Trigonometry word problems with diagrams and illustrations.

Here are the step-by-step solutions for each of the four problems shown.

Problem 1: The Tree Height

Goal: Find the total height of the tree ($CR$).

1. Identify the Triangle: We have a right-angled triangle $ADR$.
* The angle at $A$ is $32^\circ$.
* The horizontal distance $AD$ (which is the same as $CB$) is $20\text{ m}$.
* We need to find the vertical side $DR$.

2. Calculate $DR$: We use the tangent ratio because we have the adjacent side and want the opposite side.
$$ \tan(\text{angle}) = \frac{\text{Opposite}}{\text{Adjacent}} $$
$$ \tan(32^\circ) = \frac{DR}{20} $$
$$ DR = 20 \times \tan(32^\circ) $$
Using a calculator: $\tan(32^\circ) \approx 0.6249$
$$ DR = 20 \times 0.6249 = 12.498\text{ m} $$

3. Calculate Total Height: The total height of the tree is the top part ($DR$) plus the bottom part ($DC$, which is Phoebe's eye height).
$$ \text{Total Height} = 12.498 + 1.6 $$
$$ \text{Total Height} = 14.098\text{ m} $$

Rounding to one decimal place, the tree is 14.1 m tall.

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Problem 2: The Building Angle

Goal: Find the angle of elevation for a building that is $174.6\text{ m}$ high from two different distances.

We use the inverse tangent function: $\theta = \tan^{-1}\left(\frac{\text{Height}}{\text{Distance}}\right)$.

(a) From 50 metres away:
1. Set up the ratio: $\tan(\theta) = \frac{174.6}{50}$
2. Calculate the value: $\frac{174.6}{50} = 3.492$
3. Find the angle: $\theta = \tan^{-1}(3.492)$
4. Result: $\theta \approx 74.04^\circ$

(b) From 100 metres away:
1. Set up the ratio: $\tan(\theta) = \frac{174.6}{100}$
2. Calculate the value: $\frac{174.6}{100} = 1.746$
3. Find the angle: $\theta = \tan^{-1}(1.746)$
4. Result: $\theta \approx 60.20^\circ$

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Problem 3: The Leaning Tower

Goal: Find how far the object fell (the length of the tower).

1. Visualize the Triangle: Imagine a triangle formed by the tower, the ground, and an imaginary vertical line dropping straight down from the top of the tower.
* The tower is the hypotenuse (let's call its length $L$).
* The tower leans at $87.1^\circ$ to the ground.
* The object lands $3\text{ m}$ from the base. This is the horizontal distance.

2. Set up the Equation: In this right-angled triangle, the horizontal distance ($3\text{ m}$) is adjacent to the $87.1^\circ$ angle.
$$ \cos(\text{angle}) = \frac{\text{Adjacent}}{\text{Hypotenuse}} $$
$$ \cos(87.1^\circ) = \frac{3}{L} $$

3. Solve for $L$:
$$ L = \frac{3}{\cos(87.1^\circ)} $$
Using a calculator: $\cos(87.1^\circ) \approx 0.0506$
$$ L = \frac{3}{0.0506} \approx 59.29\text{ m} $$

The object fell approximately 59.3 metres.

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Problem 4: The Bermuda Triangle Area

Goal: Check if the area of the triangle is close to the claimed $500,000$ square miles.

1. Identify Given Values:
* Side $c$ (Miami to San Juan) = $954.8$ miles.
* Angle $B$ (at Bermuda) = $62.5^\circ$.
* Angle $C$ (at San Juan) = $62.5^\circ$.
* Since two angles are equal, it is an isosceles triangle. This means the side opposite Angle $C$ (Side $b$, Miami to Bermuda) is also $954.8$ miles.

2. Find the Third Angle (Angle $A$ at Miami):
The angles in a triangle add up to $180^\circ$.
$$ \text{Angle } A = 180^\circ - 62.5^\circ - 62.5^\circ $$
$$ \text{Angle } A = 180^\circ - 125^\circ = 55^\circ $$

3. Calculate the Area:
We can use the formula: $\text{Area} = \frac{1}{2} \times \text{side}_1 \times \text{side}_2 \times \sin(\text{included angle})$.
Here, we use the two known sides ($954.8$) and the angle between them ($55^\circ$).
$$ \text{Area} = \frac{1}{2} \times 954.8 \times 954.8 \times \sin(55^\circ) $$
$$ \text{Area} = 0.5 \times 911,643.04 \times 0.81915 $$
$$ \text{Area} \approx 373,368 \text{ square miles} $$

4. Conclusion: The calculated area is roughly $373,368$ square miles. This is significantly less than the claimed $500,000$ square miles.

Final Answer:
Problem 1: The tree is 14.1 m tall.
Problem 2:
(a) 74.0°
(b) 60.2°
Problem 3: It fell 59.3 m.
Problem 4: The actual area is approximately 373,368 square miles, so the claim of 500,000 square miles is incorrect (it is an overestimate).