Part (a): Finding the Missing Side

The given triangle is a right triangle, with sides labeled $AC$, $CB$, and the hypotenuse $AB$. We are given:

• $CB = 10$
• $AB = 2\sqrt{41}$

We need to find the missing side $AC$.

By the Pythagorean Theorem, for a right triangle: $AB^2 = AC^2 + CB^2$ Substituting the known values: $(2\sqrt{41})^2 = AC^2 + 10^2$ Simplifying: $4 \times 41 = AC^2 + 100$ $164 = AC^2 + 100$ Subtract 100 from both sides: $AC^2 = 64$ Taking the square root of both sides: $AC = 8$

So, the missing side $AC = 8$.

Part (b): Trigonometric Functions

Now that we have all three sides of the triangle:

• $AB = 2\sqrt{41}$ (hypotenuse)
• $AC = 8$ (opposite side to angle $\alpha$)
• $CB = 10$ (adjacent side to angle $\alpha$)

We can calculate the six trigonometric functions:

1. Sine ($\sin \alpha$): $\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AC}{AB} = \frac{8}{2\sqrt{41}}$ Rationalizing the denominator: $\sin \alpha = \frac{8}{2\sqrt{41}} \times \frac{\sqrt{41}}{\sqrt{41}} = \frac{8\sqrt{41}}{82} = \frac{4\sqrt{41}}{41}$

2. Cosine ($\cos \alpha$): $\cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{CB}{AB} = \frac{10}{2\sqrt{41}}$ Rationalizing the denominator: $\cos \alpha = \frac{10}{2\sqrt{41}} \times \frac{\sqrt{41}}{\sqrt{41}} = \frac{10\sqrt{41}}{82} = \frac{5\sqrt{41}}{41}$

3. Tangent ($\tan \alpha$): $\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{AC}{CB} = \frac{8}{10} = \frac{4}{5}$

4. Cosecant ($\csc \alpha$): $\csc \alpha = \frac{1}{\sin \alpha} = \frac{1}{\frac{4\sqrt{41}}{41}} = \frac{41}{4\sqrt{41}}$ Rationalizing the denominator: $\csc \alpha = \frac{41}{4\sqrt{41}} \times \frac{\sqrt{41}}{\sqrt{41}} = \frac{41\sqrt{41}}{164} = \frac{\sqrt{41}}{4}$

5. Secant ($\sec \alpha$): $\sec \alpha = \frac{1}{\cos \alpha} = \frac{1}{\frac{5\sqrt{41}}{41}} = \frac{41}{5\sqrt{41}}$ Rationalizing the denominator: $\sec \alpha = \frac{41}{5\sqrt{41}} \times \frac{\sqrt{41}}{\sqrt{41}} = \frac{41\sqrt{41}}{205} = \frac{\sqrt{41}}{5}$

6. Cotangent ($\cot \alpha$): $\cot \alpha = \frac{\text{adjacent}}{\text{opposite}} = \frac{CB}{AC} = \frac{10}{8} = \frac{5}{4}$

Summary of Trigonometric Functions:

1. $\sin \alpha = \frac{4\sqrt{41}}{41}$
2. $\cos \alpha = \frac{5\sqrt{41}}{41}$
3. $\tan \alpha = \frac{4}{5}$
4. $\csc \alpha = \frac{\sqrt{41}}{4}$
5. $\sec \alpha = \frac{\sqrt{41}}{5}$
6. $\cot \alpha = \frac{5}{4}$

Would you like further details or clarifications on any of the steps? Here are some related questions to explore:

1. How would the trigonometric functions change if the angle was changed to the other non-right angle?
2. What is the area of the triangle?
3. Can the trigonometric functions be expressed in terms of radians?
4. How do the values of the trigonometric functions relate to the unit circle?
5. How can you verify the results using a calculator?

Tip: When rationalizing denominators, remember to multiply both the numerator and denominator by the conjugate or the radical present in the denominator to eliminate the square root.