📐 Level 2 - Topic 13: Solving Real-World Problems with Trigonometry - Trig in Action! 🌍🚀

Example 1: Finding Height using Angle of Elevation

From a point on the ground 50 meters from the base of a tree, the angle of elevation to the top of the tree is \( 30^\circ \). Find the height of the tree.

1. Draw a Diagram: Sketch a right-angled triangle. Let the height of the tree be \( h \), the distance from the base of the tree be 50m, and the angle of elevation be \( 30^\circ \).

   (Imagine a right triangle with horizontal base 50m, vertical height 'h', and angle at the base = 30 degrees)

2. Identify the Trigonometric Ratio: We know the adjacent side (50m) and want to find the opposite side (height \( h \)). Tangent relates opposite and adjacent. So, use \( \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \).
3. Set up the equation: \( \tan(30^\circ) = \frac{h}{50} \).
4. Solve for \( h \): \( h = 50 \cdot \tan(30^\circ) \). We know \( \tan(30^\circ) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \approx 0.577 \).

   \( h = 50 \cdot \frac{\sqrt{3}}{3} \approx 50 \times 0.577 \approx 28.85 \text{ meters} \)

Solution: The height of the tree is approximately 28.85 meters.

Example 2: Finding Distance using Angle of Depression

From the top of a cliff 100 meters high, the angle of depression to a boat at sea is \( 25^\circ \). How far is the boat from the base of the cliff?

1. Draw a Diagram: Sketch a right-angled triangle. Cliff height is 100m (vertical side), angle of depression from top of cliff to boat is \( 25^\circ \). The distance of the boat from the base of the cliff is the horizontal side, let's call it \( d \).

   (Imagine a right triangle, cliff vertical 100m, horizontal distance 'd' from base of cliff to boat, angle of depression from top of cliff to boat 25 degrees. Note: Angle of depression is from horizontal at the top of the cliff down to line of sight to boat. This angle is equal to the angle of elevation from the boat to the top of the cliff, due to alternate interior angles if we draw a horizontal line at the top of the cliff and consider the line of sight as a transversal.)

2. Identify the Trigonometric Ratio: We know the opposite side (height of cliff = 100m) and want to find the adjacent side (distance \( d \)). Use tangent.
3. Set up the equation: \( \tan(25^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{100}{d} \). (Using angle of elevation from boat to cliff top, which is same as angle of depression)
4. Solve for \( d \): \( d = \frac{100}{\tan(25^\circ)} \). \( \tan(25^\circ) \approx 0.466 \).

   \( d = \frac{100}{0.466} \approx 214.59 \text{ meters} \)

Solution: The boat is approximately 214.59 meters from the base of the cliff.

Example 3: Two Angles of Elevation Problem

A flagpole is standing on top of a building. From a point on the ground, the angle of elevation to the top of the building is \( 30^\circ \), and the angle of elevation to the top of the flagpole is \( 45^\circ \). If the distance from the point to the base of the building is 20 meters, find the height of the flagpole.

1. Draw a Diagram: Sketch two nested right-angled triangles. Let \( H \) be the height of the building, and \( h \) be the height of the flagpole. Total height to top of flagpole is \( H + h \). Distance from point to base is 20m. Angle of elevation to top of building is \( 30^\circ \), to top of flagpole is \( 45^\circ \).

   (Imagine two right triangles sharing the same base of 20m. Smaller triangle has height H and angle 30 degrees. Larger triangle has height H+h and angle 45 degrees.)

2. Set up two equations:
   For the building: \( \tan(30^\circ) = \frac{H}{20} \Rightarrow H = 20 \tan(30^\circ) \).
   For the flagpole + building: \( \tan(45^\circ) = \frac{H + h}{20} \Rightarrow H + h = 20 \tan(45^\circ) \).
3. Solve for \( H \) and then \( h \):
   \( H = 20 \tan(30^\circ) = 20 \cdot \frac{\sqrt{3}}{3} \).
   \( H + h = 20 \tan(45^\circ) = 20 \cdot 1 = 20 \).
   Substitute \( H \) into the second equation to solve for \( h \): \( 20 \tan(30^\circ) + h = 20 \Rightarrow h = 20 - 20 \tan(30^\circ) = 20(1 - \tan(30^\circ)) \).

   \( h = 20 \left(1 - \frac{\sqrt{3}}{3}\right) \approx 20 (1 - 0.577) = 20 \times 0.423 \approx 8.46 \text{ meters} \)

Solution: The height of the flagpole is approximately 8.46 meters.

Example 4: Navigation Problem using Bearings

A ship sails from port P to port Q on a bearing of 060°T for a distance of 150 km. Then it sails from port Q to port R on a bearing of 140°T for a distance of 200 km. Find the distance and bearing of port R from port P.

1. Draw a Diagram: Sketch the path. Start at P, draw North line. Draw PQ at 60° from North, length 150km. From Q, draw North line again, draw QR at 140° from North (at Q), length 200km. We need to find PR and bearing of R from P.

   (Imagine drawing this on a map, starting from P, going NE to Q, then SE to R. You'll see triangle PQR and will likely need to use Law of Cosines and Law of Sines - hint for Level 2 complexity increase, but might be solvable with right triangles if broken down carefully or if problem simplifies)

2. Analyze the angles in triangle PQR: Angle NPQ = 60°. Angle NQR = 140°. Angle PQN = 180° - 140° = 40° (interior angles, straight line NQR). Angle PQR = Angle PQN + Angle NQR = 40° + (180° - 140°) = 40°. Actually, angle PQN = 180° - 140° = 40° is incorrect. Angle PQN = 180° - angle NQR = 180° - 140° = 40° if we consider reflex angle NQR, but that's not right. Let's reconsider. Angle between PQ and QR is needed. Angle NPQ is 60°, Angle NQR is 140°. Angle PQS (where S is North at Q) is 60° + 90° = 150° from East at P clockwise? No, simpler approach: Angle between North at P and PQ is 60°. Angle between North at Q and QR is 140°. Angle between North at P and line parallel to North at Q is 0°. So angle between PQ and line parallel to North at Q is 60°. Angle between QR and North at Q is 140°. Angle PQR = 140° - 60° = 80°? No, that's wrong direction. Let's use angles from East - more consistent in trigonometry. Bearing 060°T is 90-60 = 30° from East towards North. Bearing 140°T is 140-90 = 50° from East towards South, if East is 0°. No, True Bearing from North, North is 0°. East is 90°, South 180°, West 270°. Bearing 060°T means angle NPQ = 60°. Bearing 140°T means angle NQR = 140°. Angle PQN is 180° - 140° = 40°? Still not clear on interior angle. Let's rethink the angle PQR. Draw North at P and Q parallel. Angle NPQ = 60°, Angle NQR = 140°. Angle between PQ and North at Q (parallel to NP) is also 60° (alternate interior). Angle NQR is 140°. So, angle PQR = 140° - 60° = 80°? No, that's still not right visually. Angle between PQ direction and QR direction is needed *inside triangle PQR*. Let's consider angle *outside* at Q, angle between PQ extended and QR is 140° - 60° = 80°. So, interior angle PQR = 180° - 80° = 100°? Yes, that makes sense visually if drawn carefully. So, ∠PQR = 180° - (140° - 60°) = 180° - 80° = 100°. Or, simpler: Angle between 60° bearing direction and 140° bearing direction is simply the difference if they are in same direction rotation, but here they are not directly comparable, so 180 - (larger-smaller) rule may be better. 180 - |140-60| = 100°. Yes. So, ∠PQR = 100°.
3. Use Law of Cosines to find PR: In triangle PQR, we have PQ = 150km, QR = 200km, ∠PQR = 100°. Use Law of Cosines to find PR: \( PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR) \).

   \( PR^2 = 150^2 + 200^2 - 2(150)(200)\cos(100^\circ) \)

   \( PR^2 = 22500 + 40000 - 60000 \times (-0.1736) \approx 62500 + 10416 \approx 72916 \)

   \( PR = \sqrt{72916} \approx 270.03 \text{ km} \)

4. Use Law of Sines to find bearing of R from P: We need to find the bearing of R from P, which is angle NPR. First, find angle QPR in triangle PQR using Law of Sines: \( \frac{\sin(\angle QPR)}{QR} = \frac{\sin(\angle PQR)}{PR} \). \( \sin(\angle QPR) = \frac{QR \cdot \sin(\angle PQR)}{PR} = \frac{200 \cdot \sin(100^\circ)}{270.03} \approx \frac{200 \times 0.9848}{270.03} \approx 0.7294 \). \( \angle QPR = \arcsin(0.7294) \approx 46.83^\circ \).
5. Calculate bearing NPR: Bearing of R from P is ∠NPR = ∠NPQ + ∠QPR = \( 60^\circ + 46.83^\circ = 106.83^\circ \).

   Bearing of R from P is approximately 106.83°T.

Solution: The distance of port R from port P is approximately 270.03 km, and the bearing of port R from port P is approximately 106.83°T.

Example 5: Area of a Triangle using Sine Formula

Find the area of a triangle ABC where side \( a = 10 \text{ cm} \), side \( b = 12 \text{ cm} \), and the included angle \( C = 60^\circ \).

1. Recall the area formula using sine: Area of triangle = \( \frac{1}{2}ab\sin(C) \), where \( a, b \) are two sides and \( C \) is the angle between them.
2. Substitute the given values: \( a = 10 \text{ cm} \), \( b = 12 \text{ cm} \), \( C = 60^\circ \).

   Area = \( \frac{1}{2} \times 10 \times 12 \times \sin(60^\circ) \)

3. Evaluate: \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \).

   Area = \( \frac{1}{2} \times 10 \times 12 \times \frac{\sqrt{3}}{2} = 30\sqrt{3} \approx 30 \times 0.866 \approx 25.98 \text{ cm}^2 \)

Solution: The area of the triangle is approximately 25.98 cm\(^2\).

Example 6: Finding Unknown Side in a Triangle using Law of Cosines

In triangle ABC, \( a = 8 \text{ m} \), \( c = 7 \text{ m} \), and angle \( B = 45^\circ \). Find the length of side \( b \).

1. Use the Law of Cosines: To find side \( b \) when we know two sides and the included angle \( B \), use the Law of Cosines in the form: \( b^2 = a^2 + c^2 - 2ac\cos(B) \).
2. Substitute the given values: \( a = 8 \text{ m} \), \( c = 7 \text{ m} \), \( B = 45^\circ \).

   \( b^2 = 8^2 + 7^2 - 2(8)(7)\cos(45^\circ) \)

3. Evaluate: \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707 \).

   \( b^2 = 64 + 49 - 112 \times \frac{\sqrt{2}}{2} = 113 - 56\sqrt{2} \approx 113 - 56 \times 0.707 \approx 113 - 39.592 \approx 73.408 \)

   \( b = \sqrt{73.408} \approx 8.57 \text{ meters} \)

Solution: The length of side \( b \) is approximately 8.57 meters.