Trigonometric Functions on Right Triangles: Videos & Practice Problems

Trigonometric functions, including sine, cosine, and tangent, relate angles to side lengths in right triangles. The sine of an angle is the ratio of the opposite side to the hypotenuse, while cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. Special triangles, like the 45-45-90 and 30-60-90 triangles, have specific ratios that simplify calculations. For example, in a 45-45-90 triangle, both legs are equal, and the hypotenuse is the leg length multiplied by the square root of 2. Understanding these relationships aids in solving trigonometric problems efficiently.

Introduction to Trigonometric Functions

Trigonometric functions, commonly referred to as trig functions, are essential in understanding the relationships between angles and side lengths in right triangles. The three primary trig functions are sine, cosine, and tangent, each defined as specific ratios of the sides of a right triangle. These functions can be remembered using the mnemonic SOHCAHTOA, which stands for:

• Sine (sin): The ratio of the length of the side opposite the angle to the length of the hypotenuse. This can be expressed as: \[\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\]
• Cosine (cos): The ratio of the length of the adjacent side to the length of the hypotenuse. This is represented as: \[\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\]
• Tangent (tan): The ratio of the length of the opposite side to the length of the adjacent side, given by: \[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\]

For example, if we have a right triangle where the side opposite angle θ measures 3 units and the hypotenuse measures 5 units, the sine of θ would be calculated as:

\[\sin(\theta) = \frac{3}{5}\]

To find the cosine of θ, we would look at the adjacent side, which measures 4 units, leading to:

\[\cos(\theta) = \frac{4}{5}\]

For the tangent of θ, we take the opposite side (3 units) and divide it by the adjacent side (4 units):

\[\tan(\theta) = \frac{3}{4}\]

Additionally, it is important to note that the tangent function can also be expressed in terms of sine and cosine:

\[\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\]

This relationship can be verified by substituting the values of sine and cosine derived earlier. Understanding these functions and their relationships is crucial for solving problems involving right triangles.

When solving for trig functions, always ensure to reference the correct angle. For instance, if calculating the cosine of angle y in a triangle, identify the adjacent side relative to angle y, not angle x. This attention to detail is vital for accurate calculations.

In summary, mastering the definitions and applications of sine, cosine, and tangent through the SOHCAHTOA mnemonic will greatly enhance your ability to solve trigonometric problems effectively.

Given the right triangle below, evaluate $\cos\theta$.

• A. $\cos\theta=\frac{8}{17}$
• B. $\cos\theta=\frac{8}{15}$
• C. $\cos\theta=\frac{15}{17}$
• D. $\cos\theta=\frac{15}{8}$

Answer: (C)

Given the right triangle below, evaluate $\tan\theta$.

• A. $\tan\theta=\frac35$
• B. $\tan\theta=\frac45$
• C. $\tan\theta=\frac43$
• D. $\tan\theta=\frac34$

Answer: (D)

Fundamental Trigonometric Identities

In trigonometry, we expand our understanding by introducing three new functions: cosecant, secant, and cotangent. These functions are closely related to the primary trigonometric functions—sine, cosine, and tangent—through what are known as reciprocal identities. Essentially, each of these new functions is the reciprocal of one of the original functions.

The cosecant function is defined as the reciprocal of the sine function. Mathematically, this can be expressed as:

$$ \text{csc}(\theta) = \frac{1}{\sin(\theta)} $$

In terms of a right triangle, if the sine of an angle θ is represented as the ratio of the length of the opposite side to the hypotenuse, then the cosecant is the ratio of the hypotenuse to the opposite side. For example, if the hypotenuse is 5 and the opposite side is 3, then:

$$ \text{csc}(\theta) = \frac{5}{3} $$

Similarly, the secant function is the reciprocal of the cosine function:

$$ \text{sec}(\theta) = \frac{1}{\cos(\theta)} $$

Here, if the cosine of θ is the ratio of the adjacent side to the hypotenuse, then the secant is the ratio of the hypotenuse to the adjacent side. For instance, if the hypotenuse is 5 and the adjacent side is 4, then:

$$ \text{sec}(\theta) = \frac{5}{4} $$

Lastly, the cotangent function is the reciprocal of the tangent function:

$$ \text{cot}(\theta) = \frac{1}{\tan(\theta)} $$

In a right triangle, if the tangent of θ is the ratio of the opposite side to the adjacent side, then the cotangent is the ratio of the adjacent side to the opposite side. For example, if the opposite side is 3 and the adjacent side is 4, then:

$$ \text{cot}(\theta) = \frac{4}{3} $$

These reciprocal identities can also be expressed in terms of the original functions. For instance, since tangent is defined as:

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

It follows that:

$$ \text{cot}(\theta) = \frac{\cos(\theta)}{\sin(\theta)} $$

To apply these concepts, consider a right triangle where you need to find the secant of angle x. If the adjacent side is 5 and the hypotenuse is 13, the cosine of x is:

$$ \cos(x) = \frac{5}{13} $$

Thus, the secant of x is:

$$ \text{sec}(x) = \frac{13}{5} $$

For the cosecant of angle x, if the opposite side is 12, then the sine of x is:

$$ \sin(x) = \frac{12}{13} $$

Therefore, the cosecant of x is:

$$ \text{csc}(x) = \frac{13}{12} $$

When finding the cotangent of angle y, if the opposite side is 5 and the adjacent side is 12, the tangent of y is:

$$ \tan(y) = \frac{5}{12} $$

Thus, the cotangent of y is:

$$ \text{cot}(y) = \frac{12}{5} $$

Understanding these reciprocal identities is crucial for solving various trigonometric problems and reinforces the interconnectedness of trigonometric functions.

If $\tan\theta=\frac{12}{5}$, find the values of the five other trigonometric functions. Rationalize the denominators if necessary.

• A. $\sin\theta=\frac{12}{13},\cos\theta=\frac{5}{13},\cot\theta=\frac{5}{12},\sec\theta=\frac{13}{5},\csc\theta=\frac{13}{12}$
• B. $\sin\theta=\frac{5}{13},\cos\theta=\frac{12}{13},\cot\theta=\frac{5}{12},\sec\theta=\frac{13}{12},\csc\theta=\frac{13}{5}$
• C. $\sin\theta=\frac{12}{13},\cos\theta=\frac{5}{13},\cot\theta=-\frac{5}{12},\sec\theta=-\frac{13}{5},\csc\theta=-\frac{13}{12}$
• D. $\sin\theta=\frac{5}{13},\cos\theta=\frac{12}{13},\cot\theta=-\frac{5}{12},\sec\theta=-\frac{13}{12},\csc\theta=-\frac{13}{5}$

Answer: (A)

If $\sin\theta=\frac{\sqrt{17}}{17}$, find the values of the five other trigonometric functions. Rationalize the denominators if necessary.

• A. $\cos\theta=\frac{\sqrt{17}}{4},\tan\theta=\frac14,\cot\theta=4,\sec\theta=\sqrt{17},\csc\theta=\frac{4\sqrt{17}}{17}$
• B. $\cos\theta=\frac{\sqrt{17}}{4},\tan\theta=-\frac14,\cot\theta=-4,\sec\theta=\sqrt{17},\csc\theta=\frac{4\sqrt{17}}{17}$
• C. $\cos\theta=\frac{4\sqrt{17}}{17},\tan\theta=-\frac14,\cot\theta=-4,\sec\theta=\frac{\sqrt{17}}{4},\csc\theta=\sqrt{17}$
• D. $\cos\theta=\frac{4\sqrt{17}}{17},\tan\theta=\frac14,\cot\theta=4,\sec\theta=\frac{\sqrt{17}}{4},\csc\theta=\sqrt{17}$

Answer: (D)

How to Use a Calculator for Trig Functions

Understanding how to evaluate trigonometric functions using a calculator is essential for solving various mathematical problems. The primary trigonometric functions include sine, cosine, and tangent, which can be accessed through specific buttons on your calculator. Before performing calculations, it is crucial to ensure that your calculator is set to the correct mode: either degrees or radians. This choice depends on the problem at hand.

For example, when calculating the sine of 37 degrees, you must switch your calculator to degree mode. You would input this as \(\sin(37^\circ)\), and upon pressing enter, you would find that \(\sin(37^\circ) \approx 0.6\) when rounded to the nearest tenth.

In contrast, when evaluating the tangent of \(\frac{2\pi}{15}\), you need to set your calculator to radian mode. The input would be \(\tan\left(\frac{2\pi}{15}\right)\), yielding an approximate value of 0.4 when rounded to the nearest tenth.

For functions like secant, which does not have a dedicated button, you can use the reciprocal identity. Since secant is defined as \(\sec(x) = \frac{1}{\cos(x)}\), to find \(\sec(50^\circ)\), you would calculate \(\frac{1}{\cos(50^\circ)}\). This results in approximately 1.6 when rounded to the nearest tenth.

When dealing with inverse trigonometric functions, such as arctangent, you can access this function by pressing the second button on your calculator followed by the tangent button. For instance, to find \(\tan^{-1}\left(\frac{3}{4}\right)\), you would input \(\tan^{-1}\left(\frac{3}{4}\right)\) in degree mode, resulting in approximately 36.9^\circ when rounded to the nearest tenth.

By mastering these steps and understanding the relationship between trigonometric functions and their inverses, you can effectively utilize your calculator to solve a wide range of problems in trigonometry.

What is a positive value of A in the interval $\left[0\degree,90\degree\right)$ that will make the following statement true? Express the answer in four decimal places.
$\sin A=0.9235$

• A. $22.5568°$
• B. $67.4432°$
• C. $22.4432°$
• D. $33.5438°$

Answer: (B)

What is the positive value of P in the interval $\left[0\degree,90\degree\right)$ that will make the following statement true? Express the answer in four decimal places.
$\cot P=5.2371$

• A. $55.8102°$
• B. $34.1898°$
• C. $10.8102°$
• D. $79.1898°$

Answer: (C)

What is the positive value of $D$ in the interval $\left[0,\frac{\pi}{2}\right)$ that will make the following statement true? Express the answer in four decimal places.
$\sec D=3.2842$

• A. 0.3094 rad
• B. 1.2614 rad
• C. 0.4760 rad
• D. 1.0934 rad

Answer: (B)

How to Use a Calculator for Trig Functions Example 1

In this example, we are tasked with finding the missing angle, denoted as θ, in a right triangle using inverse trigonometric functions, specifically the sine function. To begin, it is essential to ensure that your calculator is set to degree mode, as the final answer needs to be expressed in degrees.

We start by recalling the relationship defined by the sine function, which is given by:

sin(θ) = opposite / hypotenuse

In this case, the opposite side measures 6, and the hypotenuse measures 13. Thus, we can express this as:

sin(θ) = 6/13

To solve for θ, we apply the inverse sine function:

θ = sin^-1(6/13)

Using your calculator, press the second function key followed by the sine key to access the inverse sine function. Input the ratio 6/13 and close the parenthesis before pressing enter. This will yield the approximate value of θ.

After performing the calculation, you should find that:

θ ≈ 27.49°

This result indicates that the missing angle in the right triangle is approximately 27.49 degrees, rounded to two decimal places. Understanding how to utilize inverse trigonometric functions is crucial for solving problems involving angles in right triangles.

45-45-90 Triangles

In the study of right triangles, the special 45-45-90 triangle is particularly significant due to its unique properties and the shortcuts it offers for solving problems. In a 45-45-90 triangle, the two legs are always of equal length, which simplifies calculations significantly. When you need to find the hypotenuse, you can simply multiply the length of one leg by the square root of 2. This relationship can be expressed mathematically as:

c = a \cdot \sqrt{2}

where c is the hypotenuse and a is the length of either leg. For example, if each leg measures 5 units, the hypotenuse can be calculated as:

c = 5 \cdot \sqrt{2} = \sqrt{50}

This shortcut allows for quick solutions without the need for the Pythagorean theorem, which states that in any right triangle:

a² + b² = c²

In the case of a 45-45-90 triangle, since both legs are equal (let's say both are 5), the equation becomes:

5² + 5² = c²

which simplifies to:

25 + 25 = c²

50 = c²

c = \sqrt{50} = 5\sqrt{2}

Both methods yield the same result, but the shortcut is much more efficient. To further illustrate this, consider an example where the hypotenuse is given. If the hypotenuse measures 13, you can set up the equation:

13 = a \cdot \sqrt{2}

To find the length of the leg a, divide both sides by the square root of 2:

a = \frac{13}{\sqrt{2}}

To rationalize the denominator, multiply the numerator and denominator by the square root of 2:

a = \frac{13\sqrt{2}}{2}

Thus, both legs of the triangle will measure 13√2/2. This consistent relationship between the legs and the hypotenuse in a 45-45-90 triangle makes it a valuable tool in geometry, allowing for quick and accurate calculations in various applications.

Given the triangle below, determine the missing side(s) without using the Pythagorean theorem (make sure your answer is fully simplified).

• A. $x=81$
• B. $x=9\sqrt2$
• C. $x=2\sqrt9$
• D. $x=\sqrt{162}$

Answer: (B)

Without using a calculator, determine all values of P in the interval $\left[0\degree,90\degree\right)$ with the following trigonometric function value.
$\csc P=\sqrt2$

• A. $P=30\degree$ only
• B. $P=45°$ only
• C. $P=60°$ only
• D. $P=30°,60°$

Answer: (B)

Common Trig Functions For 45-45-90 Triangles

In the study of special right triangles, particularly the 45-45-90 triangle, we can derive important relationships involving trigonometric functions. This triangle is characterized by its angles of 45 degrees, which leads to specific ratios for the sides. The sides opposite the 45-degree angles are equal, and the hypotenuse is \(5\sqrt{2}\) if the legs are each 5 units long.

To understand the sine function, we use the formula:

\( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)

For a 45-degree angle, the opposite side measures 5, and the hypotenuse measures \(5\sqrt{2}\). Thus, we have:

\( \sin(45^\circ) = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} \)

Rationalizing the denominator gives:

\( \sin(45^\circ) = \frac{\sqrt{2}}{2} \)

Similarly, for the cosine function, we apply the same principle:

\( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \)

Using the adjacent side of 5, we find:

\( \cos(45^\circ) = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} \)

Again, rationalizing yields:

\( \cos(45^\circ) = \frac{\sqrt{2}}{2} \)

For the tangent function, we use:

\( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \)

With both sides measuring 5, we have:

\( \tan(45^\circ) = \frac{5}{5} = 1 \)

Next, we explore the reciprocal identities. The cosecant is defined as:

\( \csc(\theta) = \frac{1}{\sin(\theta)} \)

Thus:

\( \csc(45^\circ) = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} \)

Rationalizing gives:

\( \csc(45^\circ) = \sqrt{2} \)

For the secant, we have:

\( \sec(\theta) = \frac{1}{\cos(\theta)} \)

Thus:

\( \sec(45^\circ) = \sqrt{2} \)

Finally, the cotangent is:

\( \cot(\theta) = \frac{1}{\tan(\theta)} \)

Since \( \tan(45^\circ) = 1 \), we find:

\( \cot(45^\circ) = 1 \)

In summary, for the 45-45-90 triangle, the sine and cosine values are both \( \frac{\sqrt{2}}{2} \), while the cosecant and secant are \( \sqrt{2} \), and both the tangent and cotangent equal 1. Recognizing these patterns simplifies the process of solving trigonometric problems involving this special triangle.

30-60-90 Triangles

In geometry, the 30-60-90 triangle is a special type of right triangle that has specific relationships between its side lengths. Understanding these relationships is crucial as they frequently appear in various mathematical contexts, including physics and other sciences.

In a 30-60-90 triangle, the side lengths can be determined based on the shortest leg, which is opposite the 30-degree angle. The relationships are as follows:

• The hypotenuse (the longest side) is twice the length of the shortest leg. This can be expressed mathematically as:
$$\text{Hypotenuse} = 2 \times \text{Short Leg}$$
• The long leg (the side opposite the 60-degree angle) is equal to the shortest leg multiplied by the square root of three. This relationship can be written as:
$$\text{Long Leg} = \text{Short Leg} \times \sqrt{3}$$

For example, if the shortest leg of a 30-60-90 triangle measures 5 units, the hypotenuse would be:

$$\text{Hypotenuse} = 2 \times 5 = 10 \text{ units}$$

And the long leg would be:

$$\text{Long Leg} = 5 \times \sqrt{3} \text{ units}$$

To solve for unknown sides in a 30-60-90 triangle, it is essential to first identify the known sides and angles. For instance, if the hypotenuse is given as 8 units, you can find the shortest leg by rearranging the hypotenuse relationship:

$$\text{Short Leg} = \frac{\text{Hypotenuse}}{2} = \frac{8}{2} = 4 \text{ units}$$

Then, using the relationship for the long leg:

$$\text{Long Leg} = 4 \times \sqrt{3} \text{ units}$$

In another scenario, if the long leg is known to be 1 unit, you can find the short leg by rearranging the long leg relationship:

$$\text{Short Leg} = \frac{\text{Long Leg}}{\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \text{ units (after rationalizing the denominator)}$$

Finally, to find the hypotenuse, you would use:

$$\text{Hypotenuse} = 2 \times \text{Short Leg} = 2 \times \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3} \text{ units}$$

These relationships and methods provide a systematic approach to solving problems involving 30-60-90 triangles, making it easier to find missing side lengths efficiently.

Given the triangle below, determine the missing side(s) without using the Pythagorean theorem (make sure your answer is fully simplified).

• A. $x=10,y=5\sqrt5$
• B. $x=3,y=4$
• C. $x=5\sqrt3,y=10$
• D. $x=5,y=5\sqrt2$

Answer: (C)

Without using a calculator, determine all values of A in the interval $\left[0,\frac{\pi}{2}\right)$ with the following trigonometric function value.
$\cos A=\frac{\sqrt3}{2}$

• A. $0$ only
• B. $\frac{\pi}{4}$ only
• C. $\frac{\pi}{6}$ only
• D. $\frac{\pi}{3}$ only

Answer: (C)

Common Trig Functions For 30-60-90 Triangles

In a 30-60-90 triangle, the relationships between the angles and the sides lead to specific trigonometric functions that can be derived using the properties of these special triangles. The sides of a 30-60-90 triangle are in the ratio of 1:√3:2, where the shortest side (opposite the 30-degree angle) is half the length of the hypotenuse, and the longer leg (opposite the 60-degree angle) is √3 times the length of the shorter leg.

To find the sine, cosine, and tangent for the 30-degree angle, we start with the definitions of these functions. The sine of 30 degrees is calculated as the ratio of the opposite side to the hypotenuse:

$$\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{10} = \frac{1}{2}.$$

The cosine of 30 degrees is the ratio of the adjacent side to the hypotenuse:

$$\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}.$$

The tangent of 30 degrees is the ratio of the opposite side to the adjacent side:

$$\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{5\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}.$$

Using reciprocal identities, we can find the cosecant, secant, and cotangent:

$$\csc(30^\circ) = \frac{1}{\sin(30^\circ)} = \frac{1}{\frac{1}{2}} = 2,$$

$$\sec(30^\circ) = \frac{1}{\cos(30^\circ)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3},$$

$$\cot(30^\circ) = \frac{1}{\tan(30^\circ)} = \frac{1}{\frac{\sqrt{3}}{3}} = \sqrt{3}.$$

Next, we calculate the trigonometric functions for the 60-degree angle. The sine of 60 degrees is:

$$\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}.$$

The cosine of 60 degrees is:

$$\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{10} = \frac{1}{2}.$$

The tangent of 60 degrees is:

$$\tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{5\sqrt{3}}{5} = \sqrt{3}.$$

Using reciprocal identities again, we find:

$$\csc(60^\circ) = \frac{1}{\sin(60^\circ)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3},$$

$$\sec(60^\circ) = \frac{1}{\cos(60^\circ)} = \frac{1}{\frac{1}{2}} = 2,$$

$$\cot(60^\circ) = \frac{1}{\tan(60^\circ)} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}.$$

Notably, there are patterns in the values of the trigonometric functions for the two angles. The sine and cosine values switch places between the two angles, as do the cosecant and secant. Similarly, the tangent and cotangent also exhibit this switching behavior. Recognizing these patterns can simplify the process of calculating trigonometric functions for 30-60-90 triangles.

Here’s what students ask on this topic:

What is SOHCAHTOA and how does it help in solving trigonometric problems?

SOHCAHTOA is a mnemonic that helps remember the relationships between angles and side lengths in right triangles for trigonometric functions. It stands for:

$\sin \left(\theta \right)=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$

$\cos \left(\theta \right)=\frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$

$\tan \left(\theta \right)=\frac{\mathrm{opposite}}{\mathrm{adjacent}}$

It simplifies solving problems by providing a clear formula for each function. For example, to find the sine of an angle, you divide the length of the opposite side by the hypotenuse. This tool is especially useful for memorizing and applying trigonometric ratios efficiently.

How do you calculate the sine, cosine, and tangent of an angle in a right triangle?

To calculate the sine, cosine, and tangent of an angle in a right triangle, use the following formulas:

$\sin \left(\theta \right)=\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$

$\cos \left(\theta \right)=\frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$

$\tan \left(\theta \right)=\frac{\mathrm{opposite}}{\mathrm{adjacent}}$

Identify the angle you are working with, then determine the opposite, adjacent, and hypotenuse sides relative to that angle. Plug the side lengths into the respective formula. For example, if the opposite side is 3 and the hypotenuse is 5, the sine of the angle is $\frac{3}{5}$.

What are reciprocal identities in trigonometry?

Reciprocal identities in trigonometry are relationships between primary trigonometric functions and their reciprocals. They are defined as:

$\csc \left(\theta \right)=\frac{1}{\sin }$

$\sec \left(\theta \right)=\frac{1}{\cos }$

$\cot \left(\theta \right)=\frac{1}{\tan }$

For example, if $\sin \left(\theta \right)=\frac{3}{5}$, then $\csc \left(\theta \right)=\frac{5}{3}$. Reciprocal identities simplify calculations and are essential for solving complex trigonometric problems.

What are the shortcuts for solving 45-45-90 triangles?

In a 45-45-90 triangle, the two legs are equal, and the hypotenuse is related to the leg length by a simple ratio. The shortcuts are:

$\mathrm{hypotenuse}=\mathrm{leg}\times \sqrt{2}$

For example, if the leg length is 5, the hypotenuse is $5\times \sqrt{2}$, which simplifies to $5\sqrt{2}$. This shortcut eliminates the need for the Pythagorean theorem, making calculations faster and more efficient.

How do you find trigonometric functions for a 30-60-90 triangle?

In a 30-60-90 triangle, the side lengths follow specific ratios: the shortest leg is opposite the 30° angle, the hypotenuse is twice the shortest leg, and the longer leg is the shortest leg multiplied by the square root of three. To find trigonometric functions:

$\sin \left(30\right)=\frac{1}{2}$

$\cos \left(30\right)=\frac{\sqrt{3}}{2}$

$\tan \left(30\right)=\frac{1}{\sqrt{3}}$

For the 60° angle, the sine and cosine values switch, and the tangent becomes $\sqrt{3}$. These patterns simplify calculations for this special triangle.