Find sinLsin𝐿 , cosLcos𝐿 , tanLtan𝐿 , sinMsin𝑀 , cosMcos𝑀 , and tanMtan𝑀 when  ℓ=12ℓ=12 ,  m=123⎯⎯√𝑚=123 , and  n=24𝑛=24 . Match each ratio to the corresponding trigonometric expression.

Given that 𝑛𝐶4𝑛𝑃2= 12, find n

Find ∠α∠𝛼 given the following values:a=𝑎= 12 b=𝑏= 10 c=𝑐= 15

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: \[ \sin M = \frac{2}{5} \]

If the ratio of the 5th5𝑡ℎ term from the beginning to the 5th5𝑡ℎ term from the end in the expansion of (2–√4+13√4)n24+134𝑛 is (6–√)5:165:1, then n𝑛 is equal toA17

Find the value of the trigonometric ratio.