Example Question - trigonometry problem
Here are examples of questions we've helped users solve.
Resolving a Trigonometry Problem Involving a Fire Escape Ladder
Claro, me complacerá ayudarte a resolver la pregunta que se muestra en la imagen. El texto en la imagen dice: "5) Una escalera de bomberos de 14,5 m de longitud se apoya en la fachada de un edificio formando un ángulo..." Parece que falta el resto de la pregunta, pero por el principio se deduce que probablemente se trata de un problema que involucra trigonometría, donde se necesita calcular la altura a la que la escalera alcanza en el edificio o el ángulo que forma con el suelo. Para resolver completamente el problema, necesitaríamos conocer al menos un dato adicional: el ángulo que forma con el suelo o la altura a la que la escalera llega en el edificio. Si tienes el resto del problema o los datos adicionales, por favor compártelos para poder proceder con la solución.
Finding the Tangent of Double Angle
Para resolver el problema, primero recordemos que en un triángulo rectángulo, la tangente de un ángulo es el cociente de la longitud del cateto opuesto sobre la longitud del cateto adyacente. El problema nos dice que \( AB = 4 \) y \( BC = 5 \), y como \( \theta \) es el ángulo en el vértice A, entonces: \( \tan(\theta) = \frac{BC}{AB} = \frac{5}{4} \) Ahora, para encontrar \( \tan(2\theta) \), podemos usar la fórmula de la tangente del ángulo doble: \( \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \) Sustituimos el valor de \( \tan(\theta) \): \( \tan(2\theta) = \frac{2 \cdot \left(\frac{5}{4}\right)}{1 - \left(\frac{5}{4}\right)^2} = \frac{2 \cdot \frac{5}{4}}{1 - \frac{25}{16}} = \frac{\frac{10}{4}}{\frac{16}{16} - \frac{25}{16}} = \frac{\frac{10}{4}}{\frac{-9}{16}} \) Al simplificar esta expresión, obtenemos: \( \tan(2\theta) = \frac{10 \cdot 16}{4 \cdot -9} = \frac{160}{-36} = -\frac{40}{9} \) Por lo tanto, el valor de \( \tan(2\theta) \) es \( -\frac{40}{9} \).
Calculating Distance Between Ships Using Law of Sines
The question is based on trigonometry, and it involves using the Law of Sines to solve for the distance between two ships spotted from an airplane. Given: The angles of depression to two ships are \( 32° \) and \( 40° \), respectively. The plane is 2 miles from the ship located at point A. To solve for the distance between the two ships, you need to find the length of the side opposite the given angles within the right triangles formed by the plane's altitude and the lines of sight to the ships. Firstly, since the plane is 2 miles from the ship at point A, that means the length of the line segment from the plane to ship A is 2 miles. Using the angle of depression and the fact the angle inside the right triangle formed at ship A is complementary to the angle of depression, we can determine that the angle at ship A is \( 90° - 32° = 58° \). Now, use the Law of Sines to set up the ratio of the sides of the large triangle, where \( x \) is the distance between the two ships: \[ \frac{\sin(58°)}{2\text{ miles}} = \frac{\sin(40°)}{x} \] To solve for \( x \), rearrange the equation to isolate \( x \): \[ x = \frac{2\text{ miles} \cdot \sin(40°)}{\sin(58°)} \] Now you can calculate the value of \( x \) by substituting the sine values: \[ x = \frac{2 \cdot \sin(40°)}{\sin(58°)} \] Without the exact sine values or a calculator, I cannot provide the numerical solution, but you can simply input the sine values for \( 40° \) and \( 58° \) into the formula to calculate \( x \). After you calculate it, round your answer to the nearest tenth of a mile as per the question's instructions.
Trigonometry Problem: Calculating Tangent and Cosine
The image contains a trigonometry problem which states: "(b) Given that \( \sin\theta = \frac{3}{5} \) and \( 90° \leq \theta \leq 180° \), find the exact value of \( \tan\theta + \cos\theta \)." Since \( \sin\theta = \frac{3}{5} \) and the angle \( \theta \) falls in the second quadrant (because \( 90° \leq \theta \leq 180° \)), we know that the cosine of \( \theta \) will be negative, and the tangent of \( \theta \) will be negative as well because the tangent of an angle is the sine divided by the cosine, and in the second quadrant the cosine is negative. We can find the cosine of \( \theta \) using the Pythagorean identity, where \( \cos^2\theta = 1 - \sin^2\theta \). \( \cos^2\theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \) Since \( \theta \) is in the second quadrant, \( \cos\theta \) will be negative: \( \cos\theta = -\sqrt{\frac{16}{25}} = -\frac{4}{5} \) Now we can find the tangent of \( \theta \): \( \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \) Now, we can find the value of \( \tan\theta + \cos\theta \): \( \tan\theta + \cos\theta = -\frac{3}{4} + (-\frac{4}{5}) = -\frac{3 \cdot 5}{4 \cdot 5} - \frac{4 \cdot 4}{4 \cdot 5} = -\frac{15}{20} - \frac{16}{20} = -\frac{31}{20} \) So, the exact value of \( \tan\theta + \cos\theta \) given \( \sin\theta = \frac{3}{5} \) and \( 90° \leq \theta \leq 180° \) is \( -\frac{31}{20} \).
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