The question in the image involves a right triangle $\triangle ABC$ where angle $C$ is given as $\alpha$, and side lengths $AB = 6$ and $BC = 4$ are specified. You are asked to determine the trigonometric ratios: $\sin \alpha$, $\cos \alpha$, and $\tan \alpha$.

Solution:

Step 1: Calculate the hypotenuse

Using the Pythagorean theorem to find the length of the hypotenuse $AC$: $AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$

Step 2: Determine trigonometric ratios

Since $\alpha$ is at angle $C$, the sides relative to angle $\alpha$ are:

• Opposite side: $AB = 6$
• Adjacent side: $BC = 4$
• Hypotenuse: $AC = 2\sqrt{13}$

Now, the trigonometric ratios are:

• $\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6}{2\sqrt{13}} = \frac{3}{\sqrt{13}}$
• $\cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{2\sqrt{13}} = \frac{2}{\sqrt{13}}$
• $\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{6}{4} = 1.5$

If you wish to rationalize the results: $\sin \alpha = \frac{3}{\sqrt{13}} \times \frac{\sqrt{13}}{\sqrt{13}} = \frac{3\sqrt{13}}{13}$ $\cos \alpha = \frac{2}{\sqrt{13}} \times \frac{\sqrt{13}}{\sqrt{13}} = \frac{2\sqrt{13}}{13}$

Thus:

• $\sin \alpha = \frac{3\sqrt{13}}{13}$
• $\cos \alpha = \frac{2\sqrt{13}}{13}$
• $\tan \alpha = 1.5$

Would you like more details or further explanation?

Here are some related questions you can explore:

1. How do you calculate the hypotenuse in any right triangle?
2. What are the definitions of the basic trigonometric functions in a right triangle?
3. How can you rationalize a denominator when working with trigonometric functions?
4. What are the identities relating $\sin$, $\cos$, and $\tan$?
5. How do you find the angle $\alpha$ if you know $\sin \alpha$?

Tip: Always check whether your triangle's side lengths match the conditions of a right triangle using the Pythagorean theorem.