3D Trigonometry旗杆高度证明问题及绘图困惑咨询

Hey there, I totally get how frustrating it is to struggle with trigonometry problems when you can’t visualize the diagram clearly—blurry images only make it worse! Let’s walk through this together, starting with how to sketch the diagram so you can see what’s going on, then proving the height formula step by step.

How to Draw the Diagram #

• First, sketch a horizontal line to represent the ground. Mark a point O on this line—this is the base of the flagpole. From O, draw a straight vertical line upward to point P; this is the flagpole itself, and we’ll call its total height $h$ (so $OP = h$).
• Next, mark a point E somewhere else on the ground—this is where the observer is standing. From E, draw a vertical line up to point A (the observer’s eye); this line is 2m tall, so $AE = 2m$, and it’s perfectly perpendicular to the ground.
• Now, draw a horizontal line from A (let’s call this line AH, pointing toward the flagpole). The angle of depression from A to O is 15°, so the angle between AH and AO (the line from the eye to the flagpole’s base) is 15°, and it slants downward from AH to AO.
• Finally, draw the line from A to P (the top of the flagpole). The problem says the flagpole subtends a 45° angle at the observer’s eye, which means the angle between AO and AP (written as ∠PAO) is 45°.

Proving the Height Formula #

Now that we have our diagram sorted, let’s work through the math:

1. First, find the horizontal distance between the observer’s eye and the flagpole (this is the same as the ground distance between E and O, let’s call this distance $d$).
   • In the right triangle AEO, the angle of depression tells us $\tan(15°) = \frac{AE}{d}$. Rearranging this gives $d = \frac{AE}{\tan(15°)}$.
   • Remember that $\tan(90° - \theta) = \frac{1}{\tan(\theta)}$, so $\frac{1}{\tan(15°)} = \tan(75°)$. Since $AE = 2m$, this means $d = 2\tan(75°)$.
2. Next, let’s find the angle of elevation from the observer’s eye to the top of the flagpole. We know ∠PAO (the angle between the eye’s view to the flagpole base and top) is 45°, and the angle of depression to the base is 15°, so the angle of elevation to the top is $45° - 15° = 30°$.
3. The vertical gap between the flagpole top and the observer’s eye is $h - 2$ (since the eye is 2m off the ground). This gap, the horizontal distance $d$, and the line AP form a right triangle.
   • In this triangle, $\tan(30°) = \frac{h - 2}{d}$. We already found $d = 2\tan(75°)$, so substitute that in:
     $$\tan(30°) = \frac{h - 2}{2\tan(75°)}$$
   • Multiply both sides by $2\tan(75°)$:
     $$h - 2 = 2\tan(30°)\tan(75°)$$
   • Add 2 to both sides to get the total height:
     $$h = 2 + 2\tan(30°)\tan(75°) = 2\left[1 + \tan(30°)\tan(75°)\right]$$

That’s exactly the formula we needed to show! I hope this breakdown makes the problem feel less overwhelming—trigonometry clicks way better once you can see all the right triangles laid out clearly.

备注：内容来源于stack exchange，提问作者user1249555