Foundations of Right Triangle Ratios

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What does the mnemonic SOH CAH TOA define for acute angle $\theta$?

$$\text{SOH: } \sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$$ $$\text{CAH: } \cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$$ $$\text{TOA: } \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}}$$

What is the primary ratio defining the tangent of an angle $\theta$ in terms of sine and cosine?

$$\tan\theta = \frac{\sin\theta}{\cos\theta}$$

In a right triangle, the side opposite angle $A$ is $3$ and the hypotenuse is $5$. What is $\sin A$?

$$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{5}$$

In a right triangle, the side adjacent to angle $B$ is $12$ and the hypotenuse is $13$. What is $\cos B$?

$$\cos B = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{12}{13}$$

In a right triangle, the opposite leg is $7$ and the adjacent leg is $24$. What is $\tan\theta$?

$$\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{7}{24}$$

Why do the values of $\sin\theta$ and $\cos\theta$ in a right triangle always stay strictly between $0$ and $1$ for an acute angle $\theta$?

Because side lengths are strictly positive and the hypotenuse is always strictly longer than either individual leg, making the ratios $\frac{\text{Opposite}}{\text{Hypotenuse}}$ and $\frac{\text{Adjacent}}{\text{Hypotenuse}}$ proper fractions strictly less than $1$.

Can $\tan\theta$ be greater than $1$ for an acute angle in a right triangle?

Yes. If the side opposite $\theta$ is longer than the side adjacent to $\theta$, then $\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} > 1$. Specifically, $\tan\theta > 1$ when $\theta > 45^{\circ}$.

State the Cofunction Identity relating $\sin\theta$ and $\cos\theta$ for an acute angle $\theta$.

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

What fundamental identity connects $\sin\theta$ and $\cos\theta$ via the Pythagorean Theorem?

$$\sin^2\theta + \cos^2\theta = 1$$

If $\cos\theta = \frac{4}{5}$ and $\theta$ is acute, evaluate $\sin\theta$ using identities.

$$\sin^2\theta = 1 - \cos^2\theta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}$$ $$\sin\theta = \frac{3}{5}$$

How do trigonometric ratios change if all side lengths of a right triangle are doubled?

Trigonometric ratios do not change at all. By AA similarity, triangles with the same acute angles are similar, so corresponding side ratios remain invariant.

In $\triangle ABC$ with $\angle C = 90^{\circ}$, $a = 6$, $b = 8$. What is the hypotenuse $c$ and $\tan A$?

$$c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$$ $$\tan A = \frac{a}{b} = \frac{6}{8} = \frac{3}{4}$$

In a right triangle with an acute angle $\theta$, how are the three sides identified relative to $\theta$?

A right triangle labeled with an acute angle theta, showing the adjacent side along the base, the opposite side as the vertical leg, and the hypotenuse opposite the right angle.

1. Hypotenuse: The side opposite the $90^{\circ}$ right angle (always the longest side). 2. Opposite: The leg directly across from $\theta$. 3. Adjacent: The leg directly forming $\theta$ alongside the hypotenuse.

In right triangle $\triangle ABC$ where $\angle C = 90^{\circ}$, how does $\sin A$ relate to $\cos B$?

Right triangle ABC with right angle C, hypotenuse c, adjacent side b to angle A, and opposite side a to angle A, illustrating that the opposite side of angle A serves as the adjacent side to angle B.

$$\sin A = \cos B = \frac{\text{a}}{\text{c}} $$ Because $\angle A$ and $\angle B$ are complementary ($A + B = 90^{\circ}$), the leg opposite $A$ is the exact same side as the leg adjacent to $B$.

Foundations of Right Triangle Ratios

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