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:

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

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

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 \( h \):

\[ h = \dfrac{20 \cdot \tan(35^\circ) \cdot \tan(45^\circ)}{\tan(45^\circ) - \tan(35^\circ)} = \mathbf{46.7 \text{ meters}} \]

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 \( x \):

\[ x = 200 \left[ \tan(77^\circ) - \tan(75^\circ) \right] = \mathbf{119.9 \text{ meters}} \]

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:

\[ \text{Angular speed} = \dfrac{356^\circ}{4 \text{ seconds}} = 89^\circ \text{ per second} \]

\[ = 89^\circ \times \dfrac{60}{1} = \mathbf{5340^\circ \text{ per minute}} \]

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

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

b) \( x = \dfrac{1}{4} \times 20 = 5 \)

Substitute into the equation: \( 5 = 20 \sin(t) \Rightarrow \sin(t) = \dfrac{1}{4} \)

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

Start with the right-hand side:

\[ \cos(x) - \sin(3x) = \cos(x) - \sin(x + 2x) \]

Expand \( \sin(x + 2x) \):

\[ = \cos(x) - \sin(x)\cos(2x) - \cos(x)\sin(2x) \]

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Now expand the product on the 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 and last terms:

\[ = \cos(x)(1 - 2\sin^2(x)) + \sin(x)\cdot 2\sin(x)\cos(x) - \cos(x)\sin(2x) - \sin(x)\cos(2x) \]

\[ = \cos(x) - 2\cos(x)\sin^2(x) + 2\cos(x)\sin^2(x) - \cos(x)\sin(2x) - \sin(x)\cos(2x) \]

The \( 2\cos(x)\sin^2(x) \) terms cancel out:

\[ = \cos(x) - \cos(x)\sin(2x) - \sin(x)\cos(2x) \]

The left-hand side has been transformed so that it matches the right-hand side. The proof is complete.

Expand and simplify the first equation using co-function identities:

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

Expand and simplify the second equation:

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

Now calculate \( \tan(x) \):

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

The addition formula for the tangent can be used backwards here:

\[ \dfrac{\tan(25^\circ) + \tan(50^\circ)}{1 - \tan(25^\circ)\tan(50^\circ)} = \tan(25^\circ + 50^\circ) \]

\[ = \tan(75^\circ) \]

\[ = \tan(45^\circ + 30^\circ) \]

Expand using special angles:

\[ = \dfrac{\tan(45^\circ) + \tan(30^\circ)}{1 - \tan(45^\circ)\tan(30^\circ)} \]

\[ = \dfrac{1 + \dfrac{\sqrt{3}}{3}}{1 - 1 \cdot \dfrac{\sqrt{3}}{3}} \]

Multiply numerator and denominator by 3:

\[ = \dfrac{3 + \sqrt{3}}{3 - \sqrt{3}} \]

Rationalize the denominator:

\[ = \mathbf{\sqrt{3} + 2} \]

Given \( \sin(s) = \dfrac{1}{4} \) and \( \sin(t) = -\dfrac{1}{2} \), and their respective quadrants, use the Pythagorean identity to 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 the tangent addition formula using sine and cosine:

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

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

Substitute the values:

\[ = \dfrac{ (\dfrac{1}{4})(\dfrac{\sqrt{3}}{2}) + (-\dfrac{\sqrt{15}}{4})(-\dfrac{1}{2}) }{ (-\dfrac{\sqrt{15}}{4})(\dfrac{\sqrt{3}}{2}) - (\dfrac{1}{4})(-\dfrac{1}{2}) } \]

\[ = \dfrac{\dfrac{\sqrt{3}}{8} + \dfrac{\sqrt{15}}{8}}{-\dfrac{\sqrt{45}}{8} + \dfrac{1}{8}} \]

Multiply by 8 and simplify \( \sqrt{45} = 3\sqrt{5} \):

\[ = \dfrac{\sqrt{3} + \sqrt{15}}{1 - 3\sqrt{5}} \]

Rationalizing the denominator gives:

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

First, reduce the angle into a familiar quadrant:

\[ \cos\left(\dfrac{13\pi}{12}\right) = \cos\left(\dfrac{\pi}{12} + \pi\right) = -\cos\left(\dfrac{\pi}{12}\right) \]

Now apply the half-angle formula since \( \dfrac{\pi}{12} \) is half of \( \dfrac{\pi}{6} \):

\[ = -\cos\left(\dfrac{1}{2} \cdot \dfrac{\pi}{6}\right) = -\sqrt{\dfrac{1}{2}\left(1 + \cos\left(\dfrac{\pi}{6}\right)\right)} \]

Substitute \( \cos(\pi/6) = \sqrt{3}/2 \):

\[ = -\sqrt{\dfrac{1}{2}\left(1 + \dfrac{\sqrt{3}}{2}\right)} \]

\[ = \mathbf{-\sqrt{\dfrac{1}{2} + \dfrac{\sqrt{3}}{4}}} \]

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 \( g(x) \) upward by 2 units. We are given the value \( f\left(\dfrac{\pi}{2}\right) = 1 \).

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

\[ 1 = a \sin\left(\dfrac{\pi}{2} - \dfrac{\pi}{3}\right) + 2 \]

\[ 1 = a \sin\left(\dfrac{\pi}{6}\right) + 2 \]

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

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

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

Final Equation:

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

The general form of the function is given by:

\[ y = A \sin(Bx - C) + D \]

From the problem, we know: Amplitude \( A = 5/3 \), Vertical Shift \( D = 4 \), and Phase Shift \( C = 0 \).

We are told that the period is \( \pi/2 \). The relationship between period and \( B \) is:

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

Solving for \( B \):

\[ B = 4 \]

Substitute \( A \), \( B \), and \( D \) into the function:

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

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

\[ a^2 = b^2 + c^2 - 2bc \cos(A) \]

\[ a^2 = 4^2 + 8^2 - 2 \cdot 4 \cdot 8 \cdot \cos(46^\circ) \]

\[ a = \sqrt{16 + 64 - 64 \cos(46^\circ)} \approx 5.963 \]

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

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

\[ \dfrac{\sin(B)}{4} = \dfrac{\sin(46^\circ)}{5.963} \]

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

\[ B = \arcsin\left( \dfrac{4 \sin(46^\circ)}{5.963} \right) \approx \mathbf{29^\circ} \text{ (rounded to the nearest degree)} \]

Note that:

\[ 15^2 = 12^2 + 9^2 \quad (225 = 144 + 81) \]

This means that the triangle in question is a right triangle. The largest angle is exactly \( \mathbf{90^\circ} \).

Let angle \( A \) be the angle opposite the side of length 9. Using SOH CAH TOA:

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

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

The third angle is:

\[ 90^\circ - 37^\circ = \mathbf{53^\circ} \]

Step 1: Use the Pythagorean identity \( \cos^2(x) = 1 - \sin^2(x) \) to write the equation entirely in terms of sine.

\[ 2(1 - \sin^2(x)) + \sin(x) = 1 \]

\[ 2 - 2\sin^2(x) + \sin(x) = 1 \]

Step 2: Move all terms to one side to form a standard quadratic equation equal to zero.

\[ 2\sin^2(x) - \sin(x) - 1 = 0 \]

Step 3: Factor the quadratic expression. Treat \( \sin(x) \) as a variable.

\[ (2\sin(x) + 1)(\sin(x) - 1) = 0 \]

Step 4: Solve each factor for \( x \) on the interval \( [0, 2\pi) \).

Case 1: \( 2\sin(x) + 1 = 0 \Rightarrow \sin(x) = -1/2 \)

\[ \Rightarrow x = \dfrac{7\pi}{6}, \dfrac{11\pi}{6} \]

Case 2: \( \sin(x) - 1 = 0 \Rightarrow \sin(x) = 1 \)

\[ \Rightarrow x = \dfrac{\pi}{2} \]

Answer: \( \mathbf{x = \dfrac{\pi}{2}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}} \)

Step 1: Combine the two fractions by finding a common denominator, which is \( \sin(x)\cos(x) \).

\[ \dfrac{\sin(3x)\cos(x) - \cos(3x)\sin(x)}{\sin(x)\cos(x)} \]

Step 2: Notice that the numerator is exactly the expanded form of the sine difference identity: \( \sin(A - B) = \sin A \cos B - \cos A \sin B \).

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

\[ = \dfrac{\sin(2x)}{\sin(x)\cos(x)} \]

Step 3: Apply the double-angle identity for sine in the numerator: \( \sin(2x) = 2\sin(x)\cos(x) \).

\[ = \dfrac{2\sin(x)\cos(x)}{\sin(x)\cos(x)} \]

Step 4: Cancel the common terms.

Answer: \( \mathbf{2} \)

Step 1: To find the area using \( \text{Area} = \dfrac{1}{2}ab\sin(C) \), we first need to find angle \( C \). We can use the Cosine Rule for this.

\[ c^2 = a^2 + b^2 - 2ab \cos(C) \]

\[ 8^2 = 5^2 + 7^2 - 2(5)(7)\cos(C) \]

\[ 64 = 25 + 49 - 70\cos(C) \]

\[ 64 = 74 - 70\cos(C) \]

\[ -10 = -70\cos(C) \Rightarrow \cos(C) = \dfrac{1}{7} \]

Step 2: Now use the Pythagorean identity to find \( \sin(C) \). Since \( C \) is an angle in a triangle, \( \sin(C) \) must be positive.

\[ \sin(C) = \sqrt{1 - \cos^2(C)} = \sqrt{1 - \left(\dfrac{1}{7}\right)^2} \]

\[ \sin(C) = \sqrt{\dfrac{48}{49}} = \dfrac{\sqrt{16 \cdot 3}}{7} = \dfrac{4\sqrt{3}}{7} \]

Step 3: Calculate the area.

\[ \text{Area} = \dfrac{1}{2}ab\sin(C) = \dfrac{1}{2}(5)(7)\left(\dfrac{4\sqrt{3}}{7}\right) \]

The 7s cancel, and \( \dfrac{1}{2} \cdot 4 = 2 \).

\[ \text{Area} = 5 \cdot 2\sqrt{3} = \mathbf{10\sqrt{3}} \]