Example Question - hypotenuse
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Right Triangle Problem
<p>Para resolver este problema, necesitamos aplicar el teorema de Pitágoras, que establece que en un triángulo rectángulo la suma de los cuadrados de los catetos es igual al cuadrado de la hipotenusa. La fórmula es:</p> <p>\[a^2 + b^2 = c^2\]</p> <p>Donde \(a\) y \(b\) son los catetos, y \(c\) es la hipotenusa. En este caso, tenemos un triángulo con catetos de longitud 6 cm y 7.5 cm.</p> <p>Aplicando el teorema de Pitágoras obtenemos:</p> <p>\[6^2 + 7.5^2 = c^2\]</p> <p>\[36 + 56.25 = c^2\]</p> <p>\[92.25 = c^2\]</p> <p>Tomando la raíz cuadrada de ambos lados:</p> <p>\[c = \sqrt{92.25}\]</p> <p>\[c = 9.6\, \text{cm}\]</p> <p>Por lo tanto, la longitud de la hipotenusa es de 9.6 cm.</p> <p>Para encontrar el valor del ángulo que no es de 90 grados y es adyacente al lado de 6 cm, podemos usar la función trigonométrica tangente, que es la relación entre el cateto opuesto y el cateto adyacente:</p> <p>\[\tan(\theta) = \frac{\text{cateto opuesto}}{\text{cateto adyacente}}\]</p> <p>En este caso, el cateto opuesto al ángulo es de 7.5 cm y el cateto adyacente es de 6 cm. Así que:</p> <p>\[\tan(\theta) = \frac{7.5}{6}\]</p> <p>\[\theta = \arctan\left(\frac{7.5}{6}\right)\]</p> <p>Usando una calculadora, encontramos que:</p> <p>\[\theta \approx 51.34^\circ\]</p> <p>Por lo tanto, el valor del ángulo \(\theta\) es aproximadamente 51.34 grados.</p>
Triangle Side Relationships
Para resolver el problema, necesitamos usar el teorema de Pitágoras para verificar la validez de las longitudes de los lados del triángulo. El teorema de Pitágoras se expresa como \( c^2 = a^2 + b^2 \), donde \( c \) es la longitud de la hipotenusa y \( a \) y \( b \) son las longitudes de los catetos. <p>Tomamos los valores dados en la imagen:</p> <p>\( a = 9 \) cm (cateto opuesto)</p> <p>\( b = 12 \) cm (cateto adyacente)</p> <p>\( c = 16 \) cm (hipotenusa)</p> <p>Aplicamos el teorema de Pitágoras:</p> \[ c^2 = a^2 + b^2 \] \[ 16^2 = 9^2 + 12^2 \] \[ 256 = 81 + 144 \] \[ 256 = 225 \] <p>Como \( 256 \neq 225 \), la relación dada no cumple con el teorema de Pitágoras. Por lo tanto, las longitudes dadas para los lados del triángulo son incorrectas.</p>
Solving a Right Triangle
<p>Paso 1: Determinar la hipotenusa \( C \)</p> <p>\( C = \sqrt{A^2+B^2} \)</p> <p>\( C = \sqrt{3^2+5^2} \)</p> <p>\( C = \sqrt{9+25} \)</p> <p>\( C = \sqrt{34} \)</p> <p>\( C \approx 5.83 \) (2 decimales)</p> <p>Paso 2: Encontrar las medidas de los ángulos \( A \) y \( B \)</p> <p>Usando las funciones trigonométricas:</p> <p>Para \( A \):</p> <p>\( \tan(A) = \frac{\text{Opuesto}}{\text{Adyacente}} = \frac{B}{A} \)</p> <p>\( \tan(A) = \frac{5}{3} \)</p> <p>\( A = \arctan\left(\frac{5}{3}\right) \)</p> <p>\( A \approx 59.04^\circ \) (2 decimales)</p> <p>Para \( B \) (sabemos que en un triángulo rectángulo \( A + B = 90^\circ \)):</p> <p>\( B = 90^\circ - A \)</p> <p>\( B \approx 90^\circ - 59.04^\circ \)</p> <p>\( B \approx 30.96^\circ \) (2 decimales)</p> <p>Paso 3: Determinar las 6 razones trigonométricas</p> <p>Sen \( A = \frac{\text{Opuesto a } A}{\text{Hipotenusa}} = \frac{B}{C} = \frac{5}{\sqrt{34}} \approx 0.857 \) (3 decimales)</p> <p>Cos \( A = \frac{\text{Adyacente a } A}{\text{Hipotenusa}} = \frac{A}{C} = \frac{3}{\sqrt{34}} \approx 0.515 \) (3 decimales)</p> <p>Tan \( A = \frac{\text{Opuesto a } A}{\text{Adyacente a } A} = \frac{B}{A} = \frac{5}{3} \approx 1.667 \) (3 decimales)</p> <p>Sen \( B = \frac{\text{Opuesto a } B}{\text{Hipotenusa}} = \frac{A}{C} = \frac{3}{\sqrt{34}} \approx 0.515 \) (3 decimales)</p> <p>Cos \( B = \frac{\text{Adyacente a } B}{\text{Hipotenusa}} = \frac{B}{C} = \frac{5}{\sqrt{34}} \approx 0.857 \) (3 decimales)</p> <p>Tan \( B = \frac{\text{Opuesto a } B}{\text{Adyacente a } B} = \frac{A}{B} = \frac{3}{5} \approx 0.600 \) (3 decimales)</p>
Demonstration of the Pythagorean Theorem
The image shows an equation representing the Pythagorean Theorem which states that in a right-angled triangle, the square of the length of the hypotenuse \((c)\) is equal to the sum of the squares of the lengths of the other two sides \((a)\) and \((b)\). The Pythagorean Theorem is expressed as: \[ a^2 + b^2 = c^2 \]
Solving for x in a Right Triangle using the Pythagorean Theorem
The image shows a right triangle with one of the angles marked as 90 degrees, indicating that it's indeed a right triangle. There are two sides labeled: one is labeled "10" and the other is labeled "2x". Since the side labeled "10" is opposite the right angle, it is the hypotenuse. To solve for "x," you can use the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the hypotenuse. The theorem is written as: \( a^2 + b^2 = c^2 \) where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides. From the image, we have: \( a = 2x \) \( c = 10 \) We are missing the length of the other side, but since the triangle's sides are marked with the same hash mark, it means they are of equal length. Therefore, \( b = 2x \) as well. Now we can insert the values into the Pythagorean theorem: \( (2x)^2 + (2x)^2 = 10^2 \) Expanding this gives us: \( 4x^2 + 4x^2 = 100 \) Combine like terms: \( 8x^2 = 100 \) Divide both sides by 8 to solve for \( x^2 \): \( x^2 = \frac{100}{8} \) \( x^2 = 12.5 \) Taking the square root of both sides gives us \( x \): \( x = \sqrt{12.5} \) \( x \approx 3.54 \) (to two decimal places). So, \( x \) is approximately 3.54.
Solving for the Length of a Side in a Right-Angled Triangle Using Trigonometry
The image shows a right-angled triangle with sides labeled, and we need to solve for side \( x \). This can be done using trigonometry. The triangle has one side of length 77, which is the opposite side to the angle marked as 68 degrees, and side \( x \) is the hypotenuse. To find side \( x \), we can use the sine (sin) function, which is defined as the ratio of the length of the opposite side over the hypotenuse in a right-angled triangle: \[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \] Here, \( \theta \) is 68 degrees, the opposite side is 77, and the hypotenuse is \( x \). Plugging these into the equation: \[ \sin(68^\circ) = \frac{77}{x} \] We now want to solve for \( x \): \[ x = \frac{77}{\sin(68^\circ)} \] Using a calculator: \[ x \approx \frac{77}{0.927} \] \[ x \approx \frac{77}{0.927} \] \[ x \approx 83.1 \] Thus, the length of side \( x \) is approximately 83.1, to the nearest tenth.
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