Questions: 3. HOCKEY The goals in a hockey rink are 6 feet wide. A hockey player is standing 25 feet directly in front of the left side of one of the goals. a. What is the maximum angle at which the player can shoot the puck to make it in the goal? 0=2 * tan^(-1)(w/(2d))=43 * 408 / pi degrees b. If the player skates forward 5 feet, how wide of a range of angles can the player shoot the puck to make it in the goal? 0=2 * tan^(-1)(0/(2a))=54.408 / 2 dog c c. Describe what happens to the maximum angle as the player gets closer to the goal. As the player skates closer to the goal, maximum angle theta increases due to the tangent function increasing. 4. KITES Suppose the string of a kite makes a 78° angle with the ground. If 45 feet of string is released, how high off the ground is the kite? h * L * sin(pi) h=4 sin(78°)=45/8+43 sqrt(3)/8+45 sqrt(3)/2 5. CURRENT The current I in amperes for a circuit at time t seconds is given by I=40 cos[60 pi(t+1/120)] a. Find the maximum current. Imax=40 amperes 6. WINDSHIELD WIPER A 12-inch windshield wiper blade rotates 120° as shown. Find the coordinates of the starting and ending points A and B of the tip of the blade relative to the pivot point O.

3. HOCKEY The goals in a hockey rink are 6 feet wide. A hockey player is standing 25 feet directly in front of the left side of one of the goals. a. What is the maximum angle at which the player can shoot the puck to make it in the goal? 0=2 * tan^(-1)(w/(2d))=43 * 408 / pi degrees b. If the player skates forward 5 feet, how wide of a range of angles can the player shoot the puck to make it in the goal? 0=2 * tan^(-1)(0/(2a))=54.408 / 2 dog c c. Describe what happens to the maximum angle as the player gets closer to the goal. As the player skates closer to the goal, maximum angle theta increases due to the tangent function increasing. 4. KITES Suppose the string of a kite makes a 78° angle with the ground. If 45 feet of string is released, how high off the ground is the kite? h * L * sin(pi) h=4 sin(78°)=45/8+43 sqrt(3)/8+45 sqrt(3)/2 5. CURRENT The current I in amperes for a circuit at time t seconds is given by I=40 cos[60 pi(t+1/120)] a. Find the maximum current. Imax=40 amperes 6. WINDSHIELD WIPER A 12-inch windshield wiper blade rotates 120° as shown. Find the coordinates of the starting and ending points A and B of the tip of the blade relative to the pivot point O.
Transcript text: 3. HOCKEY The goals in a hockey rink are 6 feet wide. A hockey player is standing 25 feet directly in front of the left side of one of the goals. a. What is the maximum angle at which the player can shoot the puck to make it in the goal? \[ 0=2 \cdot \tan ^{-1}\left(\frac{w}{2 d}\right)=43 \cdot 408 / \pi \text { degrees } \] b. If the player skates forward 5 feet, how wide of a range of angles can the player shoot the puck to make it in the goal? \[ 0=2 \cdot \tan ^{-1}\left(\frac{0}{2 a}\right)=54.408 / 2 \operatorname{dog} c \] c. Describe what happens to the maximum angle as the player gets closer to the goal. As the player skates closer to the goal, maximum angle $\theta$ increases due to the tangent function increasing. 4. KITES Suppose the string of a kite makes a $78^{\circ}$ angle with the ground. If 45 feet of string is released, how high off the ground is the kite? \[ \begin{array}{l} h \cdot L \cdot \sin (\pi) \\ h=4 \sin \left(78^{\circ}\right)=\frac{45}{8}+\frac{43 \sqrt{3}}{8}+\frac{45 \sqrt{3}}{2} \end{array} \] 5. CURRENT The current $I$ in amperes for a circuit at time $t$ seconds is given by \[ I=40 \cos \left[60 \pi\left(t+\frac{1}{120}\right)\right] \] a. Find the maximum current. \[ I_{\max }=40 \text { amperes } \] 6. WINDSHIELD WIPER A 12-inch windshield wiper blade rotates $120^{\circ}$ as shown. Find the coordinates of the starting and ending points $A$ and $B$ of the tip of the blade relative to the pivot point $O$.
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Solution

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Solution Steps

Step 1: Visualizing the problem

The hockey player is standing 25 feet from the left goalpost. The goal is 6 feet wide. We can imagine a triangle formed by the player's position and the two goalposts.

Step 2: Setting up the angles

We want to find the angle between the line connecting the player to the left goalpost and the line connecting the player to the right goalpost. This can be viewed as the sum of two angles, each formed by the line to the center of the goal and the lines to each post. The width from the center to each post is 3 feet (half of the total width).

Step 3: Calculating the maximum shooting angle

Let θ be the angle formed between the line from the player to the center of the goal and the line from the player to one of the goalposts. Then the maximum shooting angle is 2θ. We can use the arctangent function to find θ: θ = arctan(3/25) 2θ = 2 * arctan(3/25)

Calculating this value: 2θ ≈ 2 * 6.84° ≈ 13.68°

Step 4: Calculating the new angle after moving forward

The player moves 5 feet closer to the goal, so the distance is now 20 feet. We repeat the process from step 3. θ = arctan(3/20) 2θ = 2 * arctan(3/20)

Calculating this value: 2θ ≈ 2 * 8.53° ≈ 17.06°

Step 5: Describing the change in angle

As the player moves closer to the goal, the maximum angle at which they can shoot to make the puck into the goal increases. This is because the ratio of the width of the goal to the distance from the goal increases, and the arctangent function is increasing.

Final Answer

a. \(\boxed{13.68^\circ}\) b. \(\boxed{17.06^\circ}\) c. As the player gets closer to the goal, the maximum angle increases.

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