We are given a right triangle ABC with right angle at $C$, and the lengths of two sides are provided:

• $a = 9$
• $c = 18$ (hypotenuse)

We need to find the unknown side $b$ and the six trigonometric functions for angle $B$.

Step 1: Use the Pythagorean Theorem to find side $b$

The Pythagorean theorem states:

$a^2 + b^2 = c^2$

Substituting the known values of $a$ and $c$:

$9^2 + b^2 = 18^2$ $81 + b^2 = 324$ $b^2 = 324 - 81 = 243$ $b = \sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3}$

So, the length of side $b$ is $9\sqrt{3}$.

Step 2: Find the six trigonometric functions for angle $B$

We now know the sides of the triangle:

• $a = 9$
• $b = 9\sqrt{3}$
• $c = 18$

The six trigonometric functions for angle $B$ are as follows:

1. Sine of $B$ ($\sin B$): $\sin B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{c} = \frac{9}{18} = \frac{1}{2}$

2. Cosine of $B$ ($\cos B$): $\cos B = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{b}{c} = \frac{9\sqrt{3}}{18} = \frac{\sqrt{3}}{2}$

3. Tangent of $B$ ($\tan B$): $\tan B = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b} = \frac{9}{9\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

4. Cosecant of $B$ ($\csc B$): $\csc B = \frac{1}{\sin B} = \frac{1}{\frac{1}{2}} = 2$

5. Secant of $B$ ($\sec B$): $\sec B = \frac{1}{\cos B} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

6. Cotangent of $B$ ($\cot B$): $\cot B = \frac{1}{\tan B} = \frac{1}{\frac{\sqrt{3}}{3}} = \sqrt{3}$

Final Answers:

• $b = 9\sqrt{3}$
• $\sin B = \frac{1}{2}$
• $\cos B = \frac{\sqrt{3}}{2}$
• $\tan B = \frac{\sqrt{3}}{3}$
• $\csc B = 2$
• $\sec B = \frac{2\sqrt{3}}{3}$
• $\cot B = \sqrt{3}$

Would you like further explanations on any of the steps or trigonometric functions?

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Here are five related questions:

1. How would the values change if angle $A$ were the one with the right angle?
2. How do you derive the value of $\tan B$ from the other trigonometric functions?
3. What are some ways to verify that $\tan B = \frac{\sqrt{3}}{3}$ is correct?
4. What is the relationship between the secant and cosine functions?
5. How can we apply these values to solve other problems in trigonometry?

Tip: When simplifying trigonometric ratios involving square roots, it's often useful to rationalize the denominator to make the expression easier to interpret.