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Rule

Special Right Triangles

Right triangles with certain acute angle measures are considered noteworthy. These include the following angle relationships.

$3 0^{\circ}$ - $6 0^{\circ}$ - $9 0^{\circ}$
$4 5^{\circ}$ - $4 5^{\circ}$ - $9 0^{\circ}$
$6 0^{\circ}$ - $3 0^{\circ}$ - $9 0^{\circ}$

Although the first and third relationships contain the same angles, they are considered different because the reference angle in each is different — $3 0^{\circ}$ in the first and $6 0^{\circ}$ in the third. The sine, cosine, and tangents of these triangles can be useful when finding unknown side lengths.

Angle $θ$	$3 0^{\circ}$	$4 5^{\circ}$	$6 0^{\circ}$
$sin (θ)$	$\frac{1}{2}$	$\frac{2}{2}$	$\frac{3}{2}$
$cos (θ)$	$\frac{3}{2}$	$\frac{2}{2}$	$\frac{1}{2}$
$tan (θ)$	$\frac{3}{3}$	$1$	$3$

These special measures are justified below.

Derivation

Special Right Triangles 30-60-90

Consider the $3 0^{\circ}$ - $6 0^{\circ}$ - $9 0^{\circ}$ triangle. Suppose an equilateral triangle has side lengths of $1 .$

Bisecting the apex angle yields the following $3 0^{\circ}$ - $6 0^{\circ}$ - $9 0^{\circ}$ triangle.

The value of $h$ — the length of the third side — can be found using the Pythagorean Theorem. Here, $a = \frac{1}{2}, b = h,$ and $c = 1 .$

a^{2} + b^{2} = c^{2}

SubstituteValues

Substitute values

(\frac{1}{2})^{2} + h^{2} = 1^{2}

Calculate power

\frac{1}{4} + h^{2} = 1

$LHS - \frac{1}{4} = RHS - \frac{1}{4}$

h^{2} = \frac{3}{4}

$LHS = RHS$

h = \frac{3}{4}

$\frac{a}{b} = \frac{a}{b}$

h = \frac{3}{4}

Calculate root

h = \frac{3}{2}

Since, $h = \frac{3}{2},$ the $3 0^{\circ}$ - $6 0^{\circ}$ - $9 0^{\circ}$ triangle can be redrawn as follows.

Using $θ = 3 0^{\circ},$ the following relationships can be concluded.

sin (3 0^{\circ}) = \frac{1}{2} / 1 = \frac{1}{2} cos (3 0^{\circ}) = \frac{3}{2} / 1 = \frac{3}{2} tan (3 0^{\circ}) = \frac{1}{2} / \frac{3}{2} = \frac{3}{3}

Using the same triangle, the values for a $6 0^{\circ}$ - $3 0^{\circ}$ - $9 0^{\circ}$ triangle can be determined.

sin (6 0^{\circ}) = \frac{3}{2} / 1 = \frac{3}{2} cos (6 0^{\circ}) = \frac{1}{2} / 1 = \frac{1}{2} tan (6 0^{\circ}) = \frac{3}{2} / \frac{1}{2} = 3

Derivation

Special Right Triangles 45-45-90

Suppose an isosceles triangle has a hypotenuse of $1$ and base angles that measure $4 5^{\circ} .$

Because the triangle is isosceles, its legs have equal measure. Because they are unknown, $x$ can be used to represent them. The Pythagorean Theorem can be used to determine the value of $x .$

a^{2} + b^{2} = c^{2}

SubstituteExpressions

Substitute expressions

x^{2} + x^{2} = 1^{2}

Simplify power and terms

2 x^{2} = 1

$LHS / 2 = RHS / 2$

x^{2} = \frac{1}{2}

$LHS = RHS$

x = \pm \frac{1}{2}

$x > 0$

x = \frac{1}{2}

$\frac{a}{b} = \frac{a}{b}$

x = \frac{1}{2}

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

x = \frac{2}{2}

The $4 5^{\circ}$ - $4 5^{\circ}$ - $4 5^{\circ}$ triangle can be redrawn as follows.

Using $θ = 4 5^{\circ},$ the following relationships can be concluded.

sin (4 5^{\circ}) = \frac{2}{2} / 1 = \frac{2}{2} cos (4 5^{\circ}) = \frac{2}{2} / 1 = \frac{2}{2} tan (4 5^{\circ}) = \frac{2}{2} / \frac{2}{2} = 1

As was shown above, because the hypotenuse of these special right triangles is $1,$

sin (θ) = opp and cos (θ) = adj, for θ = 3 0^{\circ}, 4 5^{\circ}, 6 0^{\circ} .

Therefore, these values can be used to determine unknown side lengths.

Exercises

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