Grade 11 Trigonometry Problems with Solutions

a) Let \( P \) be the position of the passenger (see figure below).

The height \( h \) of the passenger is given by \[ h = 1 + 25 + y = y + 26 \] Here, \( y \) depends on the angle of rotation \( A \). Using trigonometry, we write: \[ \sin\left(\dfrac{\pi}{2} - A\right) = \dfrac{y}{\text{radius}} = \dfrac{y}{25} \quad \Rightarrow \quad y = 25 \cos(A) \] The angle \( A \) depends on the angular speed \( \omega \) as follows: \[ A = \omega t \] The angular speed \( \omega \) is given by: \[ \omega = \dfrac{2\pi}{36} = \dfrac{\pi}{18} \text{ radians/second} \] Substituting into the expression for height, we find: \[ h(t) = 25 \cos\left(\dfrac{\pi}{18} t\right) + 26 \] where \( t \) is in seconds and height is in meters.

b) To find the height at \( t = 45 \) seconds:

\[ h(45) = 25 \cos\left(\dfrac{\pi}{18} \cdot 45\right) + 26 = 25 \cos\left(\dfrac{45\pi}{18}\right) + 26 \] \[ = 25 \cos\left(\dfrac{5\pi}{2}\right) + 26 = 25 \cdot 0 + 26 = 26 \text{ meters} \]

Using the figure below,

we write the following equations based on the right triangles formed: \[ \tan(35^\circ) = \dfrac{h}{x} \quad \text{and} \quad \tan(45^\circ) = \dfrac{h}{x - 20} \] where \( h \) is the height of the tree.

Solving both equations for \( x \), we get: \[ x = \dfrac{h}{\tan(35^\circ)} \quad \text{and} \quad x = \dfrac{h}{\tan(45^\circ)} + 20 \] Setting the two expressions for \( x \) equal: \[ \dfrac{h}{\tan(35^\circ)} = \dfrac{h}{\tan(45^\circ)} + 20 \] Solving for \( h \): \[ h = \dfrac{20 \cdot \tan(35^\circ) \cdot \tan(45^\circ)}{\tan(45^\circ) - \tan(35^\circ)} = 46.7 \text{ meters (3 significant digits)} \]

Using the figure below, we write the following equations based on trigonometric relationships:

\[ \tan (90^\circ - 15^\circ) = \tan(75^\circ) = \dfrac{y}{200} \quad \text{and} \quad \tan((90^\circ - 13^\circ) ) = \tan(77^\circ) = \dfrac{y + x}{200} \]

Eliminate \( y \) from the two equations and solve for \( x \):

\[ x = 200 \left[ \tan(77^\circ) - \tan(75^\circ) \right] = 119.9 \text{ meters (rounded to 4 significant figures)} \]

Start with the right-hand side:

\[ \cos(x) - \sin(3x) = \cos(x) - \sin(x + 2x) \] Expand \( \sin(x + 2x) \) \[ = \cos(x) - \sin(x)\cos(2x) - \cos(x)\sin(2x) \] Now expand the product in left-hand side: \[ [\cos(x) - \sin(x)][\cos(2x) - \sin(2x)] \] \[ = \cos(x)\cos(2x) - \cos(x)\sin(2x) - \sin(x)\cos(2x) + \sin(x)\sin(2x) \] Use the identities \( \cos(2x) = 1 - 2\sin^2(x) \) and \( \sin(2x) = 2\sin(x)\cos(x) \) to transform the first two terms: \[ = \cos(x)(1 - 2\sin^2(x)) + \sin(x)\cdot 2\sin(x)\cos(x) - \cos(x)\sin(2x) - \sin(x)\cos(2x) \] \[ = \cos(x) - 2\cos(x)\sin^2(x) + 2\cos(x)\sin^2(x) - \cos(x)\sin(2x) - \sin(x)\cos(2x) \] \[ = \cos(x) - \cos(x)\sin(2x) - \sin(x)\cos(2x) \]

The left-hand side has been transformed so that it is equal to the right-hand side.

The function \( f \) is given in the form:

\[ f(x) = a \sin\left(x - \dfrac{\pi}{3}\right) + 2 \]

This represents a vertical shift of the graph of \( g(x) = a \sin\left(x - \dfrac{\pi}{3}\right) \) upward by 2 units.

Given Value: \( f\left(\dfrac{\pi}{2}\right) = 1 \)

Substitute \( x = \dfrac{\pi}{2} \) into the function:

\[ f\left(\dfrac{\pi}{2}\right) = a \sin\left(\dfrac{\pi}{2} - \dfrac{\pi}{3}\right) + 2 \] \[ f\left(\dfrac{\pi}{2}\right) = a \sin\left(\dfrac{\pi}{6}\right) + 2 \]

Using the known value \( \sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2} \), we get:

\[ 1 = a \cdot \dfrac{1}{2} + 2 \] \[ a \cdot \dfrac{1}{2} = -1 \] \[ a = -2 \]

Final Equation

The fully determined function is:

\[ f(x) = -2 \sin\left(x - \dfrac{\pi}{3}\right) + 2 \]

Expand and simplify: \[ \cos\left(\dfrac{\pi}{2} - x\right) = \cos\left(\dfrac{\pi}{2}\right)\cos(x) + \sin\left(\dfrac{\pi}{2}\right)\sin(x) = \sin(x) = -\dfrac{3}{5} \]

Expand and simplify: \[ \sin\left(x + \dfrac{\pi}{2}\right) = \sin(x) \cos\left(\dfrac{\pi}{2}\right) + \cos(x) \sin\left(\dfrac{\pi}{2}\right) = \cos(x) = \dfrac{4}{5} \]

\[ \tan(x) = \dfrac{\sin(x)}{\cos(x)} = \dfrac{-\dfrac{3}{5}}{\dfrac{4}{5}} = -\dfrac{3}{4} \]

The addition formula for the tangent can be used as follows:

\[ \dfrac{\tan(25^\circ) + \tan(50^\circ)}{1 - \tan(25^\circ)\tan(50^\circ)} = \tan(25^\circ + 50^\circ) \] \[ = \tan(75^\circ) \] \[ = \tan(45^\circ + 30^\circ) \] \[ = \dfrac{\tan(45^\circ) + \tan(30^\circ)}{1 - \tan(45^\circ)\tan(30^\circ)} \] \[ = \dfrac{1 + \dfrac{\sqrt{3}}{3}}{1 - 1 \cdot \dfrac{\sqrt{3}}{3}} \] \[ = \sqrt{3} + 2 \]

To find the length of side \( a \), we apply the cosine rule:

\[ a^2 = 4^2+ 8^2 - 2 \cdot 4 \cdot 8 \cdot \cos(46^\circ) \] which gives \[ a = \sqrt{16 + 64 - 2 \cdot 4 \cdot 8 \cdot \cos(46^\circ)} \]

Next, we use the sine rule to find angle \( B \):

\[ \dfrac{\sin(B)}{4} = \dfrac{\sin(A)}{a} \]

Solving for angle \( B \), we get:

\[ B = \arcsin\left( \dfrac{4 \sin(A)}{a} \right) \approx 29^\circ \]

(Rounded to the nearest degree)

Given \( \sin(s) = \dfrac{1}{4} \) and \( \sin(t) = -\dfrac{1}{2} \), and their respective quadrants, find \( \cos(s) \) and \( \cos(t) \).

\[ \cos(s) = -\dfrac{\sqrt{15}}{4} \quad \text{and} \quad \cos(t) = \dfrac{\sqrt{3}}{2} \]

We now expand:

\[ \tan(s + t) = \dfrac{\sin(s + t)}{\cos(s + t)} \] \[ = \dfrac{\sin(s)\cos(t) + \cos(s)\sin(t)}{\cos(s)\cos(t) - \sin(s)\sin(t)} \]

Substitute values:

\[ = -\dfrac{4\sqrt{3} + \sqrt{15}}{11} \]

Note that

\[ 15^2 = 12^2 + 9^2 \]

This means that the triangle in question is a right triangle.

Let angle \( A \) be the angle opposite the side of length 9. Then:

\[ \sin(A) = \dfrac{9}{15} \]

Angle \( A \) is approximately \( 37^\circ \) (rounded to the nearest degree).

The third angle is:

\[ 90^\circ - 37^\circ = 53^\circ \]

The function is given by:

\[ y = \dfrac{5}{3} \sin(Bx) + 4, \quad B > 0 \]

We are told that the period is:

\[ \dfrac{2\pi}{B} = \dfrac{\pi}{2} \]

Solving for \( B \):

\[ B = 4 \]

Substitute \( B = 4 \) into the function:

\[ y = \dfrac{5}{3} \sin(4x) + 4 \]

\[ \cos\left(\dfrac{13\pi}{12}\right) = \cos\left(\dfrac{\pi}{12} + \pi\right) = -\cos\left(\dfrac{\pi}{12}\right) \] \[ = -\cos\left(\dfrac{1}{2} \cdot \dfrac{\pi}{6}\right) = -\sqrt{\dfrac{1}{2}\left(1 + \cos\left(\dfrac{\pi}{6}\right)\right)} \quad \text{(half angle formula)} \] \[ = -\sqrt{\dfrac{1}{2} + \dfrac{\sqrt{3}}{4}} \]

Let \( R_1 \) and \( R_2 \) be the radii of gear 1 and gear 2. Let \( S_1 \) and \( S_2 \) be the arcs of rotation of gears 1 and 2 respectively. Since the interconnected gears have equal tangential velocity (measured in inches per second), the arcs \( S_1 \) and \( S_2 \) are equal in length.

\[ R_1 \times t_1 = R_2 \times t_2 \]

Here, \( t_1 \) and \( t_2 \) are the angles of rotation of the larger and smaller gears respectively.

\[ 10 \times t_1 = 4 \times 890^\circ \] \[ t_1 = 356^\circ \]

The angular speed is calculated as:

\[ \text{Angular speed} = \dfrac{356^\circ}{4 \text{ seconds}} = 89^\circ \text{ per second} \] \[ = 89^\circ \times \dfrac{60}{1 \times 60} = 5340^\circ \text{ per minute} \]

The ladder, the wall, and the ground form a right triangle. Hence,

a) \(\sin(t) = \dfrac{x}{20}\) or \(x = 20 \sin(t)\)

b) \(x = \dfrac{1}{4} \times 20 = 20 \sin(t)\)

Solve for \(t\): \[ t = \arcsin\left(\dfrac{1}{4}\right) \approx 14.48^\circ \]