Trigonometric Ratios of Acute Angles SAT Notes

Trigonometry connects an acute angle in a right triangle to ratios of side lengths. The triangle may change size, but every similar triangle with that angle produces the same ratios.

Name sides relative to the chosen angle

Name the sides relative to the selected angleA right triangle with base b, height a, hypotenuse c, and selected angle θ.

The six trigonometric ratios

Six right-triangle ratios for an acute angle
Function	Geometric ratio	Reciprocal partner
$\sin\theta$	$\frac{\text{opposite}}{\text{hypotenuse}}$	$\csc\theta=1/\sin\theta$
$\cos\theta$	$\frac{\text{adjacent}}{\text{hypotenuse}}$	$\sec\theta=1/\cos\theta$
$\tan\theta$	$\frac{\text{opposite}}{\text{adjacent}}$	$\cot\theta=1/\tan\theta$
$\csc\theta$	$\frac{\text{hypotenuse}}{\text{opposite}}$	Reciprocal of sine
$\sec\theta$	$\frac{\text{hypotenuse}}{\text{adjacent}}$	Reciprocal of cosine
$\cot\theta$	$\frac{\text{adjacent}}{\text{opposite}}$	Reciprocal of tangent

Find a missing side first

Find the missing leg before forming ratiosA right triangle with base 12, height ?, hypotenuse 13, and selected angle θ.

Worked example

Recover a leg and evaluate all six ratios

A right triangle has adjacent leg $$12$$ , hypotenuse $$13$$ , and acute angle $\theta$ at the left. Find the six ratios.

Missing leg
Use $$a^2+b^2=c^2$$ : $$12^2+b^2=13^2$$ , so $$b=5$$ .
Primary ratios
$\sin\theta=5/13$ , $\cos\theta=12/13$ , and $\tan\theta=5/12$ .
Reciprocals
Invert the paired ratios: $\csc\theta=13/5$ , $\sec\theta=13/12$ , and $\cot\theta=12/5$ .

The ratios are

\(5/13,12/13,5/12,13/5,13/12,12/5\)

in sine-cosine-tangent-cosecant-secant-cotangent order.

Complementary angles and cofunctions

Complementary acute angles share side ratiosA right triangle with base adjacent to θ, height opposite θ, hypotenuse hypotenuse, and selected angle θ.

Cofunction relationships

\[\sin\theta=\cos(90^\circ-\theta),\quad \cos\theta=\sin(90^\circ-\theta)\]

The same legs exchange opposite and adjacent roles at the two complementary acute angles.

Complementary pairs

Sine and cosine

$\sin\theta=\cos(90^\circ-\theta)$ .

Tangent and cotangent

$\tan\theta=\cot(90^\circ-\theta)$ .

Secant and cosecant

$\sec\theta=\csc(90^\circ-\theta)$ .

Use identities to move between ratios

Two central identities

\[\tan\theta=\frac{\sin\theta}{\cos\theta},\qquad \sin^2\theta+\cos^2\theta=1\]

For an acute angle, sine and cosine are positive, so take the positive square root when recovering one from the other.

Worked example

Find every ratio from one ratio

For acute $\theta$ , $\sin\theta=7/25$ . Find $\cos\theta$ and $\tan\theta$ .

Model a triangle
Use opposite $$7$$ and hypotenuse $$25$$ .
Find adjacent
$$a^2=25^2-7^2=576$$ , so $$a=24$$ .
Form ratios
$\cos\theta=24/25$ and $\tan\theta=7/24$ .

\cos\theta=24/25

and

\tan\theta=7/24

Create right triangles when none are shown

An altitude splits an isosceles triangle into right trianglesA right triangle with base 5, height 12, hypotenuse 13, and selected angle θ.

Mini check

Switch the selected angle

In a $$9$$ - $$40$$ - $$41$$ right triangle, the angle opposite $$9$$ is $\alpha$ . What is $\cos(90^\circ-\alpha)$ ?

$$9/41$$
$$40/41$$
$$9/40$$
$$41/9$$

Show answer and explanation

Answer: $$9/41$$

By the cofunction relationship, $\cos(90^\circ-\alpha)=\sin\alpha=9/41$ .

Key takeaways

Acute-ratio checklist

Mark the right angle and selected acute angle before naming opposite and adjacent.
Find a missing side with the Pythagorean theorem before forming ratios.
Pair sine-cosecant, cosine-secant, and tangent-cotangent as reciprocals.
Use cofunctions for complementary angles and identities when only one ratio is given.
Keep values exact unless a decimal is explicitly requested.

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