Master Grade 12 Math: Circle & Trigonometry Problems

Step 1: Determine the Radius.
Point A is on the x-axis at \( (8,0) \). Because the center of the circle is at the origin \( (0,0) \), the distance from the origin to point A is exactly the radius. Therefore, \( R = 8 \).

Step 2: Find the central angle \( \theta \).
We use the arc length formula \( s = R \theta \). We are given \( s = 20 \) and we know \( R = 8 \). Substitute these values to solve for \( \theta \) in radians:

\[ 20 = 8 \theta \implies \theta = \dfrac{20}{8} = 2.5 \text{ radians} \]

Step 3: Calculate the coordinates of P.
Use the standard circular coordinate formulas \( x = R \cos(\theta) \) and \( y = R \sin(\theta) \). Make sure your calculator is set to Radian mode!

\[ x = 8 \cos(2.5) \approx 8(-0.8011) \approx -6.40 \] \[ y = 8 \sin(2.5) \approx 8(0.5984) \approx 4.79 \]

Conclusion:
The coordinates of point \( P \) are approximately \( (-6.40, 4.79) \). (This makes logical sense as the point is in the second quadrant, where x is negative and y is positive).

Step 1: Find the radius \( R \).
The radius is the distance from the origin \( (0,0) \) to point P. We use the distance formula (or Pythagorean theorem):

\[ R = \sqrt{x^2 + y^2} = \sqrt{(4\sqrt{2})^2 + (5\sqrt{2})^2} \] \[ R = \sqrt{(16 \cdot 2) + (25 \cdot 2)} = \sqrt{32 + 50} = \sqrt{82} \]

Step 2: Find the central angle \( \theta \).
We use the tangent definition to find the angle. Since both x and y are positive, the point is in Quadrant I, so we don't need to adjust the angle.

\[ \tan(\theta) = \dfrac{y}{x} = \dfrac{5\sqrt{2}}{4\sqrt{2}} = \dfrac{5}{4} = 1.25 \] \[ \theta = \arctan(1.25) \approx 0.8960 \text{ radians} \]

Step 3: Calculate the arc length \( S \).
Using the formula \( S = R\theta \):

\[ S = \sqrt{82} \cdot \arctan(1.25) \approx 9.055 \cdot 0.8960 \approx 8.11 \text{ units} \]

Step 1: Determine the elapsed time.
From 11:10 PM to 11:50 PM, the elapsed time is exactly 40 minutes.

Step 2: Find the angular displacement.
A clock is a full circle (\( 360^\circ \) or \( 2\pi \) radians), representing 60 minutes of time. We need to find the fraction of the circle the minute hand covered.

\[ \text{Fraction of circle} = \dfrac{40 \text{ min}}{60 \text{ min}} = \dfrac{2}{3} \text{ of a full rotation} \]

Now, convert this fraction directly into radians. A full rotation is \( 2\pi \):

\[ \theta = \dfrac{2}{3} \times 2\pi = \dfrac{4\pi}{3} \text{ radians} \]

Step 3: Calculate the arc length.
The radius \( R \) is the length of the minute hand, which is 4.5 cm. Apply the arc length formula \( S = R \theta \):

\[ S = 4.5 \times \left( \dfrac{4\pi}{3} \right) = \dfrac{18\pi}{3} = \mathbf{6\pi \text{ cm}} \]

a) Find the coordinates of P:
Using the radius \( R = 5 \) and angle \( 35^\circ \), we use the standard formulas (ensure calculator is in Degree mode for this step):

\[ a = x = 5 \cos(35^\circ) \approx 4.096 \] \[ b = y = 5 \sin(35^\circ) \approx 2.868 \] \[ P(4.096, 2.868) \]

b) Find the arc length PQ:
First, we must convert the angle from degrees to radians because the formula \( S=R\theta \) requires radians.

\[ \theta \text{ in radians} = 35^\circ \times \left(\dfrac{\pi}{180^\circ}\right) = \dfrac{35\pi}{180} = \dfrac{7\pi}{36} \] \[ \text{Arc } PQ = R\theta = 5 \cdot \left(\dfrac{7\pi}{36}\right) = \dfrac{35\pi}{36} \approx 3.05 \text{ units} \]

c) Find the x-coordinate of M:
Point M is \( (c, -1) \). Since M is on the circle, its distance to the origin must equal the radius (5).

\[ x^2 + y^2 = R^2 \implies c^2 + (-1)^2 = 5^2 \] \[ c^2 + 1 = 25 \implies c^2 = 24 \implies c = \pm \sqrt{24} = \pm 2\sqrt{6} \]

Look at the diagram. Point M is located in Quadrant III, where x-coordinates must be negative. Therefore, we reject the positive root:

\[ c = -2\sqrt{6} \approx -4.899 \]

d) Find angle \( \theta \):
Let the total angle from the positive x-axis to M be \( \alpha \). According to the diagram, \( \alpha = \theta + 35^\circ \).

We find \( \alpha \) using the tangent of point M \( (-2\sqrt{6}, -1) \):

\[ \tan(\alpha) = \dfrac{y}{x} = \dfrac{-1}{-2\sqrt{6}} = \dfrac{1}{2\sqrt{6}} \]

Since the point is in the third quadrant, the calculator's \( \arctan \) function (which returns a Quadrant I angle) will be off by \( 180^\circ \). We must add \( 180^\circ \):

\[ \alpha = 180^\circ + \arctan\left(\dfrac{1}{2\sqrt{6}}\right) \approx 180^\circ + 11.54^\circ = 191.54^\circ \]

Now, subtract the \( 35^\circ \) offset to solve for \( \theta \):

\[ \theta = \alpha - 35^\circ = 191.54^\circ - 35^\circ = \mathbf{156.54^\circ} \]

a) Calculate arc length DP:
Convert the angle to radians: \( \theta = 130^\circ \times (\dfrac{\pi}{180^\circ}) = \dfrac{130\pi}{180} = \dfrac{13\pi}{18} \).

\[ \text{Arc } DP = R \cdot \theta = 4 \cdot \dfrac{13\pi}{18} = \dfrac{26\pi}{9} \approx 9.08 \text{ units} \]

b) Find coordinates of P:
Using standard trig definitions with \( R = 4 \) and \( \theta = 130^\circ \):

\[ a = 4 \cos(130^\circ) \approx -2.57 \] \[ b = 4 \sin(130^\circ) \approx 3.06 \] \[ P(-2.57, 3.06) \]

c) Find the length of line segment PD:
Point D lies on the positive x-axis at distance R, so its coordinates are \( D(4, 0) \). Use the 2D distance formula \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \):

\[ PD = \sqrt{(4 \cos(130^\circ) - 4)^2 + (4 \sin(130^\circ) - 0)^2} \] \[ PD = \sqrt{(-2.57 - 4)^2 + (3.06)^2} = \sqrt{(-6.57)^2 + 9.36} \] \[ PD = \sqrt{43.16 + 9.36} = \sqrt{52.52} \approx \mathbf{7.25 \text{ units}} \]

(Alternatively, you could use the Law of Cosines on triangle OPD: \( c^2 = a^2 + b^2 - 2ab\cos(\theta) \implies PD^2 = 4^2 + 4^2 - 2(4)(4)\cos(130^\circ) \). This yields the exact same result!)

Step 1: Find the angle \( \theta \) in radians.
Rearrange the arc length formula \( S = R\theta \) to solve for \( \theta \):

\[ \theta = \dfrac{S}{R} = \dfrac{12.5}{3} \approx 4.167 \text{ radians} \]

(Note: Since \( \pi \approx 3.14 \) and \( \frac{3\pi}{2} \approx 4.71 \), an angle of 4.167 radians falls squarely in the third quadrant. This matches our diagram).

Step 2: Find the coordinates.
Apply the coordinate formulas directly using the radian measure. Ensure your calculator is in Radian mode.

\[ a = x = 3 \cos\left(\dfrac{12.5}{3}\right) \approx 3(-0.523) \approx -1.57 \] \[ b = y = 3 \sin\left(\dfrac{12.5}{3}\right) \approx 3(-0.852) \approx -2.56 \]

Coordinates of B are \( (-1.57, -2.56) \).

Step 1: Find the angle \( \theta \).
We are given the x-coordinate (\( 4.1 \)) and the radius (\( 5 \)). We use the cosine relation:

\[ \cos(\theta) = \dfrac{x}{R} = \dfrac{4.1}{5} = 0.82 \]

Because point P is explicitly stated to be in Quadrant I, the angle \( \theta \) is acute (between 0 and \( \pi/2 \)). We can directly use the inverse cosine function to find the angle in radians:

\[ \theta = \arccos(0.82) \approx 0.6094 \text{ radians} \]

Step 2: Calculate the arc length.
Substitute the radian angle into the arc length formula:

\[ S = R \theta = 5 \cdot \arccos(0.82) \approx 5 \cdot 0.6094 \approx \mathbf{3.05 \text{ units}} \]

Step 1: Set up the equations.
The perimeter of a sector is made of two radii and the arc length \( S \). Therefore, \( P = 2R + S \). Since \( S = R\theta \), we can write the perimeter equation as:

\[ 100 = 2R + R\theta \]

The area of a sector is given by \( A = \frac{1}{2}R^2\theta \).

Step 2: Express Area in terms of one variable.
From the perimeter equation, solve for \( \theta \):

\[ R\theta = 100 - 2R \implies \theta = \dfrac{100 - 2R}{R} \]

Substitute this \( \theta \) into the Area equation to create a function solely in terms of \( R \):

\[ A(R) = \dfrac{1}{2}R^2 \left( \dfrac{100 - 2R}{R} \right) = \dfrac{1}{2}R(100 - 2R) = 50R - R^2 \]

Step 3: Maximize the Area function.
The area equation \( A(R) = -R^2 + 50R \) is a downward-facing parabola. The maximum occurs at the vertex. Using the vertex formula \( R = \frac{-b}{2a} \):

\[ R = \dfrac{-50}{2(-1)} = 25 \text{ cm} \]

(Alternatively, take the derivative: \( A'(R) = 50 - 2R = 0 \implies R = 25 \)).

Step 4: Find the corresponding angle \( \theta \).
Substitute \( R = 25 \) back into our equation for \( \theta \):

\[ \theta = \dfrac{100 - 2(25)}{25} = \dfrac{50}{25} = \mathbf{2 \text{ radians}} \]

Remarkably, for any fixed perimeter, a sector's area is always maximized when the central angle is exactly 2 radians!

Step 1: Find the length of the straight segments.
Draw a line parallel to the straight belt segment through the center of the smaller circle. This creates a right triangle with a hypotenuse equal to the distance between centers (\( d = 13 \)) and a leg equal to the difference in radii (\( r_2 - r_1 = 8 - 3 = 5 \)).
Let \( L \) be the length of one straight segment of the belt. Using the Pythagorean theorem:

\[ L^2 + 5^2 = 13^2 \implies L^2 + 25 = 169 \implies L^2 = 144 \implies L = 12 \text{ cm} \]

Since there are two straight segments (top and bottom), their combined length is \( 2 \times 12 = 24 \text{ cm} \).

Step 2: Find the central angle of the contact points.
Let \( \alpha \) be the angle in the right triangle adjacent to the centers.
\( \cos(\alpha) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{13} \). Thus, \( \alpha = \arccos(\frac{5}{13}) \approx 1.176 \text{ radians} \).

Step 3: Calculate the arc lengths around the pulleys.
Through geometric deduction, the angle where the belt touches the smaller pulley is \( 2\alpha \).
Arc on small pulley: \( S_1 = r_1(2\alpha) = 3(2 \times 1.176) = 3(2.352) = 7.056 \text{ cm} \).

The angle where the belt touches the larger pulley is the major arc, which is \( 2\pi - 2\alpha \).
Major angle: \( 2\pi - 2.352 \approx 6.283 - 2.352 = 3.931 \text{ radians} \).
Arc on large pulley: \( S_2 = r_2(3.931) = 8(3.931) = 31.448 \text{ cm} \).

Step 4: Sum the parts.
Total Length = Straight Segments + Small Arc + Large Arc
\[ \text{Total Length} \approx 24 + 7.056 + 31.448 = \mathbf{62.5 \text{ cm}} \]

Step 1: Find the points of intersection.
Substitute the linear equation into the circle equation:

\[ x^2 + (2x + 5)^2 = 25 \] \[ x^2 + (4x^2 + 20x + 25) = 25 \] \[ 5x^2 + 20x = 0 \]

Factor out \( 5x \):

\[ 5x(x + 4) = 0 \implies x = 0 \text{ or } x = -4 \]

Find corresponding y-values using \( y = 2x+5 \):
If \( x = 0, y = 5 \). Point 1: \( (0, 5) \).
If \( x = -4, y = -3 \). Point 2: \( (-4, -3) \).

Step 2: Find the central angle between these points.
Let vectors \( \vec{A} = (0, 5) \) and \( \vec{B} = (-4, -3) \). We can find the angle \( \theta \) between them using the dot product formula: \( \vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos(\theta) \).

Dot product: \( (0)(-4) + (5)(-3) = -15 \)
Magnitudes (both are the radius): \( |\vec{A}| = 5 \), \( |\vec{B}| = 5 \).

\[ -15 = (5)(5) \cos(\theta) \implies \cos(\theta) = \dfrac{-15}{25} = -\dfrac{3}{5} \] \[ \theta = \arccos\left(-\dfrac{3}{5}\right) \approx 2.214 \text{ radians} \]

Step 3: Calculate the arc length.
We know the radius is \( R = 5 \). Using \( S = R\theta \):

\[ S = 5 \cdot \arccos\left(-\dfrac{3}{5}\right) \approx 5 \cdot 2.214 \approx \mathbf{11.07 \text{ units}} \]