Let's break down each part of the problem step by step:

1. Use of cot(π/6):

• cot(θ) = adjacent / opposite
• Given: adjacent = $x$, opposite = 20 cm, and $\theta = \pi/6$ (or 60°).

We know that: $\cot(\pi/6) = \frac{\text{adjacent}}{\text{opposite}} = \frac{x}{20}$

From the standard trigonometric identity: $\cot(\pi/6) = \frac{1}{\tan(\pi/6)} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3}$

Thus, we can write: $\frac{x}{20} = \sqrt{3}$

Now, solve for $x$: $x = 20\sqrt{3}$

2. Use of sec(π/6):

• sec(θ) = hypotenuse / adjacent
• Given: hypotenuse = $x$, adjacent = 10 cm, and $\theta = \pi/6$.

We know that: $\sec(\pi/6) = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{x}{10}$

From the standard trigonometric identity: $\sec(\pi/6) = \frac{1}{\cos(\pi/6)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}$

Thus, we can write: $\frac{x}{10} = \frac{2}{\sqrt{3}}$

Now, solve for $x$: $x = \frac{2}{\sqrt{3}} \times 10 = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3}$

3. Use of cot(60°):

• cot(θ) = adjacent / opposite
• Given: adjacent = 17 cm, opposite = $x$, and $\theta = 60^\circ$.

We know that: $\cot(60^\circ) = \frac{\text{adjacent}}{\text{opposite}} = \frac{17}{x}$

From the standard trigonometric identity: $\cot(60^\circ) = \frac{1}{\tan(60^\circ)} = \frac{1}{\sqrt{3}}$

Thus, we can write: $\frac{17}{x} = \frac{1}{\sqrt{3}}$

Now, solve for $x$: $x = 17\sqrt{3}$

4. Use of cosec(45°):

• cosec(θ) = hypotenuse / opposite
• Given: hypotenuse = $x$, opposite = 11 cm, and $\theta = 45^\circ$.

We know that: $\csc(45^\circ) = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{x}{11}$

From the standard trigonometric identity: $\csc(45^\circ) = \frac{1}{\sin(45^\circ)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$

Thus, we can write: $\frac{x}{11} = \sqrt{2}$

Now, solve for $x$: $x = 11\sqrt{2}$

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Summary of Solutions:

1. $x = 20\sqrt{3}$ cm (from cot(π/6))
2. $x = \frac{20\sqrt{3}}{3}$ cm (from sec(π/6))
3. $x = 17\sqrt{3}$ cm (from cot(60°))
4. $x = 11\sqrt{2}$ cm (from cosec(45°))

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Would you like more details on any of the steps or have any further questions?

Here are some related questions for deeper understanding:

1. What is the relationship between trigonometric ratios and right-angled triangles?
2. How can you derive the values of trigonometric functions for specific angles like 30°, 45°, and 60°?
3. What is the difference between secant, cosecant, and cotangent functions?
4. How do the values of trigonometric ratios change in different quadrants of the unit circle?
5. How do you simplify expressions involving square roots in trigonometry?

Tip: When working with trigonometric identities, always remember the common values for sine, cosine, and tangent at key angles (like 30°, 45°, 60°) to help speed up calculations.