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Trigonometry problem illustrating angles of elevation to the top of a building from two different distances.

Diagram showing a building with height y, an observer initially 50 feet away at a 30° angle of elevation, then moving 50 feet closer to see a 40° angle of elevation.

Diagram showing a building with height y, an observer initially 50 feet away at a 30° angle of elevation, then moving 50 feet closer to see a 40° angle of elevation.

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We are given a problem involving angles of elevation and right triangles, and we need to find the height $ y $ of the building.

---

Given:


- Initially, the observer is at a distance from the building such that the angle of elevation to the top is $ 30^\circ $.
- After moving 50 feet closer, the angle becomes $ 40^\circ $.
- Let $ x $ be the distance from the closer position to the base of the building.
- So, the original distance was $ x + 50 $ feet.
- The height of the building is $ y $ (unknown).

We'll use trigonometry, specifically the tangent function, since we're dealing with right triangles.

---

Step 1: Set up equations using tangent



In a right triangle:
$$
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
$$

#### From the second position (closer):
$$
\tan(40^\circ) = \frac{y}{x} \quad \Rightarrow \quad y = x \cdot \tan(40^\circ)
\tag{1}
$$

#### From the first position (farther):
$$
\tan(30^\circ) = \frac{y}{x + 50} \quad \Rightarrow \quad y = (x + 50) \cdot \tan(30^\circ)
\tag{2}
$$

---

Step 2: Set equations (1) and (2) equal



Since both equal $ y $:
$$
x \cdot \tan(40^\circ) = (x + 50) \cdot \tan(30^\circ)
$$

Now plug in approximate values:
- $ \tan(40^\circ) \approx 0.8391 $
- $ \tan(30^\circ) \approx 0.5774 $

So:
$$
x \cdot 0.8391 = (x + 50) \cdot 0.5774
$$

Expand the right-hand side:
$$
0.8391x = 0.5774x + 28.87
$$

Subtract $ 0.5774x $ from both sides:
$$
(0.8391 - 0.5774)x = 28.87
$$
$$
0.2617x = 28.87
$$

Solve for $ x $:
$$
x = \frac{28.87}{0.2617} \approx 110.3 \text{ feet}
$$

---

Step 3: Find $ y $



Use equation (1):
$$
y = x \cdot \tan(40^\circ) \approx 110.3 \cdot 0.8391 \approx 92.55 \text{ feet}
$$

---

Final Answer:


$$
\boxed{y \approx 92.6} \text{ feet}
$$

The height of the building is approximately 92.6 feet.

---

🔍 Summary of Steps:


1. Used two right triangles formed by the observer's positions.
2. Wrote expressions for height $ y $ using tangent.
3. Equated the two expressions and solved for $ x $.
4. Substituted back to find $ y $.

Let me know if you'd like an exact symbolic solution or more decimal precision!
Parent Tip: Review the logic above to help your child master the concept of sohcahtoa word problems worksheet.
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