To solve for side \( AB \) in the given right triangle, we can use the sine function, which relates the opposite side to the hypotenuse. Given: - The angle \( \angle BCA = 20^\circ \). - The side opposite the angle \( \angle BCA \) is \( BC = 7 \). - The hypotenuse is \( AB \). We need to find \( AB \). Step 1: Write the sine function for the given angle. \[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \] Step 2: Substitute the given values into the sine function. \[ \sin(20^\circ) = \frac{7}{AB} \] Step 3: Solve for \( AB \). \[ AB = \frac{7}{\sin(20^\circ)} \] Step 4: Calculate the value using a calculator. \[ \sin(20^\circ) \approx 0.3420\] \[ AB = \frac{7}{0.3420} \approx 20.47\] Therefore, the length of side \( AB \) is approximately \( 20.47 \) (rounded to the nearest hundredth).

Find all angles, 0, degrees, is less than or equal to, theta, is less than, 360, degrees0 ∘ ≤θ<360 ∘ , that solve the following equation.sine, theta, equals, minus, one halfsinθ=− 21​

Given a right triangle, solve for θ.

To solve for the trigonometric ratios \(\sin M\), \(\cos M\), and \(\tan M\) in the given right triangle, we need to identify the sides relative to angle \(M\). Given: - \(MK = 4\) (opposite side to angle \(M\)) - \(ML = 10\) (hypotenuse) - \(KL = 2\sqrt{21}\) (adjacent side to angle \(M\)) The trigonometric ratios are defined as follows: - \(\sin M = \frac{\text{opposite}}{\text{hypotenuse}}\) - \(\cos M = \frac{\text{adjacent}}{\text{hypotenuse}}\) - \(\tan M = \frac{\text{opposite}}{\text{adjacent}}\) Now, let's calculate each ratio: 1. \(\sin M\): \[ \sin M = \frac{MK}{ML} = \frac{4}{10} = \frac{2}{5} \] 2. \(\cos M\): \[ \cos M = \frac{KL}{ML} = \frac{2\sqrt{21}}{10} = \frac{\sqrt{21}}{5} \] 3. \(\tan M\): \[ \tan M = \frac{MK}{KL} = \frac{4}{2\sqrt{21}} = \frac{2}{\sqrt{21}} = \frac{2\sqrt{21}}{21} \] So, the ratios are: - \(\sin M = \frac{2}{5}\) - \(\cos M = \frac{\sqrt{21}}{5}\) - \(\tan M = \frac{2\sqrt{21}}{21}\) For the specific question in the image, the answer for \(\sin M\) is: \[ \sin M = \frac{2}{5} \]

To solve this problem, we need to use properties of circles and angles. Let's go through each part step by step. ### Given: - \( \angle CBD = 42^\circ \) - \( \angle OBE = 20^\circ \) - \( BC \) is a tangent to the circle at \( B \) ### To find: (i) \( \angle BOE \) (ii) \( \angle OED \) (iii) \( \angle BFE \) ### Solution: #### (i) \( \angle BOE \) Since \( BC \) is a tangent to the circle at \( B \), \( \angle OBE \) is the angle between the radius \( OB \) and the tangent \( BC \). This angle is given as \( 20^\circ \). The angle subtended by the same arc at the center of the circle is twice the angle subtended at the circumference. Therefore, \( \angle BOE \) is twice \( \angle OBE \). \[ \angle BOE = 2 \times \angle OBE = 2 \times 20^\circ = 40^\circ \] #### (ii) \( \angle OED \) \( \angle OED \) is an angle at the center of the circle. To find this angle, we need to consider the angles around point \( O \). Since \( \angle BOE = 40^\circ \) and \( \angle CBD = 42^\circ \), we need to find the relationship between these angles. Notice that \( \angle OED \) is the external angle for triangle \( OBD \), and it is equal to the sum of the opposite internal angles. \[ \angle OED = \angle OBE + \angle CBD = 20^\circ + 42^\circ = 62^\circ \] #### (iii) \( \angle BFE \) To find \( \angle BFE \), we need to use the fact that \( \angle BFE \) is an angle in the alternate segment. This means it is equal to the angle subtended by the same arc at the circumference. Since \( \angle CBD = 42^\circ \), and \( \angle BFE \) is subtended by the same arc \( BD \), we have: \[ \angle BFE = \angle CBD = 42^\circ \] ### Summary of Answers: (i) \( \angle BOE = 40^\circ \) (2 marks) (ii) \( \angle OED = 62^\circ \) (2 marks) (iii) \( \angle BFE = 42^\circ \) (3 marks)

A right triangle has side lengths 8, 15, and 17 as shown below.Use these lengths to find cosB, sinB, and tanB.