Special Right Triangles: Videos & Practice Problems

Special right triangles, such as the 45-45-90 and 30-60-90 triangles, offer shortcuts for solving for unknown sides. In a 45-45-90 triangle, the hypotenuse is found by multiplying a leg by $\sqrt{2}$. For a 30-60-90 triangle, the hypotenuse is double the shortest leg, while the long leg is the short leg multiplied by $\sqrt{3}$. Trigonometric functions reveal patterns, with sine and cosine values switching between angles, enhancing problem-solving efficiency.

45-45-90 Triangles

In the study of right triangles, the 45-45-90 triangle is a special case that frequently appears in various mathematical contexts. This triangle is characterized by having two angles of 45 degrees and one right angle, which results in the two legs being of equal length. Understanding this triangle allows for efficient calculations, as there are specific shortcuts to determine the lengths of its sides.

For a 45-45-90 triangle, the relationship between the lengths of the legs and the hypotenuse can be expressed simply. If the length of each leg is denoted as \( a \), the hypotenuse \( c \) can be calculated using the formula:

\[ c = a \sqrt{2} \]

For example, if each leg measures 5 units, the hypotenuse can be quickly found by multiplying 5 by \( \sqrt{2} \), resulting in \( 5\sqrt{2} \).

Alternatively, the Pythagorean theorem can be applied to verify this relationship. According to the theorem, the sum of the squares of the legs equals the square of the hypotenuse:

\[ a^2 + a^2 = c^2 \]

Substituting \( a \) with 5 gives:

\[ 5^2 + 5^2 = c^2 \]

This simplifies to:

\[ 25 + 25 = c^2 \]

Thus, \( 50 = c^2 \), and taking the square root yields \( c = \sqrt{50} = 5\sqrt{2} \), confirming the shortcut's accuracy.

When solving for the sides of a 45-45-90 triangle, if the hypotenuse is known, the leg length can be derived by rearranging the shortcut formula. For instance, if the hypotenuse is 13, the equation becomes:

\[ 13 = a \sqrt{2} \]

To isolate \( a \), divide both sides by \( \sqrt{2} \):

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

Rationalizing the denominator results in:

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

Consequently, both legs of the triangle will measure \( \frac{13\sqrt{2}}{2} \), reinforcing the property that the legs of a 45-45-90 triangle are equal. Mastering these relationships and shortcuts significantly simplifies the process of solving problems involving special right triangles.

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 right triangles, the 45-45-90 triangle is particularly significant due to its unique properties and the relationships between its sides and angles. In a 45-45-90 triangle, the two legs are of equal length, and the hypotenuse can be expressed as the length of a leg multiplied by the square root of 2. This relationship allows for quick calculations of side lengths and trigonometric ratios.

When examining the sine function, which is defined as the ratio of the length of the opposite side to the hypotenuse, we can derive its value for a 45-45-90 triangle. For an angle of 45 degrees, if we denote the length of each leg as 5, the hypotenuse will be \(5\sqrt{2}\). Thus, the sine of 45 degrees is calculated as:

\[ \sin(45^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

Similarly, the cosine function, defined as the ratio of the adjacent side to the hypotenuse, yields the same result:

\[ \cos(45^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

The tangent function, which is the ratio of the opposite side to the adjacent side, simplifies to:

\[ \tan(45^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{5} = 1 \]

From these calculations, a clear pattern emerges: both the sine and cosine of 45 degrees equal \(\frac{\sqrt{2}}{2}\), while the tangent equals 1. This consistency extends to the reciprocal trigonometric functions. The cosecant, which is the reciprocal of sine, is calculated as:

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

Likewise, the secant, the reciprocal of cosine, also results in:

\[ \sec(45^\circ) = \frac{1}{\cos(45^\circ)} = \sqrt{2} \]

Finally, the cotangent, the reciprocal of tangent, is simply:

\[ \cot(45^\circ) = \frac{1}{\tan(45^\circ)} = 1 \]

In summary, for the 45-45-90 triangle, the sine and cosine functions yield \(\frac{\sqrt{2}}{2}\), while the tangent and cotangent both equal 1. The cosecant and secant also equal \(\sqrt{2}\). Recognizing these relationships can significantly simplify the process of solving problems involving trigonometric functions in this special triangle.

30-60-90 Triangles

In the study of right triangles, the 30-60-90 triangle is a special case that frequently appears in various mathematical contexts, including geometry and physics. Understanding the relationships between the side lengths of this triangle is essential for solving problems efficiently.

In a 30-60-90 triangle, the side lengths are related to the shortest leg, which is opposite the 30-degree angle. The relationships can be summarized as follows: the hypotenuse is twice the length of the shortest leg, and the long leg (opposite the 60-degree angle) is equal to the shortest leg multiplied by the square root of 3. These relationships can be expressed mathematically as:

Hypotenuse \( = 2 \times \text{(short leg)} \)

Long leg \( = \text{(short leg)} \times \sqrt{3} \)

For example, if the shortest leg measures 5 units, the hypotenuse would be \( 5 \times 2 = 10 \) units, and the long leg would be \( 5 \times \sqrt{3} \) units.

To apply these relationships, consider a scenario where the hypotenuse is known. If the hypotenuse is 8 units, you can find the shortest leg by rearranging the hypotenuse formula:

Short leg \( = \frac{\text{Hypotenuse}}{2} = \frac{8}{2} = 4 \text{ units} \

Next, to find the long leg, use the relationship for the long leg:

Long leg \( = 4 \times \sqrt{3} \text{ units} \

In another example, if the long leg is given as 1 unit, you can find the short leg using the relationship:

Short leg \( = \frac{\text{Long leg}}{\sqrt{3}} = \frac{1}{\sqrt{3}} \

Rationalizing the denominator gives:

Short leg \( = \frac{\sqrt{3}}{3} \text{ units} \

Finally, to find the hypotenuse, use the short leg value:

Hypotenuse \( = 2 \times \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3} \text{ units} \

By mastering these shortcuts and relationships, solving for the sides of a 30-60-90 triangle becomes a straightforward process, allowing for quick calculations in various mathematical applications.

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 the study of special right triangles, particularly the 30-60-90 triangle, understanding the associated trigonometric functions is essential. The key angles in this triangle are 30 degrees and 60 degrees, and each has specific sine, cosine, and tangent values that can be derived from the triangle's side lengths.

For a 30-degree angle, the sine function is defined as the ratio of the length of the opposite side to the hypotenuse. In a 30-60-90 triangle, the opposite side to the 30-degree angle measures 5, while the hypotenuse measures 10. Thus, the sine of 30 degrees is calculated as:

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

The cosine function, which is the ratio of the adjacent side to the hypotenuse, gives us:

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

For the tangent function, which is the ratio of the opposite side to the adjacent side, we find:

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

Rationalizing the denominator results in:

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

Using reciprocal identities, we can derive the cosecant, secant, and cotangent functions. The cosecant is the reciprocal of sine:

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

The secant is the reciprocal of cosine:

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

Rationalizing gives:

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

Finally, the cotangent is the reciprocal of tangent:

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

For the 60-degree angle, the sine, cosine, and tangent functions can be derived similarly. The sine of 60 degrees is:

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

The cosine of 60 degrees is:

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

And the tangent is:

$$\tan(60^\circ) = \frac{5\sqrt{3}}{5} = \sqrt{3}$$

Using reciprocal identities again, we find:

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

Rationalizing gives:

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

For secant:

$$\sec(60^\circ) = 2$$

And for cotangent:

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

Rationalizing gives:

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

Notably, a pattern emerges 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, and the tangent and cotangent. This symmetry is a hallmark of the relationships within the 30-60-90 triangle, making it easier to remember and apply these functions in various mathematical contexts.

Here’s what students ask on this topic:

What are the side length relationships in a 45-45-90 triangle?

In a 45-45-90 triangle, the two legs are of equal length. If each leg is denoted as a, then the hypotenuse is a√2. This relationship can be derived from the Pythagorean theorem, but it serves as a useful shortcut for quickly solving these triangles. For example, if each leg is 5 units, the hypotenuse would be 5√2 units.

How do you find the missing sides of a 30-60-90 triangle?

In a 30-60-90 triangle, the side lengths have specific relationships. The hypotenuse is twice the length of the shortest leg, and the longer leg is the shortest leg multiplied by √3. If the shortest leg is a, then the hypotenuse is 2a and the longer leg is a√3. For example, if the shortest leg is 4 units, the hypotenuse is 8 units, and the longer leg is 4√3 units.

What are the trigonometric functions for a 45-45-90 triangle?

For a 45-45-90 triangle, the trigonometric functions are consistent due to the equal angles. The sine and cosine of 45 degrees are both √2/2. The tangent of 45 degrees is 1. The reciprocal functions are: cosecant (csc) is √2, secant (sec) is √2, and cotangent (cot) is 1. These values simplify solving trigonometric problems involving 45-45-90 triangles.

What are the trigonometric functions for a 30-60-90 triangle?

In a 30-60-90 triangle, the trigonometric functions vary based on the angle. For the 30-degree angle, sine is 1/2, cosine is √3/2, and tangent is √3/3. For the 60-degree angle, sine is √3/2, cosine is 1/2, and tangent is √3. The reciprocal functions follow: cosecant (csc) of 30 degrees is 2, secant (sec) of 30 degrees is 2/√3, and cotangent (cot) of 30 degrees is √3. These patterns help in quickly solving trigonometric problems.

How do you solve for the hypotenuse in a 45-45-90 triangle?

To solve for the hypotenuse in a 45-45-90 triangle, you can use the relationship that the hypotenuse is the leg length multiplied by √2. If each leg is a, then the hypotenuse c is given by c = a√2. For example, if each leg is 7 units, the hypotenuse would be 7√2 units.