Trigonometry practice problems involving real-world scenarios like river crossings, plane navigation, and wind drift.

Here are the step-by-step solutions for the problems shown in your worksheet.

1) Jim's Swim

Problem: The river is 35 m wide. The current pushes him 20 m downstream. What angle did Jim travel at ($\alpha$)?

* Step 1: Identify the sides of the triangle relative to angle $\alpha$.
* The side opposite to $\alpha$ is the width of the river: 35 m.
* The side adjacent (next) to $\alpha$ is the distance downstream: 20 m.
* Step 2: Choose the correct trigonometric ratio. Since we have Opposite and Adjacent, we use Tangent ($\tan$).
$$ \tan(\alpha) = \frac{\text{Opposite}}{\text{Adjacent}} $$
* Step 3: Plug in the numbers.
$$ \tan(\alpha) = \frac{35}{20} = 1.75 $$
* Step 4: Calculate the angle using the inverse tangent function ($\tan^{-1}$).
$$ \alpha = \tan^{-1}(1.75) \approx 60.255^\circ $$

Answer: Angle $\alpha \approx 60.3^\circ$

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2) Rakesh's Swim

Problem: The current pushes him 18 m downstream. His angle of travel is $42^\circ$. How wide is the river?

* Step 1: Identify the sides relative to the $42^\circ$ angle.
* Opposite side = Width of the river (Unknown, let's call it $w$).
* Adjacent side = Distance downstream (18 m).
* Step 2: Use the Tangent ratio.
$$ \tan(42^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{w}{18} $$
* Step 3: Solve for $w$.
$$ w = 18 \times \tan(42^\circ) $$
$$ w = 18 \times 0.9004 \approx 16.21 \text{ m} $$

Answer: The river is approximately 16.2 m wide.

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3) Hei's Swim

Problem: Total swim is 43 m (hypotenuse). Angle of travel is $56^\circ$. Find the width and the downstream distance.

* Part A: How wide is the river?
* The width is the side opposite the $56^\circ$ angle.
* Use Sine ($\sin$): $\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$
* $\sin(56^\circ) = \frac{\text{Width}}{43}$
* $\text{Width} = 43 \times \sin(56^\circ)$
* $\text{Width} = 43 \times 0.829 \approx 35.65 \text{ m}$

* Part B: How far is she pushed downstream?
* The downstream distance is the side adjacent to the $56^\circ$ angle.
* Use Cosine ($\cos$): $\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$
* $\cos(56^\circ) = \frac{\text{Downstream}}{43}$
* $\text{Downstream} = 43 \times \cos(56^\circ)$
* $\text{Downstream} = 43 \times 0.559 \approx 24.04 \text{ m}$

Answer: The river is 35.7 m wide, and she is pushed 24.0 m downstream.

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4) Jo's Swim

Problem: River is 58 m wide. Waterfall is 38 m away. Angle of travel is $33^\circ$. Does she make it across?

* Step 1: We need to find out how far downstream the current pushes her while she crosses the 58 m width. Let's call this distance $x$.
* Step 2: Relative to the $33^\circ$ angle:
* Opposite side = Width (58 m)
* Adjacent side = Downstream distance ($x$)
* Step 3: Use Tangent.
$$ \tan(33^\circ) = \frac{58}{x} $$
* Step 4: Rearrange to solve for $x$.
$$ x = \frac{58}{\tan(33^\circ)} $$
$$ x = \frac{58}{0.6494} \approx 89.3 \text{ m} $$
* Step 5: Compare with the waterfall distance.
She travels 89.3 m downstream, but the waterfall is only 38 m away. Since $89.3 > 38$, she goes over the waterfall.

Answer: No, Jo does not make it across. She travels ~89 m downstream, which is past the 38 m waterfall.

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6) Plane Runway Approach

Problem: Runway is 5 km North. Wind pushes plane 3 km West. At what angle (from North) should the pilot fly?

* Step 1: Visualize the triangle.
* The desired path is North (Adjacent side).
* The wind correction is West (Opposite side).
* We need the angle $\alpha$ between the North line and the flight path.
* Step 2: Use Tangent.
$$ \tan(\alpha) = \frac{\text{Opposite (West)}}{\text{Adjacent (North)}} $$
$$ \tan(\alpha) = \frac{3}{5} = 0.6 $$
* Step 3: Calculate the angle.
$$ \alpha = \tan^{-1}(0.6) \approx 30.96^\circ $$

Answer: The pilot should fly at an angle of approximately 31.0° from North (towards the West).

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7) Plane Distance from Runway

Problem: Plane is 15 km from runway. It is 5.5 km perpendicular (sideways) from the centerline. At what angle towards the crosswind is it off course?

* Step 1: Identify the triangle parts.
* Hypotenuse (distance to runway) = 15 km.
* Opposite side (perpendicular distance) = 5.5 km.
* We want the angle $\alpha$ at the plane's position.
* Step 2: Use Sine.
$$ \sin(\alpha) = \frac{\text{Opposite}}{\text{Hypotenuse}} $$
$$ \sin(\alpha) = \frac{5.5}{15} \approx 0.3667 $$
* Step 3: Calculate the angle.
$$ \alpha = \sin^{-1}(0.3667) \approx 21.51^\circ $$

Answer: The angle is approximately 21.5°.

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8) Plane Engine Speed

Problem: Engine speed 200 m/s forward. Crosswind 50 m/s sideways. Find actual distance traveled in 1 second and the angle.

* Part A: Actual distance traveled (Resultant Velocity)
* This forms a right-angled triangle with legs 200 and 50. We need the hypotenuse ($c$).
* Use Pythagoras theorem: $a^2 + b^2 = c^2$
* $200^2 + 50^2 = c^2$
* $40000 + 2500 = 42500$
* $c = \sqrt{42500} \approx 206.16 \text{ m/s}$
* In 1 second, it travels 206.16 meters.

* Part B: Angle relative to heading
* Opposite side (wind) = 50.
* Adjacent side (engine) = 200.
* $\tan(\alpha) = \frac{50}{200} = 0.25$
* $\alpha = \tan^{-1}(0.25) \approx 14.04^\circ$

Answer: The plane travels 206.2 m in one second at an angle of 14.0°.

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9) Crosswind Speed

Problem: Engine speed 270 m/s. Plane is angled 15° to maintain heading. What is the speed of the crosswind?

* Step 1: Understand the geometry. To stay on a straight line despite the wind, the plane points into the wind.
* The engine speed (270 m/s) is the Hypotenuse of the velocity triangle.
* The crosswind speed is the side Opposite the 15° angle.
* Step 2: Use Sine.
$$ \sin(15^\circ) = \frac{\text{Crosswind Speed}}{270} $$
* Step 3: Solve for Crosswind Speed.
$$ \text{Crosswind Speed} = 270 \times \sin(15^\circ) $$
$$ \text{Crosswind Speed} = 270 \times 0.2588 \approx 69.88 \text{ m/s} $$

Answer: The speed of the crosswind is approximately 69.9 m/s.