I still need to know how we get from sin(.909)=.788=65? ### Step 2Here, consider:\[ y = 0.9090 \]\[ y = \sin(x) \]\[ \sin^{-1} (0.9090) = x \]Substituting the value in the above equation then,\[ \text{arc} \sin(y) = \sin^{-1} (0.9090) = x + 2k\pi \]\[ = 65.36^\circ + 2 \times \pm1 \times 180^\circ \]\[ = 65.36^\circ \pm 360^\circ \]Taking the negative sign for clockwise and positive for anticlockwise then,If the angle moves anticlockwise, then \( 360^\circ \) is considered as \( 0^\circ \).Hence, two values of \(\text{arc} \sin(0.9090)\) are \(-294.63^\circ, 65.36^\circ\).The angle is measured as anticlockwise; hence the value of \(\text{arc} \sin(0.9090)\) is \(65.36^\circ\). **Title: Solving a Right Angle Triangle Using the Pythagorean Theorem and Trigonometry****Step 1: Applying the Pythagorean Theorem**Given:- \( AB = c = 12 \)- \( AC = b \)- \( BC = a = 5 \)Using the Pythagorean Theorem for the right angle triangle:\[(\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2\]\[c^2 = a^2 + b^2\]Substituting the known values:\[12^2 = 5^2 + b^2\]Solving for \( b \):\[b^2 = 144 - 25 = 119\]\[b = \sqrt{119}\]**Step 2: Calculating the Sine of Angle B**According to the given figure:\[\sin B = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{b}{c}\]\[\sin B = \frac{\sqrt{119}}{12}\]Calculating the approximate value:\[\sin B = \frac{10.90}{12} \approx 0.9090\]Finding angle B:\[B = \sin^{-1}(0.9090) \approx 65.36^\circ\]**Answer: \( B = 65.36^\circ \)**

Name: I still need to know how we get from sin(.909)=.788=65? ### Step 2Here
Uploaded: 2024-03-20T06:46:55.000Z
Channel: Bartleby
Description: I still need to know how we get from sin(.909)=.788=65? ### Step 2Here, consider:\[ y = 0.9090 \]\[ y = \sin(x) \]\[ \sin^{-1} (0.9090) = x \]Substituting the value in the above equation then,\[ \text{arc} \sin(y) = \sin^{-1} (0.9090) = x + 2k\pi \]\