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Home Question Bank Math Trigonometry Archive 2024 November 07

Trigonometry Questions from Nov 07,2024

Browse the Trigonometry Q&A Archive for Nov 07,2024, featuring a collection of homework questions and answers from this day. Find detailed solutions to enhance your understanding.

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2) Desde lo alto de un globo se observa un pueblo A con un ángulo de \( 50^{\circ} \), y otro B, situado al otro lado y en linea recta, con un ángulo de \( 60^{\circ} \). Sabiendo que el globo se encuentra a una distancia de 6 kilómetros del pueblo A y a 4 del pueblo B, calcula la distancia entre los pueblos A y B. Q4. Use the given information for a), b) and c): \( \sin (\alpha)=-\frac{5}{13}, \pi<\alpha<\frac{3 \pi}{2} \) and \( \tan (\beta)=\frac{3}{4}, 0<\beta<\frac{\pi}{2} \) \( \begin{array}{ll}\text { a. Find } \sin (\alpha-\beta) & \text { b. Find } \sin (2 \beta) \\ \text { c. Find } \cos (2 \alpha) & \text { c. Find } \cos (\alpha+\beta)\end{array} \) Q3. What is the exact value of \( \cos 38^{\circ} \cos 68^{\circ}+\sin 38^{\circ} \sin 68^{\circ} \) ? Q2. Find the exact value of \( \cos \left(165^{\circ}\right)\left[\mathrm{HINT}: 165^{\circ}=120^{\circ}+45^{\circ}\right] \) Solve the following triangle using either the Law of Sines or the Law of Cosines. \( \mathrm{A}=13^{\circ}, \mathrm{a}=6, \mathrm{~b}=8 \) Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. (Round to two decimal places as needed.) A. There is only one possible solution for the triangle. The measurements for the remaining angles B and C and side c are as follows. \( \mathrm{B} \approx \square^{\circ} \) B. \( \mathrm{C}^{\circ} \approx \square^{\circ} \) There are two possible solutions for the triangle. \( \mathrm{B}_{1} \approx \square^{\circ} \) The triangle with the larger angle B has \( \mathrm{B}_{2} \approx \square^{\circ} \) C. There are no possible solutions for this triangle. 1. Una persona que mide 145 cm en su vision esta en el segundo piso del colegio que esta a 3 m de altura y observa que frente a el hay 2 balones alineados la persona mide que al balon mas cercano hay un ángulo de depresion de \( 53^{\circ} \) y el mas lejano de \( 28^{\circ} \) dicual es la distancia entre los balones y la distancia de cada balon al observador? \( \frac { \tan y + \cot y } { \sin y \sec y } \equiv \csc ^ { 2 } y \) Solve the equation over the interval \( [0,2 \pi) \) \( \sin x \cos x=\frac{\sqrt{2}}{4} \) Q1. Verify (Use known identities e.g., reciprocal, quotient, Pythagorean, even/odd etc. Also see if you can use common denominator, conjugate, or factoring, refer to your guided note) \( \begin{array}{ll}\text { a. }(1+\sin (x)) \cdot(1+\sin (-x))=\cos ^{2}(x) & \text { b. } \frac{\cot (\theta)}{\csc (\theta)}=\cos (\theta) \\ \text { Hint: Recall } \sin (-x)=-\sin (x) & \text { d. } \frac{\sec (\theta)}{\cos (\theta)}-\frac{\tan (\theta)}{\cot (\theta)}=1 \\ \text { c. } \frac{\tan ^{2}(\theta)}{\sec \theta}=\sec (\theta)-\cos (\theta) & \\ \text { e. } 1+\frac{\tan ^{2}(\theta)}{\sec (\theta)+1}=\sec (\theta) & \text { f. } \frac{\sec ^{2}(x)-\csc c^{2}(x)}{\sec (x)+\csc (x)}=\sec (x)-\csc (x)\end{array} \) Suppose you are solving a trigonometric equation for solutions over the interval \( [0,2 \pi) \), and your work leads to \( 2 x=\frac{2 \pi}{3}, 2 \pi, \frac{8 \pi}{3} \). What are the corresponding values of \( x \) ? \( x=\square \) (Simplify your answer. Type an exact answer in terms of \( \pi \). Use a comma to separate answers as needed.)
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