(c) In the sketch alongside Q points on a flat surface at a vertical tower. The ang top of the lower ( P ) is $$ \beta $$. (1) Show that $$ \mathrm{PR}= $$, (2) Given that $$ \alpha=3 $$ tower is $$ 30 \mathrm{~m}, ~ $$ the foot of the to size of $$ \beta $$.

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To solve the problem, we need to clarify the relationships between the points and angles mentioned. Let's break it down step by step.

### Known Conditions
1. There is a vertical tower.
2. Point $$ P $$ is at the top of the lower part of the tower.
3. The angle of elevation from point $$ Q $$ to point $$ P $$ is $$ \beta $$.
4. The angle of elevation from point $$ Q $$ to the top of the tower is $$ \alpha = 3 $$ (presumably in degrees).
5. The height of the tower is $$ 30 $$ m.

### Step 1: Show that $$ \mathrm{PR} = $$
To show that $$ \mathrm{PR} $$ relates to the angles and height, we can use trigonometric relationships.

Assuming:
- $$ h $$ is the height of point $$ P $$ from the ground.
- $$ d $$ is the horizontal distance from point $$ Q $$ to the base of the tower.

Using the tangent function for angle $$ \beta $$:
$$
\tan(\beta) = \frac{h}{d} \quad \text{(1)}
$$

Using the tangent function for angle $$ \alpha $$:
$$
\tan(\alpha) = \frac{30}{d} \quad \text{(2)}
$$

From equation (2), we can express $$ d $$:
$$
d = \frac{30}{\tan(\alpha)} \quad \text{(3)}
$$

Substituting equation (3) into equation (1):
$$
\tan(\beta) = \frac{h}{\frac{30}{\tan(\alpha)}} \implies h = 30 \cdot \frac{\tan(\beta)}{\tan(\alpha)} \quad \text{(4)}
$$

### Step 2: Given $$ \alpha = 3 $$ degrees, find the size of $$ \beta $$
Now we need to find $$ \beta $$ given $$ \alpha = 3 $$ degrees and the height of the tower is $$ 30 $$ m.

Using equation (4):
$$
h = 30 \cdot \frac{\tan(\beta)}{\tan(3)}
$$

Since $$ h $$ is the height of point $$ P $$, we can express it in terms of $$ \beta $$:
$$
h = 30 \cdot \frac{\tan(\beta)}{\tan(3)}
$$

To find $$ \beta $$, we need to know the height $$ h $$ of point $$ P $$. If we assume $$ h $$ is a certain value, we can solve for $$ \beta $$.

Let's assume $$ h $$ is a fraction of the total height of the tower. For example, if $$ h = 15 $$ m (half the height of the tower), we can substitute this value into the equation and solve for $$ \beta $$.

Now, let's calculate $$ \tan(3) $$ and then find $$ \beta $$ using the assumed height $$ h = 15 $$ m.

### Calculation
1. Calculate $$ \tan(3) $$.
2. Use the equation to find $$ \beta $$.

Let's perform the calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$\tan\left(3\right)$$
Solve the equation $$ 15 = 30 * (\tan(beta) / \tan(3)) $$.
Solve the equation by following steps:
- step0: Solve for $$beta$$:
  $$15=30\left(\frac{\tan\left(beta \right)}{\tan\left(3\right)}\right)$$
- step1: Find the domain:
  $$15=30\left(\frac{\tan\left(beta \right)}{\tan\left(3\right)}\right),beta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
- step2: Simplify:
  $$15=\frac{30\tan\left(beta \right)}{\tan\left(3\right)}$$
- step3: Swap the sides:
  $$\frac{30\tan\left(beta \right)}{\tan\left(3\right)}=15$$
- step4: Multiply both sides:
  $$\frac{30\tan\left(beta \right)}{\tan\left(3\right)}\times \frac{\tan\left(3\right)}{30}=15\times \frac{\tan\left(3\right)}{30}$$
- step5: Calculate:
  $$\tan\left(beta \right)=15\times \frac{\tan\left(3\right)}{30}$$
- step6: Calculate:
  $$\tan\left(beta \right)=\frac{\tan\left(3\right)}{2}$$
- step7: Use the inverse trigonometric function:
  $$beta =\arctan\left(\frac{\tan\left(3\right)}{2}\right)$$
- step8: Add the period:
  $$beta =\arctan\left(\frac{\tan\left(3\right)}{2}\right)+k\pi ,k \in \mathbb{Z}$$
- step9: Check if the solution is in the defined range:
  $$beta =\arctan\left(\frac{\tan\left(3\right)}{2}\right)+k\pi ,k \in \mathbb{Z},beta \neq \frac{\pi }{2}+k\pi ,k \in \mathbb{Z}$$
- step10: Find the intersection:
  $$beta =\arctan\left(\frac{\tan\left(3\right)}{2}\right)+k\pi ,k \in \mathbb{Z}$$
### Step 1: Calculate $$ \tan(3) $$

The value of $$ \tan(3) $$ is approximately:
$$
\tan(3) \approx -0.142547
$$

### Step 2: Solve for $$ \beta $$

Using the equation derived earlier:
$$
15 = 30 \cdot \left( \frac{\tan(\beta)}{\tan(3)} \right)
$$

Substituting $$ \tan(3) $$:
$$
15 = 30 \cdot \left( \frac{\tan(\beta)}{-0.142547} \right)
$$

Rearranging gives:
$$
\tan(\beta) = 15 \cdot \left( \frac{-0.142547}{30} \right)
$$

Calculating $$ \tan(\beta) $$:
$$
\tan(\beta) \approx 15 \cdot \left( \frac{-0.142547}{30} \right) \approx -0.071153
$$

To find $$ \beta $$, we can use the arctangent function:
$$
\beta \approx \tan^{-1}(-0.071153) + k\pi, \quad k \in \mathbb{Z}
$$

This gives us the general solution for $$ \beta $$:
$$
\beta \approx -0.071153 + k\pi, \quad k \in \mathbb{Z}
$$

### Conclusion
1. We derived the relationship for $$ \mathrm{PR} $$ in terms of angles and height.
2. Given $$ \alpha = 3 $$ degrees and assuming $$ h = 15 $$ m, we found that:
   $$
   \beta \approx -0.071153 + k\pi, \quad k \in \mathbb{Z}
   $$

If you have a specific height for $$ P $$ or any other conditions, please provide them for a more precise calculation of $$ \beta $$.
To solve the problem: 1. **Show that $$ \mathrm{PR} = $$**: - Use trigonometric relationships to express $$ \mathrm{PR} $$ in terms of the given angles and height. 2. **Find the size of $$ \beta $$ given $$ \alpha = 3^\circ $$ and the tower height is $$ 30 $$ m**: - Assume $$ h = 15 $$ m (half the tower height). - Calculate $$ \tan(\beta) $$ using the derived relationship. - Use the arctangent function to find $$ \beta $$. Final answer: $$ \beta \approx -0.071153 + k\pi, \quad k \in \mathbb{Z} $$ If a specific height for $$ P $$ is provided, a more precise value of $$ \beta $$ can be determined.

(c) In the sketch alongside Q points on a flat surface at a vertical tower. The ang top of the lower ( P ) is \( \beta \). (1) Show that \( \mathrm{PR}= \), (2) Given that \( \alpha=3 \) tower is \( 30 \mathrm{~m}, ~ \) the foot of the to size of \( \beta \).

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