(30^\circ\) angle
+* \(a\): side opposite the \(30^\circ\) angle
-- \(b\): side opposite the \(60^\circ\) angle
+* \(b\): side opposite the \(60^\circ\) angle
-- \(c\): hypotenuse
+* \(c\): hypotenuse
 ===Using trigonometric identities===
@@ Line 18: / Line 18: @@
 ==== 1. Hypotenuse <math>AC</math> ====
-\(\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{<math>BC</math>}{<math>AC</math>}\)
+\(\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{AC}\)
-Since <math>sin(30^\circ)</math>= <math>1/2</math>:
+And:
-<math>\boxed{\text{AB/AC} = 1/2  \text{ or } \textbf{2AB=AC}}</math>
+\(\sin(30^\circ) = \frac{1}{2} = \frac{AB}{AC}\)
+<math>\boxed{\frac{{AB}}{{AC}} = \frac{1}{2}  \text{ or } \bf{2AB=AC}}</math>
 ==== 2. Side <math> AB </math> ====
-\(\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{b}{c}\)
+\(\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BC}{AC}\)
 Since \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\):
-\(\frac{\sqrt{3}}{2} = \frac{b}{c} \implies b = c \cdot \frac{\sqrt{3}}{2} = 2a \cdot \frac{\sqrt{3}}{2} = a\sqrt{3}\)
+<math>\frac{\sqrt{3}}{2} = \frac{BC}{AC} \implies BC = AC \cdot \frac{\sqrt{3}}{2} \implies \boxed{\bf{BC = AB \cdot \sqrt{3}}}</math>
+In conclusion, the side lengths are:
+\begin{array}{l}
+\text{Opposite } 30^\circ: AB \\
+\text{Opposite } 60^\circ: AB\sqrt{3} \\
+\text{Hypotenuse }    : 2AB
+\end{array}
+Therefore, the side ratios are <math>\boxed{1:\sqrt{3}:2}</math>
-Thus, the side lengths are:
-- Opposite \(30^\circ\): \(a\)
-- Opposite \(60^\circ\): \(a\sqrt{3}\)
-- Hypotenuse: \(2a\)
-Therefore, the side ratios are \(1 : \sqrt{3} : 2\).
 {{stub}}

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