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Write the ratios for sin M, cos M, and tan M. M sin M = (Type an integer or a simplified fraction. Type an exact answer, using radicals as needed.) K 9 3√7 12

Write the ratios for sin M, cos M, and tan M. M sin M = (Type an integer or a simplified fraction. Type an exact answer, using radicals as needed.) K 9 3√7 12

Trigonometry (MindTap Course List)
Trigonometry (MindTap Course List)
8th Edition
ISBN:9781305652224
Author:Charles P. McKeague, Mark D. Turner
Publisher:Charles P. McKeague, Mark D. Turner
Chapter2: Right Triangle Trigonometry
Section: Chapter Questions
Problem 6GP
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### Trigonometric Ratios in Right Triangle In this educational exercise, we are asked to determine the trigonometric ratios for sin M, cos M, and tan M for a right triangle \( \triangle MKL \) where \(\angle MKL\) is the right angle. #### Diagram Explanation The right triangle \( \triangle MKL \) is presented with: - \( K \) as the vertex at the right angle. - \( M \) and \( L \) as the other two vertices. The side lengths are labeled as: - \( MK = 9 \) - \( KL = 12 \) - \( ML = 3\sqrt{7} \) To find the trigonometric ratios, we use the following definitions: - **Sine (sin)** of an angle is the ratio of the length of the opposite side to the hypotenuse. - **Cosine (cos)** of an angle is the ratio of the length of the adjacent side to the hypotenuse. - **Tangent (tan)** of an angle is the ratio of the length of the opposite side to the adjacent side. Given triangle \( \triangle MKL \): - The hypotenuse (\(ML\)) = \(3\sqrt{7}\) - The side opposite to \(\angle M\) (\(K L\)) = 12 - The side adjacent to \(\angle M\) (\(MK\)) = 9 #### Finding the Ratios: - \( \sin M = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{KL}{ML} = \frac{12}{3\sqrt{7}} = \frac{4}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{4\sqrt{7}}{7} \) - \( \cos M = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{MK}{ML} = \frac{9}{3\sqrt{7}} = \frac{3}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{3\sqrt{7}}{7} \) - \( \tan M = \frac{\text{opposite}}{\text{adjacent}} = \frac{KL}{
Transcribed Image Text:### Trigonometric Ratios in Right Triangle In this educational exercise, we are asked to determine the trigonometric ratios for sin M, cos M, and tan M for a right triangle \( \triangle MKL \) where \(\angle MKL\) is the right angle. #### Diagram Explanation The right triangle \( \triangle MKL \) is presented with: - \( K \) as the vertex at the right angle. - \( M \) and \( L \) as the other two vertices. The side lengths are labeled as: - \( MK = 9 \) - \( KL = 12 \) - \( ML = 3\sqrt{7} \) To find the trigonometric ratios, we use the following definitions: - **Sine (sin)** of an angle is the ratio of the length of the opposite side to the hypotenuse. - **Cosine (cos)** of an angle is the ratio of the length of the adjacent side to the hypotenuse. - **Tangent (tan)** of an angle is the ratio of the length of the opposite side to the adjacent side. Given triangle \( \triangle MKL \): - The hypotenuse (\(ML\)) = \(3\sqrt{7}\) - The side opposite to \(\angle M\) (\(K L\)) = 12 - The side adjacent to \(\angle M\) (\(MK\)) = 9 #### Finding the Ratios: - \( \sin M = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{KL}{ML} = \frac{12}{3\sqrt{7}} = \frac{4}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{4\sqrt{7}}{7} \) - \( \cos M = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{MK}{ML} = \frac{9}{3\sqrt{7}} = \frac{3}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{3\sqrt{7}}{7} \) - \( \tan M = \frac{\text{opposite}}{\text{adjacent}} = \frac{KL}{
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